# Statistics Course in Psychology

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Lesson 1 Introduction Outline Statistics Descriptive versus inferential statistics Population versus Sample Statistic versus Parameter Simple Notation Summation Notation Statistics What are statistics? What do you thing of when you think of statistics? Can you think of some examples where you have seen statistics used? You might think about where in the real world you see statistics being used, or think about how statistics in used in your major. Statistics are divided into two main areas: descriptive and inferential statistics. Descriptive statistics- These are numbers that are used to consolidate a large amount of information. Any average, for example, is a descriptive statistic. So, batting averages, average daily rainfall, or average daily temperature are good examples of descriptive statistics. Inferential statistics- inferential statistics are used when we want to draw conclusions. For example when we want to determine if some treatment is better than another, or if there are differences in how two groups perform. A good book definition is using samples to draw inferences about populations. More on this once we define samples and populations. Population- Any set of people or objects with something in common. Anything could be a population. We could have a population of college students. We might be interested in the population of the elderly. Other examples include: single parent families, people with depression, or burn victims. For anything we might be interested in studying we could define a population. Very often we would like to test something about a population. For example, we might want to test whether a new drug might be effective for a specific group. It is impossible most of the time to give everyone a new treatment to determine if it worked or not. Instead we commonly give it to a group of people from the population to see if it is effective. This subset of the population is called a sample. When we measure something in a population it is called a parameter. When we measure something in a sample it is called a statistic. For example, if I got the average age of parents in single-family homes, the measure would be called a parameter. If I measured the age of a sample of these same individuals it would be called a statistic. Thus, a population is to a parameter as a sample is to a statistic. This distinction between samples and population is important because this course is about inferential statistics. With inferential statistics we want to draw inferences about populations from samples. Thus, this course is mainly concerned with the rules or logic of how a relatively small sample from a large population could be tested, and the results of those tests can be inferred to be true for everyone in the population. For example, if we want to test whether Bayer asprin is better than Tylonol at relieving pain, we could not give these drugs to everyone in the population. Its not practical since the general population is so large. Instead we might give it to a couple of hundred people and see which one works better with them. With inferential statistics we can infer that what was true for a few hundred people is also true for a very large population of hundreds of thousands of people. When we write symbols about populations and samples they differ too. With populations we will use Greek letters to symbolize parameters. When we symbolize a measure from a sample (a statistic) we will use the letters you are familiar with (Roman letters). Thus, if I measure the average age of a population Id indicate the value with the Greek letter mu ( =24). While if I were to measure the same value for a subset of the population or a sample then I would indicate the value with a roman letter ( X =24). Simple Notation You might thing about descriptive statistics as the vocabulary of the "language" of statistics. If this is true then summation notation can be thought of as the alphabet of that language. Notation and summation notation is just a short hand way of representing information we have collected and mathematical operation we want to perform. For example, if I collect data on a variable, say the amount of time (in minutes) several people spent waiting at a bus stop, I can represent that group of numbers with the variable X. The variable X represents all of the data that I collected. Amount of Time X 5.0 11.1 8.9 3.5 12.3 15.6 With subscripts I can also represent an individual data point within the variable set we have labeled X. For example the third data point, 8.9, is the X3 data point. The fifth data point X5 is the number 12.3. Very often when we want to represent ALL of the data points in a variable set we will use X by itself, but we may also add the subscript i. Whenever you the subscript i, you can assume that we are referring to all the numbers for the variable X. Thus, Xi is all of the numbers in the data set or: 5,11.1,8.9,3.5,12.3,15.6. There are other common symbols we will use besides X. Sometimes we will have two data sets to deal with and refer to one distribution as X and the other distribution as Y. It is also necessary for many formulas to know how many data points are in a data set. The symbol for the number of data points in a set is N. For the data set above the number of data points or N = 6. In addition, we will use the average or mean value a good deal. We will indicate the mean, as noted above, differently for the population () than for the sample ( X ). Summation Notation Another common symbol we will use is the summation sign ( ). This symbol does not represent anything about our data itself, but instead is an operation we must perform. Whenever you see this symbol it means to add up whatever appears to the right of the sign. Thus, X or Xi tells us to add up all of the data points in our data set. For our example above it would be: 5 + 11.1 + 8.9 + 3.5 + 12.3 + 15.6 = 56.4. You will see the summation sign with other mathematical operations as well. For example X2 tells us to add all the squared X values. Thus, for our example: X2 = 52 + 11.12 + 8.92 + 3.52 + 12.32 + 15.62 -or- 25 + 123.21 + 79.21 + 12.25 + 151.29 + 243.36 = 634.32. A few more examples of summation notation are in order since the summation sign will be central to the formulas we write. The following examples should give you a better idea about how the summation sign is used. Be sure you recall the order of operations needed to solve mathematical expressions. You will find a review on the web page or you can click here: http://faculty.uncfsu.edu/dwallace/sorder.html For the examples below we will use a new distribution. X = 1 2 3 4 Y = 5 6 7 8 ( )22 XX For this expression we are saying that the sum of the squared Xs is not equal to the sum of the Xs squared. Notice here we want to perform the operation in parentheses first, and then the exponents, and then the addition. Thus: ( )22 XX ( )22222 43214321 ++++++ 1 + 4 + 9 + 16 (10)2 30 100 For the next expression we show, like in algebra, that the law of distribution applies to the summation sign as well. Again, what is important is to get a feel for how the summation sign works in equations. YXYX +=+ )( (1+5)+(2+6)+(3+7)+(4+8) = (1+2+3+4)+(5+6+7+8) 6 + 8 + 10 + 12 = 10 + 26 36 = 36 Lesson 2 Scales of Measure Outline Variables -measurement versus categorical -continuous versus discreet -independent and dependent Scales of measure -nominal, ordinal, interval, ratio Variables A variable is anything we measure. This is a broad definition that includes most everything we will be interested in for an experiment. It could be the age or gender of participants, their reactions times, or anything we might be interested in. Whenever we measure a variable, it could be a measurement (quantitative) difference or a categorical (qualitative) difference. You should know both terms for each type. Measurement variables are things to which we can assign a number. It is something we can measure. Examples include age, height, weight, time measurement, or number of children in a household. These examples are also called quantitative because they measure some quantity. Categorical variables are measures of differences in type rather than amount. Examples include anything categorize such as race, gender, or color. These are also called qualitative variables because there is some quality that distinguishes these objects. Another dimension on which variables might differ is that they may be either continuous or discreet. A continuous variable is a variable that can take on any value on the scale used to measure it. Thus, a measure of 1 or 2 is valid, as well as 1.5 or 1.25. Any division on any unit on the scale produces a valid possible measure. Examples include things like height or weight. You could have an object that weighed 1 pound or 1.5 pounds or 1.25 pounds. All are possible measures. Discreet variables, on the other hand, can assume only a few possible values on the scale used to measure it. Divisions of measures are usually not valid. Thus, if I measure the number of television sets in your home it could be 1 or 2 or 3. Divisions of these values are not valid. So, you could not have 1.5 televisions or 1.25 televisions in your home. You either have a television or you dont. Another way to keep this difference in mind is that with a continuous variable is a measure of how much. A discreet variable is a measure of how many. Scales of Measure whenever we measure a variable it has to be on some type of scale. The following scales are delivered in order of increasing complexity. Each scale presented is in order of increasing order. Nominal scales These are not really scales as all, but are instead numbers used to differentiate objects. Real world examples of these variables are common. The numbers are just labels. So, social security numbers, the channels on your television, and sports team jerseys are all good examples of nominal variables. Ordinal Scales Ordinal scales use numbers to put objects in order. No other information other than more or less is available from the scale. A good example is class rank, or any type of ranking. Someone ranked at four had a higher GPA than someone ranked as five, but we dont know how much better four is than five. Interval Scales- Interval scales contain an ordinal scale (objects are in order), but have the added feature that the distance between scale units is always the same. Class rank would not qualify because we dont know how much better one unit is than another, but with interval there is the same distance from one unit to the next anywhere we are on the scale. Examples include temperature (in Fahrenheit or Celsius), or altitude. For temperature you know that the difference in ten degrees is the same no matter how hot or cold it might be. Ratio Scales Ratio scales contain an interval scale (equal intervals between units on the scale), but have the added feature that there is a true zero point on the scale. This zero point is necessary for ratio statements to have meaning. Examples include height or weight or measures of amount of time. Notice that it is not valid to have a measure below zero on any of these scales. Something could not weigh a negative amount. These scales are much more common than interval scales because if a scale usually has a zero point. In fact scientist invented the Kelvin temperature scale so that they would have a measure of temperature on a ratio scale. Again, in order to make ratio statements such as something is twice or half of another then it must be a variable on a ratio scale. Lesson 3 Data Displays Outline Frequency Distributions Grouped Frequency Distributions -class interval and frequency -cumulative frequency -relative percent -cumulative relative percent -interpretations Histograms/Bar Graphs Frequency Distributions We often form frequency distributions as a way to abbreviate the values we are dealing with in a distribution. With frequency distributions we will simply record the frequency or how many values fall at a particular point on the scale. For example, if I record the number of trips out of town (X) a sample of FSU students makes, I might end up with the following data: 0 2 5 3 2 4 3 1 0 2 6 0 4 7 0 1 2 4 3 5 4 3 1 6 1 0 5 3 Instead of having a jumbled set of numbers, we can record how many of each value (f) there are for the entire x-distribution. Below is a simple frequency distribution where the X column represents the number of trips, and the corresponding value for f indicates how many people in the sample gave us that particular response. X f 0 5 1 4 2 4 3 5 4 4 5 3 6 2 7 1 From the graph we can see that five people took no trips out of town, four people took one trip out of town, four people took two trips out of town, and so on. It is important not to confuse the f-value and the x-value. The f-values are just a count of how many. So, you can reverse the process as well. It might also be helpful in some examples to go from a frequency distribution back to original data set, especially if it causes confusion. In the following example I start with a frequency distribution and go backward to find all the original values in the distribution. X f 0 2 1 3 2 4 3 3 4 2 What is the most frequent score? The answer is two because we will have four twos in our distribution: 0 0 1 1 1 2 2 2 2 3 3 3 4 4 Grouped Frequency Distributions The above examples used discreet measures, but when we measure a variable it is often on a continuous scale. In turn, there will be few values we measure that are at the exact same point on the scale. In order to build the frequency distribution we will group several values on the scale together and count any of measurements we observe in that range for the frequency. For example, if we measure the running time of rats in a maze we might obtain the following data. Notice that if I tried to count how many values fall at any single point on the scale my frequencies will all be one. 3.25 3.95 4.61 5.92 6.87 7.12 7.58 8.25 8.69 9.56 9.67 10.24 10.95 10.99 11.34 11.59 12.34 13.45 14.53 14.86 We will begin by forming the class interval. This will be the range of value on the scale we include for each interval. There are many rules we could use to determine the size of the interval, but for this course I will always indicate how big the interval should be. In the end, we want to construct a display that has between 5 and 15 intervals. Thus: Class Interval 0-2 3-5 6-8 9-11 12-14 Once we have the class interval, we will count how many values fall within the range of each interval. Since there is a gab in each class interval, we will be actually counting any values that would get rounded down or up into a particular interval. For example, with the above data the value 8.26 would be rounded down into the 6-8 class interval. The value 8.69 would be rounded up into the 9-11 class interval. We will include a column to indicate the real limits of the class interval. These are the limits of the interval, including any rounded values. Real Limits Class Interval f -.5-2.5 0-2 0 2.5-5.5 3-5 3 5.5-8.5 6-8 5 8.5-11.5 9-11 7 11.5-14.5 12-14 5 Notice that my real limits cover half the distance of the gap between each class interval. Most of the time this value will be 0.5 since most scales will have one unit values and 0.5 is half the distance. So, real limits have no gap, but the class intervals do. If a value falls exactly on one of the real limits we could randomly choose its group. Cumulative Frequency Once we have formed the basic grouped frequency distribution above, we can add more columns for more detailed information. The first of these is the cumulative frequency column. With this column we will keep a running count of the frequency column as we move down the class interval. Real Limits Class Interval f Cum. f -.5-2.5 0-2 0 0 2.5-5.5 3-5 3 3 5.5-8.5 6-8 5 8 8.5-11.5 9-11 7 15 11.5-14.5 12-14 5 20 So, at the first interval we have zero frequency, so cumulatively we have zero values. For the second interval we have three, so cumulatively we have three. For the third interval we have five values, so cumulatively we have 8. That includes the five for the third interval, plus the three from the previous intervals. We continue this process until the last interval. Notice that when we reach the last interval we have all the values in the distribution represented. So, the bottom cumulative frequency is N or the total number of values in the distribution (20 here). Relative Percent Another column will tell us the proportion of total values that fall at each interval. That is, we will express the frequency (column) as a percentage of the total. To convert the frequency to a percentage take the frequency (f) and divide by the number of values (N). This will give us the proportion of values for that particular interval. Move the decimal over two places (or multiply by 100) to change the proportion into a percent. Thus: Real Limits Class Interval f Cum. f Rel % -.5-2.5 0-2 0 0 0 2.5-5.5 3-5 3 3 15 5.5-8.5 6-8 5 8 25 8.5-11.5 9-11 7 15 35 11.5-14.5 12-14 5 20 25 Cumulative Relative Percent For a final column we will keep a running count of the relative percent column in the same way we did with the cumulative frequency. Keep in mind we are counting relative percentages now as we move down the display. Real Limits Class Interval f Cum. f Rel % Cum. Rel. % -.5-2.5 0-2 0 0 0 0 2.5-5.5 3-5 3 3 15 15 5.5-8.5 6-8 5 8 25 40 8.5-11.5 9-11 7 15 35 75 11.5-14.5 12-14 5 20 25 100 Notice that we can keep a running count of the relative percent column, but we could also obtain the same numbers by computing the percentage for each cumulative frequency as well. Interpretations The data display gives a good deal of information about where values in the sample fall. One good piece of information is about percentiles. A percentile is the percentage at or below a certain score. You often get percentile information when you get your SAT or ACT test scores back. Percentile information is found in the cumulative relative percentage column. Each value in that column tells us the percentage of the distribution at that point or less on the scale. Since we will be rounding values down into a certain interval based on the real limits, then we will indicate where on the scale a certain percentile is based on its corresponding upper real limit. For example, what score corresponds with the 75th percentile? The answer is 11.5 because any values of 11.5 or less are within the bottom 75% of the distribution. Similarly, what percentile is associated with a score of 8.5? We would use the cumulative relative percent that corresponds to 8.5, which is 40%. So, the score 8.5 corresponds with the bottom 40% or 40th percentile of the distribution. Other interpretations from the table can be made as well. For example, we might be interested in how many people fall at a particular interval, or at or below a certain interval. How many scored between 3 and 5? The answer is a found in the frequency column, or three. How many scored 8.5 or less? The answer for this question is in the cumulative frequency column, or eight. Histograms/Bar Graphs We can also take the frequency information in our frequency or grouped frequency distribution and form a graph. In the graph we will form a simple x-y axis. On the x-axis we will place values from our scale, and on the y-axis we will plot the frequency for each point on the scale. For grouped frequency distributions, we will use the midpoint of each interval to indicate different points on the scale. We will continue with our previous example, but notice I have created a new column that indicates the center or midpoint of each interval. We will use this value to graph the display. Real Limits Class Interval MP f Cum. f Rel % Cum. Rel. % -.5-2.5 0-2 1 0 0 0 0 2.5-5.5 3-5 4 3 3 15 15 5.5-8.5 6-8 7 5 8 25 40 8.5-11.5 9-11 10 7 15 35 75 11.5-14.5 12-14 13 5 20 25 100 Note that the bars are touching. The bars touch like this when we are dealing with continuous data rather than discreet data. When the scale measures discreet values we call it a bar graph, and the lines do not touch. For example, if I measured the number of democrats, republicans, and independents in a sample, we would use a bar graph if we wanted to create a data display. 0100200300400500600Dem Rep IndPartyFrequencyLesson 4 Measures of Central Tendency Outline Measures of a distributions shape -modality and skewness -the normal distribution Measures of central tendency -mean, median, and mode Skewness and Central Tendency Measures of Shape With frequency distribution you can an idea of a distributions shape. If we trace the outline of the edges of the frequency bars you can idea about the shape. From this point on, I will draw these shapes to illustrate different point throughout the semester. Keep in mind what you are looking at is a line indicating the frequency or how many values in a distribution lie at a particular point on the scale: just like a histogram. Modality measures the number of major peaks in a distribution. A single major peak is unimodal, and is the most common modality. Two major peaks is a bi-modal distribution. You could also have multi-modal distributions. Skewness measures the symmetry of a distribution. A symmetric distribution is most common, and is not skewed. If the distribution is not symmetric, and one side does not reflect the other, then it is skewed. Skewness is indicated by the tail or trailing frequencies of the distribution. If the tail is to the right it is a positive skew. If the tail is to the left then it is a negatively skewed distribution. For example, a positively skewed distribution would be: 1, 1, 2, 2, 2, 3, 3, 3, 9, 10. The outliers are on the high end of the scale. On the other hand a negatively skewed distribution might be: 1, 2, 9, 9, 9, 10, 10, 10, 11, 11, 11. Here the outliers are on the low end of the scale. The normal distribution is one that is unimodal and symmetric. Most things we can measure in nature exhibit a normal distribution of values. Regression toward the mean is an idea that states values will tend to cluster around the mean with few values toward the trailing ends or tails of the distribution. As a result, most things we measure will tend to have a normal shape. Think about measures of height. There are very few people that are extremely tall or extremely short, but most tend to cluster around the average. With I.Q. scores, measures of weight, or most anything we can measure, the same pattern will repeat. Since most things we measure have more values close to the mean, we end up with mostly normally shaped distributions. Measures of Central Tendency Knowing where the center of a distribution is tells us a lot about a distribution. Thats because most of the scores in a distribution will tend to cluster about the center. Measures of central tendency give us a number that describes where the center lies (and most scores as well). Mean The mean, or average score, is the arithmetic center of the distribution. You can find the mean by adding all the scores ( X ) together and dividing by the number of values you added together (N). X 1 2 3 4 5 15=X N = 5 For the Population: = XN = 515 = 3 For the Sample: x = Xn= 515 = 3 Note that we calculate the mean the same way for both the sample and the population we symbolize them differently. Other statistics will differ in how they are computed for the sample versus the population. Most students are familiar with these measures of central tendency, but there are several properties that may be new to you. 1) The first property of the mean is that it is the most reliable and most often used measure of central tendency. 2) The second property of the mean is that it need not be an actual score in the distribution. 3) The third property is that it is strongly influenced by outliers. 4) The fourth property is that the sum of the deviations about the mean must always equal zero. The last two properties need further explanation. An outlier is an extreme score. It is a score that lies apart from most of the rest of the distribution. If there are several outliers in a distribution it will often result in skewed shape to the distribution. Outliers tend to pull central tendency measures with them. Thus a distribution of values 1, 2, 3, 4, 5 has an average of 3. Three does a good job of describing where most of the scores in this distribution lie. However, if there is an outlier, say by substituting 25 for the 5 in the above distribution, then the mean changes a great deal. The new distribution 1, 2, 3, 4, 25 has a mean of 7. Seven is not really close to most of the other values in the distribution. Thus, the mean is a poor measure of the center when we have outliers, or a skewed distribution. A deviation is just a difference. A deviation from the mean is the difference between a score and the mean. So, when we say the sum of the deviations about the mean must always equal zero is just a way of saying that there are just as many differences between values above the mean and the center as there are differences between values below the mean and the center. Thus, for a simple distribution 1, 2, 3, 4, 5 the average is 3. Lets use population symbols and say = 3. The deviations are the differences between the score and the mean. X X- 1 1-3 = -2 2 2-3 = -1 3 3-3 = 0 4 4-3 = 1 5 5-3 = 2 = 3 )( X = 0 Now if we add these deviations we will always get zero, no matter what original values we use. This concept will be important when we consider standard deviation because we will need to look at differences between values in our distribution and the mean. Median The median is the physical center of the distribution. It is the value in the middle when the values of the distribution are arranged sequentially. The distribution: 1, 1, 2, 2, 3, 4, 5, 5, 5, 6, 7 has a median value of 4 because there are five values above this point and five values below this point in the distribution (1, 2, 2, 3, 4, 5, 5, 5, 6, 7). If you have an even set of numbers then there will be two values that are at the center, and you average these two values together in order to determine the median. For example, if we take out one of the numbers in the distribution so that we have 1, 1, 2, 2, 3, 4, 5, 5, 5, 6 then the two values in the center are 3 and 4 (1, 1, 2, 2, 3, 4, 5, 5, 5, 6). The average is 3.5 and that is the median. The median is resistant to outliers. That is, outliers will generally not affect the median and it will not be affected as much as the mean. It is possible the median might move slightly in the direction of the skew or outliers in the distribution. Mode The mode is the most frequent value in the distribution. It is simply the value that appears most often. For example in the distribution: 1, 1, 2, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7 there is only one mode (4). But, in the distribution: 1, 1, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6 there are two modes (3 and 4). If there is only one of each value then there is no mode. The mode is not affected by outliers. Since it is only determined by one point on the scale, other values will have no effect. Skewness and Central Tendency We have already discussed how each measure is affected by outliers or skewed distribution. Lets consider this information further. In a positively skewed distribution the outliers will be pulling the mean down the scale a great deal. The median might be slightly lower due to the outlier, but the mode will be unaffected. Thus, with a negatively skewed distribution the mean is numerically lower than the median or mode. The opposite is true for positively skewed distributions. Here the outliers are on the high end of the scale and will pull the mean in that direction a great deal. The median might be slightly affected as well, but not the mode. Thus, with a positively skewed distribution the mean is numerically higher than the median or the mode. Lesson 5 Measures of Dispersion Outline Measures of Dispersion - Range - Interquartile Range - Standard Deviation and Variance Measures of Dispersion Measure of central tendency give us good information about the scores in our distribution. However, we can have very different shapes to our distribution, yet have the same central tendency. Measures of dispersion or variability will give us information about the spread of the scores in our distribution. Are the scores clustered close together over a small portion of the scale, or are the scores spread out over a large segment of the scale? Range The range is the difference between the high and low score in a distribution. Simply subtract the two numbers to find the range. So, in the distribution: 1, 3, 5, 9, 11 the range is 11 1 = 10. Remember to subtract the two numbers to give one number for the final answer. Interquartile Range (IQR) The interquartile range is the range of the middle 50% of a distribution. Because any outliers in our distribution must be on the ends of the distribution, the range as a measure of dispersion can be strongly influenced by outliers. One solution to this problem is to eliminate the ends of the distribution and measure the range of scores in the middle. Thus, with the interquartile range we will eliminate the bottom 25% and top 25% of the distribution, and then measure the distance between the extremes of the middle 50% of the distribution that remains. To actually compute some IQRs we would need to use calculus. Instead, of that possibility we will use a method that will yield a consistent, and somewhat accurate answer. Before we compute the value, lets learn some new definitions. A quartile is a quarter or 25% of the distribution. When we compute the IQR we will want to find each of the quartiles. The first quartile is the same as the 25th percentile because 25 percent of the distribution is at or below that point. The second quartile is the same thing as the 50th percentile and the median. The third quartile is the same as the 75th percentile. The IQR is the found by eliminating the values that lie between the bottom end and the first quartile (bottom 25%). We will also eliminate the values between the third quartile and the top of the distribution. We then subtract the new low and high score of the left over middle part of the distributions. So, IQR = Quartile 3 Quartile 1 or IQR = 75th percentile 25th percentile. To compute the IQR first arrange your numbers from lowest to highest and 1) find the median. The median is the 50th percentile and second quartile. Its a starting point for us to find the other quartiles. 2) Next find the median of the bottom half of the distribution (ignoring the top half). This value is the 25th percentile or first quartile because we have taken the bottom 50% and cut it in half. 3) Find the median of the top half of the distribution just like we did for the bottom. This value is the 75th percentile or third quartile. 4) Next subtract the upper and lower medians you found in step 2 and 3. In the following example the median is 8 because it is the average of the two middle numbers. The value 8 is the 50th percentile or second quartile, though we will not use this number in the computation. 1 2 5 6 7 9 10 12 15 19 Once we find the median we can divide the distribution into two halves 1 2 5 6 7 9 10 12 15 19 Bottom 50% Top 50% The median of the bottom 50% is 5. The median of the top 50% is 12. So, IQR = 12 5 = 7 Bottom 25% Top 25% 1 2 5 6 7 9 10 12 15 19 Standard Deviation and Variance While the interquartile range eliminates the problem of outliers it creates another problem in that you are eliminating half of your data. Generally, this is not acceptable because of the difficulty in collecting data in the first place. The solution to both problems is to measure variability from the center of the distribution. Both standard deviation and variance measure how far on average scores deviate or differ from the mean. In this way, we use all the values in our data to compute variability, and outliers will not have undue influence. Middle 50% To compute standard deviation and variance we first start by finding the deviation about the mean. Recall that we did the same thing when discussing properties of the mean. Ill use the same example with the simple distribution 1, 2, 3, 4, 5. First we find the mean and the deviations about the mean. What we want to do is add up these deviations and find out how far on average the scores deviate from the mean. The problem we run into is that whenever we add the deviations (in order to find the average of the deviations) they will always sum to zero. How can we get an average if the sum is always zero? X X- 1 1-3 = -2 2 2-3 = -1 3 3-3 = 0 4 4-3 = 1 5 5-3 = 2 = 3 )( X = 0 One solution is to square all of the deviations. When we square all the numbers the negative values will all become positive and we can then add the deviations without getting zero. X X- (X )2 1 1-3 = -2 4 2 2-3 = -1 1 3 3-3 = 0 0 4 4-3 = 1 1 5 5-3 = 2 4 = 3 )( X = 0 2)( X = 10 Once we add the squared deviations we have a measure of overall variability in the distribution. The sum of the average squared deviations is called the sums of squares, and will be used in almost everyone formula we learn this semester. Please refer back to this section if formulas give you problems later on in the course. Once we divide these squared sums we will get the average squared deviation or variance. In this example it is 10/5 = 2. Since we are in squared units and not the same units as our scale we can take the square root of the variance in order to get the standard deviation. The standard deviation is the average deviation about the mean. For our example we take the square root of 2 and find 1.41 is the standard deviation. The formula that contains all these operations is as follows. Note that 2 is just the symbol we use for population variance and is the symbol we use to denote population standard deviation. ( )22N = X = 2 population variance population standard deviation When dealing with a sample a minor change to the formula is made, and instead of subtracting the numerator by N, we divide by n 1. Try the numbers in the above example to compute the sample variance and standard deviation (variance is 2.5, standard deviation is 1.58). ( )221= nXXS s = s2 sample variance sample standard deviation Please review the animated demonstration on variance and standard deviation for another example of how the population formula works. In addition an alternative formula for these same computations is presented. Although the formula detailed here is the best for understanding the concept, the one presented on the web page will be easier to use in the long run. Both appear in the homework packet formula section as well. See http://faculty.uncfsu.edu/dwallace/ssandrd1.html Lesson 6 Z-Scores Outline Linear Transformation -effect of addition/subtraction -effect of multiplication/division Z-Transformation Z-Distribution -properties Using Z-scores to compare values Linear Transformation Anytime we change a distribution by using a constant we perform a linear transformation. For example if I measure the heights of everyone registered in this course, but then found the tape measure I was using was missing the first two inches (so it started at inch two instead of zero), what would I have to do to find the true heights of everyone? If you think about it you will see that I must subtract two inches from each measurement to get the true heights (because the start position was too high). This example of a linear transformation is one in which we simply shift the numbers up on the same scale. Notice that even though all the numbers move, the relationship between values is not affected. X X + 2 55 57 57 59 58 510 510 60 You will need to know how linear transformations affect the mean and standard deviation of a distribution as well. How does adding or subtracting a constant affect the mean and standard deviation? How does multiplying and dividing a constant affect the mean and standard deviation? When adding or subtracting a constant from a distribution, the mean will change by the same amount as the constant. The standard deviation will remain unchanged. This fact is true because, again, we are just shifting the distribution up or down the scale. We do not affect the distance between values. In the following example, we add a constant and see the changes to the mean and standard deviation. X X +5 1 6 2 7 3 8 4 9 5 10 = 3 = 8 = 1.41 = 1.41 The effect is a little different when we multiply or divide by a constant. For these transformations the mean will change by the same amount as the constant, but this time the standard deviation will change too. That is because when we multiply numbers together, for example, we change the distance between values rather than just shifting them up or down the scale. In the following example, we multiply a constant and see the changes to the mean and standard deviation. X X * 5 1 5 2 10 3 15 4 20 5 25 = 3 = 15 = 1.41 = 7.91 Z-Transformation The z-transformation is a linear transformation, just like those we have discussed. Transforming a raw score to a Z-score will yield a number that expresses exactly how many deviations from the mean a score lays. Here is the formula for transforming a raw score in a population to a Z-score: z = X Notice that the distance a score lies from the mean is now relative to how much scores deviate in general from the mean in the population. Regardless of what the raw score values are in the population, when we use the Z-transformation we obtain a standard measure of the distance of the score from the mean. Anytime Z=1, the raw score that produced the Z is exactly one standard deviation from the mean for that population. Anytime Z=1.5, the raw score that produced the Z is exactly 1.5 standard deviations from the mean for that population. Think for a minute about what it means to know how many standard deviations from a mean a score lays. Consider our simple distribution example. X 1 2 3 4 5 = 3 = 1.41 What z-score will we expect the value 3 to have in this example? That is, how many standard deviations from the mean is 3? The answer is that it is at the mean, so it is zero standard deviations from the mean and we will get a z-score of zero for the original value of three. Now consider the value 1 in the distribution. What z-score will we expect to get for this score? Will it be less than one standard deviation or more than one standard deviation away from the mean? You can estimate the z-score by counting from the mean. One standard deviation is 1.41 units. Counting down from the mean the value 2 is one unit from the mean. Thats a little less than one standard deviation. We have to go down 1.41 units from the mean before we reach one standard deviation. So, when we get to 1 on the scale, we are two units from the mean and a little more than one standard deviation below the mean. Lets transform the simple distribution into a distribution of z-scores by plugging each value into the z-formula: X Z-Tranformation Z 1 ==41.131z -1.42 2 z = X = -.71 3 z = X = 0 4 z = X = .71 5 z = X = 1.42 = 3 = 0 = 1.41 = 1 The value of 1 is 1.42 standard deviations below the mean. The value 2 is .71 standard deviations below the mean, and so on. Notice that the mean is at zero, so any scores below the mean in the original distribution will always have a negative z-score and any score above the mean will have a positive z-score. Properties of the z-distribution. Also notice in the above example that we had to compute the mean and standard deviation of the simple x-distribution in order to compute the z-score. We can compute the mean and standard deviation of the resulting z-distribution as well. The mean of the z-distribution will always be zero, and the standard deviation will always be one. These facts make sense because the mean is always zero standard deviations away from the mean, and units on the z-distribution are themselves standard deviations. Using Z-scores to Compare Values Since z-scores reflect how far a score is from the mean they are a good way to standardize scores. We can take any distribution and express all the values as z-scores (distances from the mean). So, no matter scale we originally used to measure the variable, it will be expressed in a standard form. This standard form can be used to convert different scales to the same scale so that direct comparison of values from the two different distributions can be directly compared. For example, say I measure stress in the average college sophomore on a scale between 0 and 30 and find the mean is 15. Another researcher measures stress with the same population, but uses a scale that ranges from 0 to 300 with a mean of 150. Who has more stress, a person with a stress score of 15 from the first distribution or a person with a stress score of 150 from the second group? The value of 150 is a much larger number than 15, but both are at the mean of their own distribution. Thus, they will both have the same z-score of zero. Both values indicate an average amount of stress. Consider another example. Lets say that Joan got an x = 88 in a class that had a mean score of 72 with a standard deviation of 10 ( = 72, = 10). In a different class lets say Bob got a x = 92. The mean for Bobs class, however, was 87 with a standard deviation of 5 ( = 87, = 5). Who had the higher grade relative to their class? If you think about it for a second you will know that Joans score of 88 is much higher relative to the average of 72 compared to Bobs score of 92 to the average of 87. We can easily compare the values, however, if we simply compute the z-score for each. Joan 6.11016107288 ===Z Joan has the higher score because she is 1.6 standard deviations above the mean, and Bobs score is only 1 standard deviation above the mean. Bob 15558792 ===ZLesson 7 Z-Scores and Probability Outline Introduction Areas Under the Normal Curve Using the Z-table Converting Z-score to area -area less than z/area greater than z/area between two z-values Converting raw score to Z-score to area Converting area to Z-score to raw score Introduction/Area Under the Curve Please note that area, proportion and probability are represented the same way (as a decimal value). Some examples require that you convert the decimal value to a percentage. Just move the decimal place to the right two places to turn the decimal into a percentage. Start this section by reviewing the first two topics in the above outline on the web page. Find the Z-score animated demonstrations or click here http://faculty.uncfsu.edu/dwallace/sz-score.html Using the Z-table Knowing the number of standard deviations from the mean gives us a reliable way to know how likely a score is for a population. There is a table of z-scores that gives the corresponding areas or probabilities under the curve. You will need the z-table in Appendix B of your text for this discussion. See page A-24 through A-26 in your text. The table shows the z-score in the left column, and then the proportion or area under the curve in the body, and finally there is a third column that shows the proportion or area under the curve in the tail of the distribution. Whenever we compute a z-score it will fall on the distribution at some point. The larger portion is the body, and the smaller portion is the tail. If the z-score is positive, then the body will be the area that is below that z-score, and the tail will be the area that is above that z-score. If the z-score is negative, then the body will be the area that is above that z-score, and the tail will be the area that is below that z-score. Converting a Z-Score to an Area Finding areas below a z-score What area lies below a z-score of +1? If we look this z-score up on the table we find that the area in the body is .8413 and the area in the tail is.1587. Since the z-score is positive and we want the area below the z-score, then we will want to look at the body area. So, .843 is the proportion in the population that have a z = 1 or less. Now lets consider the situation if the z-score is negative. What area lies below a z-score of -1.5? You will not find negative values on the table. The distribution is symmetric, so if we want to know the area for a negative value we just look up the positive z-score to find the area. We will use a different column on the table, and that is why we must consider whether z is positive or negative when using the table. If we look this z-score (1.5) up on the table we find that the area in the body is .9332, and the area in the tail is .0668. Since the z-score is negative and we want the area below that point we will be using the tail area. So, .0668 is the proportion in the population below a z-score of -1.5. Finding areas above a z-score The process for this type of problem is the same as what we have already learned. The only difference is in which column we will be using to answer the question. What area lies above a z-score of +1? If we look this z-score up on the table we find that the area in the body is .8413 and the area in the tail is.1587. Since the z-score is positive and we want the area above the z-score, then we will want to look at the tail area. So, .1587 is the proportion in the population that have a z = 1 or more. Now lets consider the situation if the z-score is negative. What area lies above a z-score of -1.5? Again, you will need to look up the positive z-value for 1.5. If we look this z-score up on the table we find that the area in the body is .9332, and the area in the tail is .0668. Since the z-score is negative and we want the area above that point we will be using the body area. So, .9332 is the proportion in the population above a z-score of -1.5. Finding areas between two z-scores When we have two different z-scores and want to find the area between them, we first must consider if both values are on the same side of the mean, or if one value is positive and the other negative. For our table, if the values are either both positive zs or both negative zs, we can find the tail area for both z-scores and subtract the two areas. You could just as easily find the two body areas for both z-scores and subtract them as well. For example, what is the area between Z = 1 and Z = 1.5? Since both scores are positive and we want the area between them, we will look up the tail area and subtract the two table values. Note that you never subtract z-scores, only areas from the table. On the other hand, if you have one positive and one negative z-score then you must use the body area for either one of the z-scores, and the tail area for the other. Once you get the two areas off the table, then you subtract the two areas. For example, what is the area between Z = -1 and Z = 1.5? Since one score is positive and the other negative, the area we are looking for will cross the mean. Use the body area for one value and the tail area for the other. Once you get these values off the table, subtract them to find the area in between. Converting a Raw Score to a Z-score and then into an Area These problems are exactly the same as the others we have been working. You must still find areas above/below/and between two z-scores, but now you must first compute the z-value using the z-formula before using the table. For example lets look at IQ scores for the population with a mean of 100 and standard deviation of 15 ( = 100, = 15). What proportion of the population will have an IQ of 115 or less? I first must compute Z. It is equal to z = 1. Now the question becomes what proportion of scores lie above z = 1? Please review other examples of this type of problem on the web-page. Find the link to Z-scores and probability or click here http://faculty.uncfsu.edu/dwallace/sz-score2.html Converting an Area to a Z-score and then into a Raw Score For these problems we will be doing the same process we have been doing, but everything will be done in reverse order. We will start with a given area or proportion. You then use the z-table to find the area. However, when you use the z-table for these problems you must look up the area in either the body or tail column and then trace it back to find the z-score. Once we get the z-score we will plug in the values we know and solve for X in the z-score equation. For example, IQ scores for the population with a mean of 100 and standard deviation of 15 ( = 100, = 15), what score cuts off the top 10% of the distribution? Notice that these questions are always asking what the score is for a certain point. We are solving for X now. Prior examples were all asking for an area or proportion. The first step to solving this type of problem is to find the Z-score. We wont be computing z, but instead finding it from the table. Since we want the top 10%, we will be looking for the area on the table where the tail is .10 and the body is .90. You can look in either column, and it might help to draw the distribution in order to be sure you are using the right column. Make sure you are not using the z-column at this point. We find the z-score that leaves the tail at the top of the distribution equal to .10 is Z = 1.28. Always use the z-score closest to the area of interest. Once we have that number, we can plug in what we know into the z-formula and solve for X. Alternatively, if you have trouble with algebra, you can use the following formula: X = Z + Special note for values in the lower 50% of the distribution: Whenever we want to find a z-score for a value below the mean, we must remember to make the value negative. Recall that the z-table only gives positive z-values. If the value is below the mean, then you must remember to insert the negative sign before doing the computation. For example, if I were looking for the IQ score for the bottom 10% of the distribution, in the above example, then I would have look up a tail area of .10 or body area of .90. The z-score we need is -1.28 even though the table shows only the positive value. Please review other examples of this type of problem on the web-page. Find the link to Z-scores and probability or click here http://faculty.uncfsu.edu/dwallace/sz-score2.html Lesson 8 Probability Outline Probability of an Event Probability of Single Events Probability of Multiple Events -without replacement -mutually exclusive events Conditional Probability Probability of an Event There are three classes of events: Impossible Events--------------Possible Events-----------Certain Events P = 0 P = 0 to 1 P = 1 The probability (P) of an impossible event is zero because there is zero chance of it happening. A certain event has a probability of one or 100% because it will always happen. Most of the events we will be interested in are possible events. These probabilities will always have a value between zero and one. Always leave your answer in decimal form, instead of fractional form. Simply divide out any fraction you have by dividing the top number by the bottom number. So, is 0.25. A simple experiment we could run to examine probabilities is to roll a six-sided die. What is the probability of rolling a 5 on a single die roll? Most people know it is 1/6 or .167 because there is only one side that is a 5 and there are six sides on the die. You can determine the probability of any even in this manner. Probability itemsTotalcriteriainitemsP__#___#)( = So, we were only looking for one side in the last problem, out of a total of six sides. Another experiment we could use to look at probabilities is drawing cards from a standard deck. If I draw a card from a deck of cards what is the probability it is a heart? P 25.0415213)( ===Heart Note that a standard deck of cards has 52 cards with 13 hearts/13 clubs/13 diamonds/13 spades. Since diamonds and hearts are red cards and the rest are black, there are 26 red and 26 black cards. Probability of Single Events An individual event is a single event. With single events we are measuring the likelihood of a one thing happening. We might be interested in different outcomes, but we are still just going to roll the die once or draw a single card from a deck. For example, what is the probability that I roll a 5 or a 6 on a single die roll or P (5 or 6)? With single events you will see this or connector and you will add the two individual probabilities. So: P (5 and 6) = 1/6 + 1/6 = .167 + .167 = .334 What is the probability I draw a Heart or Club with a single draw from a standard deck of cards? P (Heart or Club) P(Heart or Club) = P(Heart) + P(Club) = 13/52 + 13/52 = .25 + .25 = .5 Probability of Multiple Events With multiple events we will be interested more than one outcome be realized. So, we will roll the die more than once or draw more than one card from a deck. For example, what is the probability of rolling a 5 and a 6 on two die rolls. To get both a five and a six I will have to roll the die more than once. When you see this and connector you will multiply individual probabilities. P (5 and 6) = P(5) * P(6) = 1/6 * 1/6 = .167 * .167 = .028 We will only be dealing with independent events in this section, or events that do not affect the outcome of other events. With the card experiment, then, we will not look at multiple draws where one draw could affect the probability of a separate event. For example, what is the probability of drawing a Heart and a Club from standard deck? P(Heart and Club) = P(Heart) * P(Club) = 13/52 * 13/52 = .25 * .25 = .062 Without Replacement Although we will focus on independent events like the last example, we will also consider what happens to probabilities in situations in which there is no replacement. The above examples assumed that once we drew a card from the deck that it was replaced before another draw was made. Notice that when figuring how many total events there were we used 52 every time because we assumed each draw was from a fresh deck. If the problem, however, specifies that there is no replacement then we must take this into account when figuring the probabilities. For example, what is the probability of drawing a Heart and a Club from a deck without replacement? When we count how many cards are left for the Club draw, there will be one less card in the deck because we already had to draw the Heart from the deck. Thus: P(Heart and Club) = P (Heart) * P (Club) = 13/52 * 13/51 = .25 * .255 = .064 We might also have to subtract a value from the numerator as well as the denominator. Try to find the probability of drawing three red cards from a deck without replacement. (Answer: 0.1176) Mutually Exclusive Events Mutually exclusive events are events that cannot happen together. For example, being a freshman and a sophomore are mutually exclusive. You are either one or the other but not both. For mutually exclusive events the probability the two events will occur together must always equal zero. Conditional Probability With conditional probabilities we will consider the probability of an event given that some other event has already happened. Thus, these are not independent events, and the rules we learned above will not apply. For these problems frequency data (or counts) will be given in a contingency table. This table will display the frequencies for different combinations of events. For example, consider the probability of having a computer or not, and living the U.S. or elsewhere. In U.S. Not in U.S. Computer 30 15 No Computer 10 20 Before we consider conditional probabilities, lets look at some of the types of questions we have already examined. For example, what is the probability that choosing someone from our sample will yield a person with a computer? To answer this question we will need to add up the total for each row and column in the table: In U.S. Not in U.S. Total Computer 30 15 45 No Computer 10 20 30 total 40 35 Since there are a total of 45 people in our sample with a computer out of 75 total people, there is a 0.6 probability that a random draw will yield a person with a computer. Now find the probability that a random draw will yield someone with a computer that is living in the U.S. Instead of looking in the total column for this type of problem, we will use one of the original values. There are 30 people living in the U.S. that also have a computer. So, 30 out of the total of 75 people or 0.4 live in the U.S. and have a computer. For a single event in a table like this one, use the values in the margin or the totals, and divide by the total number in the sample. For a combined event, use the original table values out of the total. For conditional probabilities we will restrict our sample to those items given to have already have happened. For example we might know the probability of having a computer and living in the U.S. for the entire sample. We could make this conditional by saying what is the probability of having a computer given that we know the person is living in the U.S.? With the second question we are not asking the probability of picking a person at random from the total, but instead we are restricting our sample to just those that live in the U.S. For conditional probabilities the total or denominator is the value given. For this example it is the total for those living in the U.S. or 40. We want to know what proportion have a computer out of these 40 people living in the U.S. Since 30 of those in the U.S. have a computer out of 40 the probability is 30/40 = .75 In this same example what is the probability someone does not live in the U.S. given they have a computer? We can write: P (not in U.S. | computer) Where the first probability is the one we are interested in, the vertical line means given and the second probability computer is what is given. Again, our new total is those with a computer or 45. We want to know the proportion out of these that are not in the U.S. or 15. So, the conditional probability is 15/45 = .33 Lesson 9 Hypothesis Testing Outline Logic for Hypothesis Testing Critical Value Alpha () -level .05 -level .01 One-Tail versus Two-Tail Tests -critical values for both alpha levels Logic for Hypothesis Testing Anytime we want to make comparative statements, such as saying one treatment is better than another, we do it through hypothesis testing. Hypothesis testing begins the section of the course concerned with inferential statistics. Recall that inferential statistics is the branch of statistics in which we make inferences about populations from samples. Up to this point we have been mainly concerned with describing our distributions using descriptive statistics. Hypothesis testing is all about populations. Although we will start using just one value from the population and eventually a sample of values in order to test hypotheses, keep in mind that we will be inferring that what we observe with our sample is true for our population. We will want to see if a value or sample comes from a known population. That is, if I were to give a new cancer treatment to a group of patients, I would want to know if their survival rate, for example, was different than the survival rate of those who do not receive the new treatment. What we are testing then is whether the sample patients who receive the new treatment come from the population we already know about (cancer patients without the treatment). Again, even though we are talking about a sample, we infer that the sample is just part of an entire population. The population is either the one we already know about, or some new population (created by the new treatment in this example). Logic 1) To determine if a value is from a known population, start by converting the value to a z-score and find out how likely the score is for the known population. 2) If the value is likely for the known population then it is likely that it comes from the known population (the treatment had no effect). 3) If the value is unlikely for the known population then it is probably does not come from the population we know about, but instead comes from some other unknown population (the treatment had an effect). 4) A value is unlikely if it is less than 5% likely for the known population. Any value that occurs 5% or more of the time for the known population is likely and part of the known population. The 5% cut-off point is rather arbitrary, and it will change as we progress. For now we will use it as a starting point to illustrate several concepts. Lets look at a simple example. Say the earth has been invaded by aliens that look just like humans. The only way to tell them apart from humans is to give them an IQ test since they are quite a bit smarter than the average human. Lets say the average human IQ (the known population) is = 100 = 15. We want to know if Bob is an alien. Bob has an IQ score of 115? Is it likely that he comes from the known population and is human, or does he come from a different alien population? To answer the question, first compute the z-score. 1151515100115 ===Z Next, find out how likely this z-score is for the population. For hypothesis testing we will always be interested in whether the value is extreme for the population, or unlikely. Thus, we will be looking in the Tail Column when deciding if the value is unlikely. So, the likelihood of observing an IQ of 115 or more is .1587. Since the probability is not less than 5% or .05 we have to assume Bob comes from the general population of humans. Say Neil has an IQ of X = 30. Is it likely Neil comes from the general population? Again we first compute the z-score, and then find how likely it is to get that value or one more extreme. 2153015100130 ===Z Next, find out how likely this z-score is for the population. So, the likelihood of observing an IQ of 130 or more is .0228. Since the probability is less than 5% or .05 we have to assume Neil comes from a different population than the one we know about (the general population of humans). Critical Value Critical Values are a way to save time with hypothesis testing. We dont really have to look up the probability of getting a particular value in order to verify it is less than 5% likely. The reason for this fact is that the z-score that marks the point where a value becomes unlikely does not change on the z-scale. That is, there is only one z-score at the that is 5% likely. Any z-score beyond that point is less than 5% likely. Thus, I dont have to look up each particular area when I compute my z-score. Instead I only have to verify that the z-score I computed is more extreme than the one that is 5% likely. So, we can stop once we compute the z-score without reference to the z-table. What z-score will be exactly 5% likely for any population? This is the z-score we will make comparisons against. Use the z-table to determine the z-score that cuts off the top 5% of scores. Finding the critical value is important, and will be one of the steps that must be performed anytime we conduct a hypothesis test. Since we will be doing hypothesis testing from this point on, many points on subsequent exams will come from just knowing the critical value. So, Z = 1.64 is the critical value. Any value more extreme than 1.64 is unlikely, and all other events will be likely for the known population. If Z=2.1, or Z=1.88 you conclude that the value is unlikely, and so must be part of a different population. If Z = 1.5, or Z =0.43 you conclude that the value is likely, and so must be part of the known population. Note that we were only working on one side of the distribution in the above problem. If we were interested in a value that was below the mean, instead of above it, then we would flip our decision line to the other side. Since the distribution is symmetric, the numbers will not change. Alpha Alpha is the probability level we set before we say a value is unlikely for a known population. The critical value we just found is only one that we will use. It assumes that a value must be less than 5% likely to be unlikely, and therefore part of a different population. Alpha was .05 ( = .05) for that example. Sometimes researchers want to be very sure before they decide a value is different. Thus, we will also use an alpha level of .01 or 1% as well. If alpha is .01, then a value must be less than 1% likely before it is said to be unlikely for a known population. If alpha is 1% then the critical value will be different than the one we found above. Alpha is given in every problem, but you must use that information to determine the critical value. What z-score will be exactly 1% likely for any population? This is the z-score we will make comparisons against when alpha is set to 1% ( = .01). Use the z-table to determine the z-score that cuts off the top 1% of scores just like the last example. Use the Tail column and find .01. You could also look in the Body Column and find .99. We will use Z = 2.33 when Alpha is set to the 1% level. Also note that when we are interested in determining if values below the mean are unlikely our critical value will be negative. One-Tail versus Two-Tail Tests Another factor that will affect our critical value is whether we are performing a one or a two-tail test. The critical values we have looked at so far were for one-tail tests because we were only looking at one tail of the distribution at a time (either on the positive side above the mean or the negative side below the mean). With two-tail tests we will look for unlikely events on both sides of the mean (above and below) at the same time. I will discuss how to determine if a problem is a one or a two tail test in a later lesson, but lets go ahead and find the critical values for two-tailed test the same way we did the one-tail tests above. Lets begin with an alpha level of 5%. We still want 5% of our events to be unlikely and 95% of our events to be likely for the known population. Now, however, we want to be looking for unlikely events in both directions at the same time. So, we will split the unlikely block into two parts, each half the total 5% area. What z-scores will then mark the middle 95% of our distribution? You will have to look up an area of .025 in the Tail Column of the z-table. The process for finding the two-tail critical values when alpha is set to .01 is the same. This time we will want 99% of our values in the middle, leaving only .005 or half of one percent on each side. Can you find the critical value on the z-table (answer: z = 2.58). So, we have learned four critical values. 1-tail 2-tail = .05 1.64 1.96/-1.96 = .01 2.33 2.58/-2.58 Notice that you have two critical values for a 2-tail test, both positive and negative. You will have only one critical value for a one-tail test (which could be negative). Lesson 10 Steps in Hypothesis Testing Outline Writing Hypotheses -research (H1) -null (H0) -in symbols Steps in Hypothesis Testing -step1: write the hypotheses -step2: find critical value -step3: conduct the test -step4: make a decision about the null -step5: write a conclusion Writing Hypotheses Before we can start testing hypotheses, we must first write the hypotheses in a formal way. We will be writing two hypotheses: the research (H1) and the null (H0) hypothesis. The research hypothesis matches what the researcher is trying to show is true in the problem. The null is a competing hypothesis. Although we would like to directly test the research hypothesis, we actually test the null. If we disprove the null, then we indirectly support the research hypotheses since it competes directly with the null. We will discuss this fact in more detail later in the lesson. Again, the research hypothesis matches the research question in the problem. Lets take a look at a sample problem: Suppose some species of plants grows at 2.3 cm per week with a standard deviation of 0.3 ( = 2.3 = 0.3). I take a sample plant and genetically alter it to grow faster. The new plant grows at 3.2 cm per week (X = 3.2). Did the genetic alteration cause the plant to grow faster than the general population? Set alpha = .05. Lets focus on writing hypotheses, rather than any other steps we have learned for now. In order to write the research hypothesis look at what the researcher is trying to prove. Here we are trying to show that the genetically altered plant grows at a faster rate than unaltered plants. Thats what we want the research hypothesis to say. However, when you write your hypotheses, be sure to include three elements: 1) explicitly state the populations you wish to compare. For now, one will be a treatment population and the other will always be the general population. 2) State the dependent variable. We have to be explicit about the scale on which we expect to find differences. 3) State the type or direction of the effect. Are we predicting the treatment population will be greater or less than the general population (1-tail)? Or, are we looking for differences in either direction at the same time (2-tail)? The above problem is one-tail since we are looking for a growth rate higher than the average. Look for words that indicate a direction in the problem for one-tail test (e.g. higher/lower, more/less, better/worse). It would be two-tailed if the problem had stated that we expected a different growth rate than the general population. Different could be higher or it could be different because it is lower. The current example is easy to translate into a hypothesis, but check the homework packet because the wording is not always so obvious. For the research hypotheses (denoted by H1 ): H1: The population of genetically altered plants grows faster than the general population. You could vary the wording a bit, as long as you include the three elements. Notice that we state both the treatment population and the population we will compare that to, the general population. Growth rate is the dependent variable, and we indicate the direction by saying it will grow faster. The null hypothesis (denoted by H0) is a competing hypothesis. Its basically the opposite of the research hypothesis. In general it states that there is not effect for our treatment or no differences in our populations. For this example: H0: The population of genetically altered plants grows at the same or lower rate as the general population. Ive included the same or lower wording for the one-tail test because we want to cover all the possible outcomes of the test. We only want to show that the treatment population grows faster. If they end up growing slower it wont support the research hypothesis, so we include left-over elements with the null. For two-tail tests, substitute different for the word faster in the research hypothesis. The two-tail null would say the groups are do not differ. In Symbols We can also write the hypothesis in notational form. We will restate both the null and research hypotheses in symbols we have been using for our formulas. Thus: H1: gen.alt. > 2.3 H0: gen.alt < 2.3 Notice that we represent the treatment population with a mu (). We do this because we want to make inferences about the population, not the single value sample I am using to test the hypothesis. Our inferences will be that the entire population the plant comes from grows at a faster rate. The value of 2.3 is the general population mean we are comparing against. Although it is represented with a mu in the problem, we dont the symbol because we know the exact value for that population. For two-tail test we simply change the direction arrows to equal/not-equal signs (an = sign for the null and / sign for the research hypothesis). Steps in Hypothesis Testing Now we can put what we have learned together to complete a hypothesis test. The steps will remain the same for each subsequent statistic we learn, so it is important to understand how one step follows from another now. Lets continue with the example we have already started: Suppose some species of plants grows at 2.3 cm per week with a standard deviation of 0.3 ( = 2.3 = 0.3). I take a sample plant and genetically alter it to grow faster. The new plant grows at 3.2 cm per week (X = 3.2). Did the genetic alteration cause the plant to grow faster than the general population? Set alpha = .05. Step 1: Write the hypotheses in words and symbols H1: The population of genetically altered plants grows faster than the general population. H0: The population of genetically altered plants grows at the same or lower rate as the general population. H1: gen.alt. > 2.3 H0: gen.alt < 2.3 Step 2: Find the critical value for the test Since alpha is .05, and it is a one-tail test because we think our treatment will produce plants that grow faster than the general population: Zcritical =1.64 Step 3: Run the test Here we find out how likely the value is by computing the z-score. 33.09.03.03.22.3 ===Z Step 4: Make a decision about the Null Reject the Null or Fail to Reject the Null (retain the null) are the only two possible answers here. Since the value we computed for the z-test is more extreme than the critical value, we reject the Null. Graphically, though not required for the answer, we have: Note that we are testing the Null. It is either proven or disproven. We never prove the research hypothesis, even thought that is our intent. Instead, if we disprove the null, we indirectly support the research hypothesis. This is true, because you will notice that our decision is based on the statistical test that the treatment value is not likely to have come from the same population. We infer it is a different population, but we actually prove that it is not from the same population. It may seem like a matter of semantics, but indulge me on this one. Step 5: Write a conclusion For this example, we conclude: The population of genetically altered plants grows at a different rate than the general population. Although we have a conclusion in step 4, write a conclusion here in plain language without any statistical jargon. What did our test show? If you reject the null, then the then there was a difference (treatment had an effect). The research hypothesis is your conclusion (you can simply restate it from Step 1). If you fail to reject the null, then the null hypothesis is your conclusion (again, you can just rewrite it from Step 1). Lesson 11 Hypothesis Testing with a Sample of Values We have looked at the basics of hypothesis testing using the z-formula we had already learned. However, we never test a hypothesis based on one individual from a population. Instead, we will want to have a sample of values to test against the population. The formula we will want to use has a minor change from the one we have been using. xXz = , where x = n Notice that there is sample mean now in the numerator instead of just a single x-value. Often this will be given just like the x-value in prior problems, but now you may also have to compute it from the sample. Compute x first for the denominator by dividing the standard deviation by the square root of the given sample size (n). Once you get that number plug it in as the denominator in the z-score formula. The rest of this lesson is devoted to the theory behind the changes we make when moving from tests with a single x-value to tests with samples of x-values. There are no computational additions for the exam other than the formula change above. However, you should be concerned with understanding the conceptual meaning of this lesson. At a minimum you should be able to recognize the rules of the Central Limit Theorem for the exam (detailed below). The lesson continues on the web page. Take notes on the sampling distribution page, the standard error page, and the standard error with hypothesis testing page. Links to these pages are provided below. It is important to review each one. Again, these lessons contain conceptual information for the most part, however, the last page is devoted to the computations you will perform for the exams. Sampling Distributions Standard Error Standard Error and Z-score Hypothesis Testing Lesson 12 Errors in Hypothesis Testing Outline Type I error Type II error Power Examples in the real world Anytime we make a decision about the null it is based on a probability. Recall that we reject the null when it tests a value that is unlikely for the known population. Extreme values are unlikely for the population, but not impossible. So, there is always some chance that our decision is in error. Note that we will never know whether we know we have made an error or not with our hypothesis test. When running a test, I only know what my decision is about the test, and not the true state of reality. Thus, this discussion on errors is strictly theoretical. Type I Errors Whenever a value is less than 5% likely for the known population, we reject the null, and say the value comes from some other population. Notice that we are saying the value is really from another population distribution out there that we dont know about. However, some of the time the value really does come from the known population. Notice that even though the value represented is beyond the critical value it still lies under the curve for the normal population. We reject any values in this range, even though they really are part of the known population. When we reject the null, but the value really does come from the known population a Type I error has been committed. A Type I error, then, happens when we reject the null when we really should have retained it. Note that a Type I error can only occur when we reject the null. The part of the distribution that remains under the curve for the known population but is beyond our critical value in the region of rejections is alpha (). When we set alpha we are setting the probability of making a Type I error. Type II Errors Whenever a value is more than 5% likely for the known population, we retain the null, and say the value comes from the known population. But, some of the time the value really does come from a different unknown population. Notice in this situation the value is below the critical value, so we retain the null. However, the value is still under the unknown population distribution, and may in fact come from the unknown population instead. Thus, when I retain the null, when I should really have rejected it I commit a Type II error. The probability of making a Type II error is equal to beta and not strictly defined by alpha. Although we know the probability of a Type I error because we set alpha, a Type II error takes in a few more factors than that. You can see the region of Beta () below. Notice that it is the area below the critical value, but that is still part of the other unknown distribution. Power Power is the probability of correctly rejecting the null hypothesis. That is, it is the probability of rejecting the null when it is really false. Again, we never really know if the null is false or not in reality. Power is another way of talking about Type II errors. Such errors have been recognized as a problem in the behavioral sciences, so it is important to be aware of such concepts. However, we will not be computing power in this course. You can ignore the power demonstration on the web page for that reason. An easy way to remember all these concepts might be to put them in a table, much like your textbook does. Examples of Errors in the Real World Another way to think about Type I and Type II errors is to think of them in terms of false positives and false negatives. A Type I error is a false positive, and a Type II error is a false negative. A false positive is when a test is performed and shows an effect, when in fact there is none. For example, if a doctor told you that you were pregnant, but you were not then it would be a false positive result. The test shows a positive result (what you looking for is there), but the test if false. A false negative is when a test is performed and shows no effect, when in fact there is an effect. The opposite situation of the above example would apply. A doctor tells you that you are not pregnant, when if fact you are pregnant. The test shows a negative result (what you are looking for is not there), but the test is false. Lets look at another example. A sober man fails a blood alcohol test. What type of error has been committed (if any)? For this type of problem you will get two pieces of information. First, whether the test was positive or negative. The test is positive if what you are looking for is found. It is negative if the test shows what you are looking for is not there. The second piece of information is whether the test is in error or not (false or true test). Thus, for this example, the test is positive because if you fail a blood alcohol test it is showing that there is alcohol in your system. You are positive for alcohol in that case. Since the man is sober, the test is false. So, here we have a false positive test or Type I error. Lesson 13 Hypothesis Testing with the t-test Statistic Outline Unknown Population Values The t-distribution -t-table Confidence Intervals Unknown Population Values When we are testing a hypothesis we usually dont know parameters from the population. That is, we dont know the mean and standard deviation of an entire population most of the time. So, the t-test is exactly like the z-test computationally, but instead of using the standard deviation from the population we use the standard deviation from the sample. The formula is: t = X sx , where sx = sn The standard deviation from the sample (S), when used to estimate a population in this way, is computed differently than the standard deviation from the population. Recall that the sample standard deviation is S and is computed with n-1 in the denominator (see prior lesson). Most of the time you will be given this value, but in the homework packet there are problems where you must compute it yourself. The t-distribution There are several conceptual differences when the statistic uses the standard deviation from the sample instead of the population. When we use the sample to estimate the population it will be much smaller than the population. Because of this fact the distribution will not be as regular or normal in shape. It will tend to be flatter and more spread out than population distribution, and so are not as normal in shape as a larger set of values would yield. In fact, the t-distribution is a family of distributions (like the z-distribution), that vary as a function of sample size. The larger the sample size the more normal in shape the distribution will be. Thus, the critical value that cuts off 5% of the distribution will be different than on the z-score. Since the distribution is more spread out, a higher value on the scale will be needed to cut off just 5% of the distribution. The practical results of doing a t-test is that 1) there is a difference in the formula notation, and 2) the critical values will vary depending on the size of the sample we are using. Thus, all the steps you have already learned stay the same, but when you see that the problem gives the standard deviation from the sample (S) instead of the population (), you write the formula with t instead of z, and you use a different table to find the critical value. The t-table Critical values for the t-test will vary depending on the sample size we are using, and as usual whether it is one-tail or two-tail, and due to the alpha level. These critical values are in the Appendices in the back of your book. See page A27 in your text. Notice that we have one and two-tail columns at the top and degrees of freedom (df) down the side. Degrees of freedom are a way of accounting for the sample size. For this test df = n 1. Cross index the correct column with the degrees of freedom you compute. Note that this is a table of critical values rather than a table of areas like the z-table. Also note, that as n approaches infinity, the t-distribution approaches the z-distribution. If you look at the bottom row (at the infinity symbol) you will see all the critical values for the z-test we learned on the last exam. Confidence Intervals If we reject the null with our hypothesis test, we can compute a confidence interval. Confidence intervals are a way to estimate the parameters of the unknown population. Since our decision to reject the null means that there are two populations instead of just the one we know about, confidence intervals give us an idea about the mean of the new unknown population. See the Confidence Interval demonstration on the web page or click here http://faculty.uncfsu.edu/dwallace/sci.html for the rest of the lesson. Lesson 14 Independent Samples t-test Outline No Population Values Changes in Hypotheses Changes if Formula -standard error Pooled Standard Error -weighted averages Critical Values -df Sample Problem No Population Values With the independent samples t-test we finally reach the point where we have no population values. This fact is important because when we test hypotheses we are usually testing an idea and a population that we know nothing about. Think about the kinds of scientific discoveries you hear about often. New treatments for diseases, new drugs, or new techniques for improving depression all involve testing a population created by the treatment or drug or technique. So, with the independent samples t-test we will compare two sample values directly. Note that we are still making the inference about the populations from which the samples are drawn. Changes in Hypotheses All hypotheses from this point on in the course will be two-tailed. In addition, since we no longer no any population values we will use mu to represent both populations. So for example, H0: diet = placebo H1: diet placebo Formula Changes Recall the formula for the t-test we have been using: t = X sx , where sx = sn The numerator will now have two sample values )( 21 XX instead of one sample and one population. The denominator, recall, is the standard error (the standard deviation divided by the square root of the sample size). Our standard error (denominator) was: sx = sn Remember that the standard error measures variability we expect to see among samples. Now that we have two samples we will want to include the estimate of variability from both. Thus, we will have to take into account the standard deviations and sample sizes of both samples. We will compute the standard error separately for each sample and then add them together. Because of the formula we will develop, it will be easier if we switch from using the standard deviation to the variance. In this way we can eliminate the radical in the denominator. The two formulas are equivalent: ns = ns 2 Since we are adding the two separate standard errors together we have: 22212121 nsnss XX += Notice that we now denote the combined standard error with (21 XXs ). Again, its just a way to symbolize the final value we will divide into the numerator. 21)( 21XXsXXt= where sX 1 X 2 =s12n1+ s22n2 Pooled Standard Error Note that the formulas I present in this section differ from your text! The above formula is useful when our sample sizes are the same. However, in situations where our sample sizes are different, we cannot simply add the two standard errors together. Instead we have to give more weight to the larger sample. Weighted Averages Lets say I have one sample with N = 20 S = 1.5 Another sample has: N = 100 S = 15 Although this example is extreme, you can see that you would not want to simply average the two groups together in order to get the average of S. If we did that we would have the average of two groups rather than the average of all 120 people. Simple average of two groups: 15 + 1.5 = 16.5 25.825.16 = Weighted average of 120 people: 75.12120153012030150020100)20(5.1)100(15 ==+=++ For the weighted average we are multiplying each variance times the sample size to get a sum of all 120 people, and in the final step we divide by the total number of people. You can see that if I have 100 people with such a large variance that the average of those people plus 20 more of them with a small standard deviation should yield a value closer to the larger group (12.5) than the smaller group (8.25). When we have unequal sample sizes we will want to use a similar process to average or pool the variances from our two samples. Below is the formula that does just that. Notice that we are doing the same process we used for the weighted average above. We multiply the variances times the sample size and divide by the total number of people. The value n-1 or degrees of freedom is used to represent the sample size. ( ) ( )211212222112++=nnsnsns p Here 2ps is the symbol we use for the pooled variance. Once we compute that value we plug it into the same formula we used with equal sample sizes, but now denote the variance as pooled. 21)( 21XXsXXt= sX 1 X 2 =sp2n1+ sp2n2 Critical Values We will use the same table to find the critical values as we did with the one-sample t-test. However degrees of freedom are now computed from two samples, so: df = n1 + n2 2 Sample Problem A new program of imagery training is used to improve the performance of basketball players shooting free-throw shots. The first group did an hour imagery practice, and then shot 30 free throw basket shots with the number of shots made recorded. A second group received no special practice, and also shot 30 free throw basket shots. The data are below. Did the imagery training make a difference? Set alpha = .05. X1 X2 15 5 17 6 20 10 25 15 26 18 27 20 646.2566.21211===nSX 646.3933.12222===nSX Step 1: Write the hypotheses in words and symbols H1: The population receiving imagery practice will make a different number of baskets than the population receiving no imagery practice. H0: The population receiving imagery practice will make a different number of baskets than the population receiving no imagery practice. H1: imagery. no imagery H0: imagery = no imagery Step 2: Find the critical value for the test Since alpha is .05, and it is a two-tail: tcritical = 2.228 Step 3: Run the test Since we have equal sample sizes (ns) for each group we can use the first (shorter) formula: 21)( 21XXsXXt= where sX 1 X 2 =s12n1+ s22n2 All the values are given above, so you just have to plug and compute. 28.382.1057.624.4646.39646.2521==+=+=XXs 84.228.333.928.3)33.1266.21( ===t Note that we could have used the longer formula here as well because it will work for equal or unequal sample sizes. Step 4: Make a decision about the Null Reject the Null since the value we computed in Step 3 is more extreme than the critical value in Step 2, we reject the idea that they are from the same population. Step 5: Write a conclusion The population of players with imagery training made a different number of baskets compared to those with no training. Lesson 15 ANOVA (analysis of variance) Outline Variability -between group variability -within group variability -total variability -F-ratio Computation -sums of squares (between/within/total) -degrees of freedom (between/within/total) -mean square (between/within) -F (ratio of between to within) Example Problem Note: The formulas detailed here vary a great deal from the text. I suggest using the notation I have outlined here since it will coincide more with what we have already done, but you might look at the text version as well. Use whatever method you find easiest to understand. Variability Please read about this topic on the web page by locating the ANOVA demonstration or you can click here: http://faculty.uncfsu.edu/dwallace/sanova.html Between group variability and within group variability are both components of the total variability in the combined distributions. What we are doing when we compute between and within variability is to partition the total variability into the between and within components. So: Between variability + within variability = total variability Hypothesis Testing Again, with ANOVA we are testing hypotheses that involve comparisons of two or more populations. The overall test, however, will indicate a difference between any of the groups. Thus, the test will not specify which two, or if some or all of the groups differ. Instead, we will conduct a separate test to determine which specific means differ. Because of this fact the research hypothesis will state simply that at least two of the means differ. The null will still state that there are no significant differences between any of the groups (insert as many mus as you have groups). H0: 1=2=3 Critical values are found using the F-table in your book. The table is discussed in the example below. Computation How do we measure variability in a distribution? That is, how do we measure how different scores are in the distribution from one another? You should know that we use variance as a measure of variability. With ANOVA or analysis of variance, we compute a ratio of variances: between to within variance. Recall that variance is the average square deviation of scores about the mean. We will compute the same value here, but as the definition suggests, it will be called the mean square for the computations. So, we are computing variance. Recall that when we compute variance we first find the sum of the square deviations, and then divide by the sample size (n -1 or degrees of freedom for a sample). s 2 =X2 X( )n2n 1 freedomofreesSquaresofSums__deg__ When we compute the Mean Square (variance) in order to form the F-ratio, we will do the exact same thing: compute the sums of squares and divide by degrees of freedom. Dont let the formulas intimidate you. Keep in mind that all we are doing is finding the variance for our between factor and dividing that by the variance for the within factor. These two variances will be computing by finding each sums of squares and dividing those sums of squares by their respective degrees of freedom. Sums of Squares We will use the same basic formula for sums of squares that we used with variance. While we will only use the between variance and within variance to compute the F-ration, we will still compute the sums of squares total (all values) for completeness. Total Sums of Squares SSTOT = XTOT2 XTOT( )2NTOTNote that it is the same formula we have been using. The subscript (tot) stands for the total. It indicates that you perform the operation for ALL values in your distribution (all subjects in all groups). Within Sums of Squares ( ) ( ) ( )+++= kkkWITHIN nXXnXXnXXSS222222212121 ... Notice that each segment is the same formula for sums of squares we used in the formula for variance and for the total sums of squares above. What is different here is that you consider each group separately. So, the first segment with the subscript 1 means you compute the sum of squares for the first group. Group two is labeled with a 2, but notice that after that we have group k instead of a number. This notation indicates that you continue to find the sums of squares as you did for the first two groups for however many groups you have in the problem. So, k could be the third group, or if you have four groups then you would do the same sums of squares computation for the third and fourth group. Between Sums of Squares ( ) ( ) ( ) ( )TOTTOTkkBETWEEN NXnXnXnXSS22222121 ... +++= We have the same k notation here. Again, you perform the same operation for each separate group in your problem. However, with this formula once we compute the value for each group we must subtract an operation at the final step. This operation is half the sums of squares we computed for the sums of squares total. Degrees of Freedom Again, we will first compute the sums of squares for each source of variance, divide the values by degrees of freedom in order to get the two mean square values we need to form the F-ratio. Degrees of freedom, however, is different for each source of variability. Total Degrees of freedom N 1 this N value is the total number of values in all groups Within Degrees of freedom KNdfwithin = K is the number of categories or groups, N is still the total N Between Degrees of freedom 1= KdfBetween We will also use degrees of freedom to locate the critical value on the F-table (see page A-29 for alpha .05 and A-30 for alpha .01). The numerator of the F-ratio is the between factor, so we will use the degrees of freedom between along the top of your table. The denominator of the F-ratio is the within subjects factor, so will use degrees of freedom within along the left margin of the table. Mean Square Now we divide each sums of squares by the respective mean square. Dont let the formulas intimidate you. All we are doing is matching up degrees of freedom with the Sums of squares to get the mean square (variance) Within Mean Square Between Mean Square BetweenBetweenBeteween dfSSMS = WithinWithinWITHIN dfSSMS = F-ratio The final step is to divide our between by within variance to see if the effect (between) is large compared to the error (within). WithinBetweenMSMSF = Example A therapist wants to examine the effectiveness of 3 therapy techniques on phobias. Subjects are randomly assigned to one of three treatment groups. Below are the rated fear of spiders after therapy. Test for a difference at = .05 Therapy A Therapy B Therapy C 5 3 1 2 3 0 5 0 1 4 2 2 2 2 1 1x = 18 2x = 10 3x = 5 21x = 74 22x =26 23x = 7 STEP 1: State the null and alternative hypotheses. H1 at least one mean differs H0: 1 = 2 = 3 STEP 2: Set up the criteria for making a decision. That is, find the critical value. You might do this step after Step 3 since that is where you compute the critical value. 1= KdfBetween = 3-1 = 2 KNdfwithin = = 15-3 = 12 Fcritical = 3.88 STEP 3: Compute the appropriate test-statistic. Although in this example I have given the summary values, for some problems you might have to compute the sum of x, and sum of squared xs yourself. ( )TOTTOTTOTTOT NXXSS22 = = ( ) 4.346.7210715108910715331072====TOTSS ( ) ( ) ( )+++= kkkWITHIN nXXnXXnXXSS222222212121 ... ( ) ( ) ( ) + + =5575102651874222WITHINSS + + =5257510026532474WITHINSS ( ) ( ) ( )5720268.6474 ++=WITHINSS 2.17262.9 =++=WITHINSS ( ) ( ) ( ) ( )TOTTOTkkBETWEEN NXnXnXnXSS22222121 ... +++= ( ) ( ) ( ) ( )153355510518 2222 ++=BETWEENSS 15108952551005324 ++=BETWEENSS 2.176.725208.64 =++=BETWEENSS Note that anytime you compute two of the Sums of Squares you can derive the third one without computation because Between + Within = Total dftot = N 1 1= KdfBetween KNdfwithin = 14115 ==totdf 213 ==Betweendf 12315 ==withindf BetweenBetweenBeteween dfSSMS = WithinWithinWITHIN dfSSMS = WithinBetweenMSMSF = 6.822.17 ==BeteweenMS 43.1122.17 ==WITHINMS 643.16.8 ==F Once we have computed all the values, very often we place them in a source table (below). Putting the values in a table like this one may make it easier to think about the statistic. Notice that once we get the Sums of Squares on the table, we will divide those values by the df in the next column. Once we get the two mean squares we divide those to get F. Source SS df MS F Between (group) 17.2 _ 2__ __8.6_ _6_ Within (error) _17.2_ _12__ __1.43_ Total _34.4_ _14__ STEP 4: Evaluate the null hypothesis (based on your answers to the above steps). Reject the null STEP 5: Based on your evaluation of the null hypothesis, what is your conclusion? There is at least one group that is different from at least one other group. Lesson 16 Post-hoc Tests Outline Tukeys HSD Post-hoc test -differences between means -studentized range statistic (q) -honestly significant difference (HSD) Example Tukey Problem Magnitude of the Effect -eta-square -omega-square Tukeys HSD Post-hoc test A post-hoc test is needed after we complete an ANOVA in order to determine which groups differ from each other. Do not conduct a post-hoc test unless you found an effect (rejected the null) in the ANOVA problem. If you fail to reject the null, then there are no differences to find. For the Tukeys post-hoc test we will first find the differences between the means of all of our groups. We will compare this difference score to a critical value to see if the difference is significant. The critical value in this case is the HSD (honestly significant difference) and it must be computed. It is the point when a mean difference becomes honestly significantly different. nMSqHSD Within= Note that q is a table value, and n is the number of values we are dealing with in each group (not total n). The Mean Square value is from the ANOVA you already computed. To find q or the studentized range statistic, refer to your table on page A-32 of your text. On the table k or the number of groups is found along the top, and degrees of freedom within is down the side. Cross index the row and column to find the value you need to put in the formula above. Example This example is a continuation of the ANOVA problem we did in the last lesson. Here I show the groups, but have computed the average or mean of each group. Therapy A Therapy B Therapy C 5 3 1 2 3 0 5 0 1 4 2 2 2 2 1 =1X 3.6 =2X 0 =3X 1 The first step is to compute all possible differences between means: 6.306.321 == XX 6.216.331 == XX 11032 == XX We will only be concerned with the absolute difference, so, you can ignore any negative signs. Next we compute HSD. nMSqHSD Within= = =543.177.399.1)53(.77.3286.77.3 == Now we will compare the difference scores we computed with the HSD value. If the difference is larger than the HSD, then we say the difference is significant. Groups 1 and 2 differ Groups 1 and 3 differ Groups 2 and 3 do not differ Magnitude of the Effect It is a common misconception that the size of the F-ratio you compute directly indicates how strongly the relationship is between the independent and dependent variable. However, a separate computation is needed to get a true idea of the strength of the relationship. Our test indicates only that there is difference in the treatments, and it is not a measure of a relationship between variables. Magnitude of the effect is the amount of variability that our independent variable can account for in the dependent variable. The easiest measure of the effect is eta-square (2). It measures the proportion of our between factor variability to the total variability. If we know what proportion of variability of the total is due to our treatment, we will have a rough idea of how strong the relationship might be. totalBetweenSSSS=2 Notice it is just a ratio of treatment effect variability to total variability For our example: 50.4.342.172 == 50% of the variability in rated fear of spiders is due to the type of therapy they received. One problem with eta-square is that it is a biased estimate and tends to overestimate the effect. A more accurate measure of the effect is omega-square (2). ( )WithintotalWithinBetweenMSSSMSkSS+= 12 ( ) 40.83.3534.1483.3586.22.1783.3543.1)2(2.1743.14.3443.1132.172 ====+= Lesson 17 Pearsons Correlation Coefficient Outline Measures of Relationships Pearsons Correlation Coefficient (r) -types of data -scatter plots -measure of direction -measure of strength Computation -covariation of X and Y -unique variation in X and Y -measuring variability Example Problem -steps in hypothesis testing -r2 Note that some of the formulas I use differ from your text. Both sets of formulas are in the homework packet, and you should use the formulas you feel most comfortable using. Measures of Relationships Up to this point in the course our statistical tests have focused on demonstrating differences in effects of a dependent variable by an independent variable. In this way, we could infer that by changing the independent variable we could have a direct affect on the independent variable. With the statistics we have learned we can make statements about causality. Pearsons Correlation Coefficient (r) Types of data For the rest of the course we will be focused on demonstrating relationships between variables. Although we will know if there is a relationship between variables when we compute a correlation, we will not be able to say that one variable actually causes changes in another variable. The statistics that reveal relationships between variables are more versatile, but not as definitive as those we have already learned. Although correlation will only reveal a relationship, and not causality, we will still be using measurement data. Recall that measurement data comes from a measurement we make on some scale. The type of data the statistic uses is one way we will distinguish these types of measures, so keep it in mind for the next statistic we learn (chi-square). One feature about the data that does differ from prior statistics is that we will have two values from each subject in our sample. So, we will need both an X distribution and Y distribution to express two values we measure from the same unit in the population. For example, if I want to examine the relationship between amount of time spent studying for an exam (X) in hours and the score that person makes on an exam (Y) we might have: X Y 2 65 3 70 3 75 4 70 5 85 6 85 7 90 Scatter plots An easy way to get an idea about the relationship between two variables is to create a scatter plot of the relationship. With a scatter plot we will graph our values on an X, Y coordinate plane. For example, say we measure the number of hours a person studies (X) and plot that with their resulting correct answers on a trivia test. (Y). X Y 0 0 1 1 1 2 2 3 3 5 4 5 5 6 Plot each X and Y point by drawing and X,Y axis and placing the x-variable on the x-axis, and the y-variable on the y-axis. So, when we are at 0 on the X-axis for the first person, we are at 0 on the y-axis. The next person is at 1 on the X-axis and 1 on the Y-axis. Plot each point this way to form a scatter plot. 012345670 2 4 6Number of Hours StudyingNumber of Correc Answers In the resulting graph you can see that as we increase values on the x-axis, it corresponds to an increase in the y-axis. For a scatter plot like this one we say that the relationship or correlation is positive. For positive correlations, as values on the x-axis increase, values on y-increase also. So, as the number of hours of study increases, the number of correct answers on the exam increases. The opposite is true as well. If one variable goes down the other goes down as well. Both variables move in the same direction. Lets look at the opposite type of effect. In this example the X-variable is number of alcoholic drinks consumed, and the Y-variable is number of correct answers on a simple math test. 0246810120 2 4 6 8Number of DrinksNumber of Correct Answers This scatter plot represents a negative correlation. As the values on X increase, the values on Y decrease. So, as number of drinks consumed increases, number of correct answers decreases. The variables are moving in opposite directions. Measures of Strength Scatter plots gave us a good idea about the measure of the direction of the relationship between two variables. They also give a good idea of how strongly related two variables are to one another. Notice in the above graphs that you could draw a straight line to represent the direction the plotted points move. 0246810120 2 4 6 8Number of DrinksNumber of Correct Answers The closer the points come to a straight line, the stronger the relationship. We will express the strength of the relationship with a number between 0 and 1. A zero indicates no relationship, and a one indicates a perfect relationship. Most values will be a decimal value in between the two numbers. Note that the number is independent of the direction of the effect. So, we may express a -1 value indicated a strong correlation because of the number and a negative relationship because of the sign. A value of +.03 would be a weak correlation because the number is small, and it would be a positive relationship because the sign is positive. Here are some more examples of scatter plots with estimated correlation (r) values. A B C Graph A represents a strong positive correlation because the plots are very close together (perhaps r = + .85). Graph B represents a weaker positive correlation (r = +.30). Graph C represents a strong negative correlation (r = -.90). Computation When we compute the correlation it will be the ratio of covariation in the X and Y variable, to the individual variability in X and the individual variability in Y. By covariation we mean the amount that X and Y vary together. So, the correlation looks at the how much the two variables vary together relative to the amount they vary individually. If the covariation is large relative to the individual variability of each variabile, then the relationship and the value of r is strong. A simple example might be helpful to understand the concept. For this example, X is population density and Y is number babies born. Individual variability in X You can think of a lot of different reasons why population density might vary by itself. People live in more densely populated areas for many reason including job opportunities, family reasons, or climate. Individual variability in Y You can also think of a lot of reasons why birth rate may vary by itself. People may be influenced to have children because of personal reasons, war, or economic reasons. Covariation of X and Y For this example it is easy to see why we would expect X and Y to vary together as well. No matter what the birth rate might happen to be, we would expect that more people would yield more babies being born. When we compute the correlation coefficient we dont have to think of all the reasons for variables to vary or covary, but simply to measure the variability. How do we measure variability in a distribution? I hope you know the answer to that question by now. We measure variability with sums of squares (often expressed as variance). So, when we compute the correlation we will insert the sums of squares for X and Y in the denominator. The numerator is the covariation of X and Y. For this value we could multiply the variability in the X-variable times the variability in the Y-variable, but see the formula below for an easier computation. ( ) ( )=2222nYYnXXnYXXYr The only new component here is the sum of the products of X and Y. Since each unit in our sample has both and X and a Y value, you will multiply these two numbers together for each unit in your sample. Then add the values you multiplied together. See the example below as well. Example Problem The following example includes the changes we will need to make for hypothesis testing with the correlation coefficient, as well as an example of how to do the computations. Below are the data for six participants giving their number of years in college (X) and their subsequent yearly income (Y). Income here is in thousands of dollars, but this fact does not require any changes in our computations. Test whether there is a relationship with Alpha = .05. # of Years of College Income X Y X2 Y2 XY 0 15 0 225 0 1 15 1 225 15 3 20 9 400 60 4 25 16 625 100 4 30 16 900 120 6 35 36 1225 210 =X 18 =Y 140 = 2X 78 = 2Y 3600 =XY 505 Notice that I have included the computation for obtaining the summary values for you for completeness. Be sure you know how to obtain all the summed values, as they will not always be given on the exam. Step 1: State the Hypotheses in Words and Symbols H1 The correlation between years of education and income is equal to zero in the population. H0: The correlation between years of education and income not equal to zero in the population. As usual the null states that there is no effect or no relationship, and the research hypothesis states that there is an effect. When we write them in symbols we will use the Greek letter rho () to indicate the correlation in the population. Thus: H1 0 H0: = 0 Step 2: Find the Critical Value Again, we will use a table to find the critical value in Appendix A of your book. Locate the table, and find the degrees of freedom for the appropriate test to find the critical value. For this test df = n 2, where n is the number of pairs of scores we have. Df = 6 2 = 4 rcritical = + 0.811 Step 3: Run the Statistical Test ( ) ( )=2222nYYnXXnYXXYr =61403600618786)140)(18(50522r =619600360063247862520505r [ ][ ]67.326636005478420505=r 95.44.898592.799985)33.333)(24(85 ====r Step 4: Make a Decision about the Null Reject the null Since the value we computed in Step 3 is larger than the critical value in Step 2, we reject the null. Step 5: Write a Conclusion There is a relationship between years spent in college and income. The more years of school, the more the subsequent income. r2 Often times we will square the r-value we compute in order to get a measure of the size of the effect. Just like with eta-square in ANOVA, we will compute the percentage of variability in Y, that is accounted for by X. For the current example r2 = .90, so 90% of the variability in income is accounted for by education. Lesson 18 Regression Outline Equation of a Line -Slope -Y-Intercept -regression line Making Predictions Error in Prediction -residual variation -standard error of the estimate Regression is very similar to correlation, but instead of measuring the relationship we will make predictions based on the relationship. Even though we are not inferring a causal relationship, we can nevertheless predict one variable if we have information about the other. We will refer to X as the predictor variable, and Y as the criterion variable. Thus, we will attempt to use X in order to predict what Y should be. Equation of a Line Recall that when we were first looking at scatter plots and we drew a line through the dots in order to indicate the direction of the effect. With regression, that is exactly what we want to do. We will compute the best fitting line for the data that we have. In a scatter plot a single line will not hit every data point, but we will construct a line that simultaneously comes as close to each data point as possible. As with similar topics in geometry, we will express the line we come up with in an equation. You may recognize Y = mx + b. We will use a similar equation to express a straight line, though our Y-value will not be the same value we have in our data, but instead be the value we would predict for Y. Since our data are scattered about, it is unlikely that the value our line would predict for Y is actually a point in our data set. Thus, our equation for the line will be slightly different: abXY += Even though we use b here, it is still the slope of the line, and a is the y-intercept. We also use Y instead of just Y to indicate that this is a predicted value for way based on the regression line rather then an actual Y-value. In order to make predictions, we will compute these regression coefficients (b and a), and plug them into our equation. We can then plug in an X-value and get out a predicted Y value ( Y ). Slope Slope is the unit change in Y for each single unit change in X. That is, since we will multiply the slope with the X-value we want to make predictions about, the predicted Y will change by the amount of the slope for each single unit of X. Slope computation is very similar to correlation. In fact, if you have already computed the correlation you can use the same values to compute slope. ( ) =nXXnYXXY22b Notice that we have the covariation of X and Y in the numerator, so the entire numerator will be the same as with r. The denominator is the sums of squares for X. This was a portion of the denominator for r as well. Y-Intercept The y-intercept is the point at which the regression line crosses the Y-axis. It is also the value we predict for Y when X = 0. Thats because we are at the Y-axis when X=0. nXbYa = Notice that we must compute the slope b before we can compute the y-intercept. Making Predictions Once we have computed the regression coefficients, we can put them into our equation for a line and start making predictions. Lets look at an example problem. For this problem we will continue with the example we used with correlation. # of Years of College Income X Y X2 Y2 XY 0 15 0 225 0 1 15 1 225 15 3 20 9 400 60 4 25 16 625 100 4 30 16 900 120 6 35 36 1225 210 =X 18 =Y 140 = 2X 78 = 2Y 3600 =XY 505 Now, however, we will compute the equation of the line that predicts income from number of years in college. First compute slope and y-intercept. 54.32485547842050563247862520505618786)140)(18(5052 ====b 7.12628.76672.631406)18(54.3140 ====a Once these values are computed, we can write the equation of the line. Y =3.54X + 12.7 Now we simply plug in an x-value that would like to make a prediction about. Note again that this equation will not yield an actual y-value, but a prediction on the line that best describes the relationship between X and Y. For example, what income would we predict for someone with five years of education? Y =3.54(5) + 12.7 Y =17.7 + 12.7 Y =30.4 Here we just insert the X-value, and compute Y. So, we expect someone with five years of education to make about 30 thousand dollars a year. Error in Predictions The amount our predicted Y-value differs from the actual Y-value in our data is error or residual variation. If we average this residual variation for all of our scores, we can get a measure of the error our equation yields. The standard error ( yyS ) is the average deviation between actual Y and predicted Y. If we first plug in each X-value from our data into the regression line to get out a predicted Y-value, then we can see how different the Y and predicted Y-values differ. # of Years of College Income X Y 0 15 12.7 1 15 16.24 3 20 23.32 Y =3.54X + 12.7 4 25 26.86 4 30 26.86 6 35 33.94 Remember the residual variation is the difference between Y and . Once we find this difference we want to add the differences to get an average. # of Years of College Income X Y Y- 0 15 12.7 2.3 1 15 16.24 -1.24 3 20 23.32 -3.32 4 25 26.86 -1.86 4 30 26.86 3.14 6 35 33.94 1.06 = )( YY 0 Notice that we have the same problem here as with standard deviation. The sum of the deviations about the mean must always equal zero. In fact, the standard error of the estimate we are calculating is the standard deviation of the regression line we computed. So, we will square the difference score we computed in the last step so that we can add the squared differences. The formula we are working toward looks like this: ( )2 2 = nYYS yy So, all we have left to do is square our residual variations, sum them, divide by df, and take the square root of the whole thing. # of Years of College Income X Y Y- (Y- )2 0 15 12.7 2.3 5.29 1 15 16.24 -1.24 1.54 3 20 23.32 -3.32 11.02 4 25 26.86 -1.86 3.46 4 30 26.86 3.14 9.86 6 35 33.94 1.06 1.12 = )( YY 0 = 2)( YY 32.47 Plugging into the formula we have: 85.212.8447.322647.32 ==== yyS Our interpretation is that on average actual Y and predicted Y differ by 2.85 units. So, any prediction we make about income using the equation will be off by about $2,850. Lesson 19 Chi-Square Outline Categorical Data Goodness of Fit Test -observed frequency -expected frequency -X2 statistic Example -hypothesis testing Categorical Data As mentioned at the start of the lesson with correlation, all of the data we have been working with so far involve measurement data. We actually took measurements from units in our sample to create our distribution. Often times, however, we will want to analyze categorical or qualitative data as well. For categorical data we will not have a measure of individual units in the sample. Instead, we will analyze frequencies or counts of people falling into different categories or groups. When analyzing categorical data we say the test is non-parametric. Thus, all the tests we have learned before this point were parametric tests. Chi-Square Goodness-of-Fit Test We will learn two different Chi-square tests. The first of these is the goodness-of-fit test. With the goodness-of-fit test we will test whether the data fit good with what we would expect if only chance factors were operating. For example, if I measured the number of insurance claims for different car types, I might have the following data: High Performance Compact Mid Size Full Size 20 14 7 9 Notice that our data is now frequency values or how many values in our sample fit into different categories. The test will tell us whether there is a difference in how many values fall at different levels of the single variable (car type). Is there a difference in number of claims for different car types? The values we observe in our sample are the observed frequencies ( 0f ). What we want to know is if they differ from the frequencies we would observe by chance. The values we would expect if there really was no difference in the number of claims made for different car types are what we call the expected frequencies ( ef ). If there really was no difference in the frequencies for each level of the variable, then we would expect equal numbers of claims for each car type. Since there a total of 50 claims in our sample, and there are 4 different levels of the variable, then we would expect 12.5 claims for each car type. Thus: High Performance Compact Mid Size Full Size 20 14 7 9 Observed 12.5 12.5 12.5 12.5 Expected What the Chi-square statistic does is to compare the values we observe to those we would expect if there was no difference. If what we observe varies a good bit from the values we would expect if there was not difference, then there must be a difference. If there really was no difference in the number of insurance claims, for this example, then we would expect the number of claims to be close to the expected frequencies. ( ) =eeofff 22 Notice that we subtract each expected value from each observed value, square the difference, and divide by the expected frequency. We then sum up all of the values we computed. Lets take a look at the example we have been working on within the context of hypothesis testing. We will continue the problem with Alpha set to .05. Step 1: Write the Hypotheses for the test. H1 of ef H0: of = ef Here we are stating that the observed frequencies are the same as the expected for the null. Step 2: Find the Critical Value Again we will use Appendix A to find the critical value, see page A-34. For our test degrees of freedom is equal to C 1, where C is the number of categories Df = 4 1 = 3 2critical = 7.81 Step 3: Run the Statistical Test We have already computed the expected values, so we just need to plug the numbers into the formula. High Performance Compact Mid Size Full Size 20 14 7 9 Observed 12.5 12.5 12.5 12.5 Expected ( ) =eeofff 22 5.12)5.129(5.12)5.127(5.12)5.1214(5.12)5.1220( 22222 +++= 5.12)5.3(5.12)5.5(5.12)5.1(5.12)5.7( 22222 +++= 5.1225.125.1225.305.1225.25.1225.562 +++= 08.898.042.218.05.42 =+++= Step 4: Make a Decision About the Null Reject the null. The value we computed in Step 3 is larger than Step 2, so we reject the null. Step 5: Conclusion Since we rejected the null we say there is a difference in the number of claims made for different car types. Lesson 20 Chi-Square (Test of Independence) Outline Measuring Independence -observed frequencies -expected frequencies -chi-square Measuring Independence The Chi-square test of independence is similar to the test we just learned in the last lesson. However, instead of measuring frequencies along only one dimension, we will measure frequencies for two variables at the same time. Our test is designed to test whether or not these two variables are independent (not related). If we reject the null and say the variables are not independent of one another, then we have established that the two variables are related. The test of independence starts with frequencies or counts we observe in our sample, or the observed frequencies. For this example, a sample of 50 people is used to record personality and color preference is measured: Blue Red Yellow Extroverted 5 20 5 Introverted 10 5 5 __ Observed Frequencies Is there a relationship between personality type and color preference? Our hypotheses will state exactly that, with the null as usual stating that there is no effect or no relationship. H1: There is a relationship between color preference and personality type (variables are not independent). H0: There is no relationship between color preference and personality type (variables are independent). Since we have two variables, our degrees of freedom will change. df = (R - 1) (C- 1) where R is the number of rows and C is the number of columns in our table. There are two rows going across and three rows going down. So, degrees of freedom for this example are: df = (2 - 1) (3- 1) = (1)(2) = 2 We will find the critical value using the same table we used in the goodness-of-fit test. In our example: 2critical = 5.99 When we start to compute the statistic, it will be similar to the goodness-of-fit test as well. However, we will need a formula to compute the expected frequencies instead of just dividing our sample size equally between groups. We will need to compute the expected value separately for each observed value in our sample. So, in our example we must compute six different expected frequencies. Note that our expected frequencies with this test are the values we expect if the null is true. Thus, the expected frequencies are the values we expect if there is no relationship or the variables are independent. nfff rce = The value n is the total or 50 here. In the numerator we have the frequencies for the c column and r row. You multiply these values together. They are the total frequency for the row and column of the individual expected value we are looking for with the computation. So, you must first add up the frequencies for each row and column: Blue Red Yellow Extroverted 5 20 5 30 Introverted 10 5 5 __ 20 15 25 10 Each value of our six observed values will have separate column frequency ( cf ) and row frequency ( rf ). For example the observed value in the first row and column (5) has a row frequency of 30 and a column frequency of 15. Blue Red Yellow Extroverted 5 20 5 30 Introverted 10 5 5 __ 20 15 25 10 So, the expected frequency is 9504505015*30 ===ef Continue finding each expected frequency in this same way for each of the observed values. Notice that when we compute a different expected frequency our row and column frequencies will change. So, for the next value (extroverted and red): Blue Red Yellow Extroverted 5 20 5 30 Introverted 10 5 5 __ 20 15 25 10 15507505025*30 ===ef Again, you continue the process until you have found all the expected frequencies. There is a shortcut for finding the rest of these values if you feel comfortable with the statistic. Once we compute these two expected frequencies all the others are determined (hence the two degrees of freedom). All of our expected frequencies must have the same row and column sums as our observed frequencies. So, once we have computed these two expected frequencies all of the other values can be found by subtracting out the row or column total. The remaining unknown expected value in the first column, then, must be the number that will make that first column add up to 15. Since the first expected value is 9, then the remaining number must be 15 9 = 6. Whether you compute each individual expected frequency or use the short cut, you will get a complete table of expected frequencies. Blue Red Yellow Extroverted 9 15 6 30 Introverted 6 10 4 __ 20 15 25 10 Expected Frequencies The process for finding the final Chi-square value is the same as before. We will find the difference between our expected and observed frequencies, square the difference, and then divide by the expected frequency for each value in our table. ( ) =eeofff 22 4)45(10)105(6)610(6)65(15)1520(9)95( 2222222 +++++= 4)1(10)5(6)4(6)1(15)5(9)4( 2222222 +++++= 4110256166115259162 +++++= 2 = 1.78+1.67+0.167+2.67+2.5+.25 2 = 9.04 Since this value is greater than our critical value, we will reject the null here and say there is a relationship. Thus, knowing someones personality type gives you information about their likely color preference. PSY 233 Homework Packet Spring 2010 2PSY 233 Formulas sample mean x = Xn population mean = XNsums of squares ( ) =22nXXSS sample variance population variance ( )221= nXXs ( )22N = X -OR- ( )1222= nnXXs ( )NNXX =222 sample standard deviation s = s2 population standard deviation = 2 z-score formula z = X Z = X + z-test formula z = X x , where x =nsingle sample t-test formula t = X sx , where sx = sn df = n 1 )( Xcrit StXCI = independent measures t-test formulas (equal sample sizes only) 21)( 21XXsXXt= where sX 1X 2= s12n1+ s22n2 df = n1 + n2 2 -for pooled variances (equal or unequal sample sizes or ns) 21)( 21XXsXXt= where sX 1X 2= sp2n1+ sp2n2, where ( ) ( )211212222112++=nnsnsns p 3 +++=212122212121112)1()1(NNNNNsNsXXt ANOVA (analysis of variance) formulas SSTOT = XTOT2 XTOT( )2NTOT ( ) ( ) ( ) ( )TOTTOTkkBETWEEN NXnXnXnXSS22222121 ... +++= ( ) ( ) ( )+++= kkkWITHIN nXXnXXnXXSS222222212121 ... dftot = N 1 1= Kdf Between KNdf within = BetweenBetweenBeteween dfSSMS = WithinWithinWITHIN dfSSMS = WithinBetweenMSMSF = totalBetweenSSSS=2 ( )WithintotalWithinBetweenMSSSMSkSS+= 12 nMSqHSD Within= 4 correlation formulas df = n 2 ( ) ( )=2222nYYnXXnYXXYr -OR- r =YXXYSSSSSP where nYXXYSPxy= regression formulas Y = bX + a ( ) =nXXnYXXY22b -OR- XXYSSSPb = nXbYa = -OR- XbYa = ( )211 2 = nnrss YYY ( )2 2= nYY goodness of fit chi-square formulas ( ) =eeofff 22 df = C - 1 Test of independence chi-square formulas nfff rce = ( ) =eeofff 22 df = (R - 1) (C- 1) 5Exam 1 Exam 1 will cover chapters 1-3 in the text, and Lesson 1-4 online. In Chapter 2 we will not be covering frequency distribution polygons on pages 38-39 of your text. 6Worksheet Chapters 1 and 2 1. The relation between a sample and a statistic is the same as the relation between a. a population and a parameter b. a dependent variable and an independent variable c. descriptive statistics and inferential statistics d. measurement data and categorical data 2. Which scale of measurement are the following examples (nominal, ordinal, interval, or ratio)? Select the best answer. 2A. numbers used to identify political affiliation: republican, democrat, independent 2B. freshman, sophomore, junior, senior, graduate, faculty member 2C. social security number (hint: the number is just a label). 2D. amount of time it takes a pain reliever to work 2E. length or width of a room 3. Are the following examples discrete or continuous variables? Amount of verbal material learned in 30 minutes Number of children in a family 4. A recent report concludes that participants on an exercise regimen of running two miles each day had a lower percentage of body fat than participants on no exercise program. 4A. What is the independent variable? 4B. What is the dependent variable? 75. A study is conducted to determine whether listening to different types of music impairs memory. Participants are given 10 minutes to memorize as many words as they can. During this 10 minute period, one group listens to hard rock, a second group listens to classical music, and a third group listens to no music at all. Each group is then given a list of 50 words to memorize. They are then given a blank piece of paper and told to write down as many words as they can remember. 5A. What is the independent variable? 5B. What is the dependent variable? 5C. Is the independent variable discrete or continuous? 6. A study was conducted to determine whether physically fit persons sleep more hours than those who are not physically fit. Two groups of people were selected. One group consisted of people who work out at least 3 times a week. The other group consisted of people that do not work out at all. For one week, subjects slept in a sleep lab and an experimenter recorded the number of hours each person slept. 6A. What is the independent variable? 6B. What is the dependent variable? 6C. Is the dependent variable discrete or continuous? 6D. Is the data collected measurement data or categorical data? 6E. What scale of measurement is the data (nominal, ordinal, interval, or ratio)? 7. Use the following data set for 7A through 7G: X Y 3 -2 4 6 5 7 2 1 7A. X 7B. Y + 2 7C. XY 7D. X2 7E. (Y)2 8 7F. (X)( Y) 7G. (X-Y) 8. Draw a positively skewed distribution. 9. Twenty FSU students were asked, "How many phone calls did you receive last night?" The numbers below are their answers. 10 7 4 6 5 2 3 0 1 11 7 6 4 4 5 3 2 2 0 3 Complete the grouped frequency distribution. Real Mid- Fre- Cumulative Relative Cumulative Interval Limits point quency Frequency Percentage Percentage 0-1 ______________ 2-3 ________ 4-5 ________ 6-7 ________ 8-9 ________ 10-11 _________ __ 10. What percentage of FSU students received between 2 and 3 phone calls? 11. How many people received less than 9 phone calls? 12. What score falls at the 70th percentile? Interpret. 13. What percentile is associated with a score of 3.5? Interpret. 9 Worksheet Chapters 2 and 3 1. A sample of 44 drivers in South Carolina reported the number of trips they took outside the county of where they lived. The data is reproduced below. 0123456789Frequency 1A. Compute the mean of the distribution. 1B. Compute the median of the distribution 1C. Compute the mode of the distribution 2. A retailer created a grouped frequency distribution for the number of weeks individuals spent paying for lay-away items. The data are reproduced below: Real Limits f 2.5-6.5 5 6.5-10.5 1 10.5-14.5 5 14.5-18.5 5 18.5-22.5 0 22.5-26.5 3 26.5-30.5 5 30.5-34.5 1 34.5-38.5 5 Create a histogram for the above data 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Number of Trips 103. A distribution of scores has a mean = 30, Median = 20, and a Mode = 10. The distribution: a. has a positive skew b. has a negative skew c. is normal d. is bimodal 4. Use the following distribution to answer the next three questions Score f 5 1 6 0 7 0 8 0 9 4 10 6 11 7 12 7 4A. The above distribution: a. has a positive skew b. has a negative skew c. is normal d. is bimodal 4B. The mode for the above distribution is: a. 7 b. 0 and 7 c. 11 and 12 d. 6, 7, and 8 4C. Which of the following numbers would be considered an outlier in the above distribution? a. 0 b. 1 c. 5 d. 7 5. The median is equivalent to: a. the 25th percentile b. the 50th percentile c. the 75th percentile d. none of the above 6. The only measure of central tendency we are certain to actually observe as a value in our data set is: a. the mean b. the median c. the mode d. all measure of central tendency must be actual values in the distribution 11Worksheet: Chapters 3 1. A sample of twenty FSU students were randomly selected and asked, "How many phone calls did you receive last night?" The numbers below are their responses. 0 1 2 0 2 4 2 3 4 5 3 4 5 6 3 10 7 6 7 11 1A. What is the mode? 1B. What is the median? 1C. What is the mean? 2. A survey asks whether participants think O. J. Simpson is innocent or guilty. Which would be the best measure of central tendency to describe this data set? a. the mean b. the median c. the mode 3. Which is the most commonly used measure of central tendency? a. the mean b. the median c. the mode 4. A survey asked Ohio University students which pizza place they preferred. The results are as follows. Pizza Place Frequency Late Night Pizza 5 Papa Johns 6 Pizza Hut 3 Little Caesers 5 4A. What is the best measure of central tendency for this data? a. the mean b. the median c. the mode 4B. Find the mode of this distribution. 4C. Which pizza place is most popular among the students surveyed? 5. How does it affect the mean when you add a constant to every score? That is, if an instructor adds 5 points to everyone's test score, how will the mean change? a. the new mean and the old mean will be the same b. the new mean will be 5 points higher than the old mean c. the new mean will be 5 points less than the old mean d. not enough information to answer this question 12 6. There are five brothers. Their mean income is $200 per week, and their median income is $170 per week. Bruce, the lowest paid, gets fired from his $100 a week job and now has an income of $0 per week. What is the median weekly income of the five brothers after Bruce lost his job? Two samples are as follows Sample A: 7 9 10 8 9 12 Sample B: 13 5 9 1 17 9 7. What is the mode for sample A? 8. What is the mean for sample B? 9. What is the mean for sample A? 10.What is the median for sample B? 11 If a distribution has a positive skew, which of the following is true... (circle one) a. the median, the mean, and the mode will all be the same b. the median will greater than the mode c. the median will be less than the mode 12 A geography exam was given to samples of high school seniors and college students. The lowest possible score on the exam is 0 and the highest possible score is 75. The data showing the test scores is below: high school seniors Sample A: 28 30 33 35 40 40 45 50 50 55 college students Sample B: 35 38 40 40 40 40 40 41 42 45 12A. What is the mean for Sample A? 12B. What is the mean for Sample B? 12C. Based on the two means, does it appear that one group is more accurate than the other? 13Exam 1: Sample Test Multiple Choice (2 points each) A researcher wants to measure the number of pounds of tin the population recycles on average every year. He randomly samples data from 100 recycling plants around the country. Since the researcher knows 70% of the recycling plants are in urban areas, 70% of the sample was specifically taken from urban areas. 1. What type of scale would be used to measure the tin? a. nominal b. ordinal c. interval d. ratio 2. The scale used to measure the tin is: a. continuous b. discreet c. qualitative d. parabolic In order to determine whether a new gene therapy will benefit colon cancer patients, a random sample of patients is given either the new gene therapy, conventional therapy, or a placebo. The number of months of survival was measured to determine therapy success. 3. The independent variable was: a. the type of therapy b. the number of months survival c. gene therapy d. colon cancer 4. The dependent variable was: a. the type of therapy b. the number of months survival c. gene therapy d. colon cancer 5. When constructing histograms from a grouped frequency distribution, what should be used to denote the points on the scale of measure? a. apparent limits b. real limits c. upper real limits d. mid-point 146. Not everything naturally follows a normal distribution, such as salaries in the U.S. The distribution of salaries in the U.S. is: a. negatively skewed because poor people represent outliers who earn significantly less than everyone else b. positively skewed because poor people represent outliers who earn significantly less than everyone else c. negatively skewed because rich people represent outliers who earn significantly more than everyone else d. positively skewed because rich people represent outliers who earn significantly more than everyone else 7. Measuring the number of times an individual eats during the day is an example of a __________________ variable. a. nominal b. qualitative c. continuous d. discreet 8. Which of the following is not a discrete variable? a. number of bars a shuffle group visited b. number of tables available c. amount of time they stayed in a bar d. number of people who passed out 9. In a positively skewed distribution, Alice scored the mean, Betty scored the median, and Claire scored the mode. Who had the highest score? a. Alice b. Betty c. Claire d. They all scored approximately the same 10. What scale of measurement is used if you know that one variable is larger than another, but you do not know how much larger? a. nominal b. ordinal c. interval d. ratio 11. If the 40th percentile on an examination is 75.5, then a. 40% of the people got a score of 75.5 b. less than 40% of the people got a score higher than 75.5 c. 40% of the people got a score of 75.5 or less d. 60% of the people got a score lower than 75.5 1512. The value of one score in a distribution is changed form X = 20 to X = 30. Which measure(s) of central tendency is/are certain to be changed? a. the mean b. the median c. the mean and the median d. the mode 13. The concept of generalizing from a few observations to an entire group is central to the area of: a. descriptive statistics b. nominal scaling c. ratio scaling d. inferential statistics 14. When a distribution has two separate and distinct medians, then a. it is positively skewed b. it is negatively skewed c. it is bimodal d. a distribution can never have more than one median 15. An example of a quantitative variable is: a. religious affiliation b. number of children in a family c. being a registered voter d. college major 16. Students voted for their preferred professors by ranking them. This is an example of measurement on a scale. a. nominal b. ordinal c. interval d. ratio Use the following data set for the next three problems: (Show your work!) (1 point each) X Y C=3 -3 -2 0 1 -4 1 1 0 17. Compute (XY) __________ 18. Compute CX __________ 19. Compute (X-Y)__________ 16Use the following population data set for the next few problems 5 10 10 12 15 15 18 18 18 20 20 25 20. Compute the mean ___________ Show Work! (2 points) 21. Compute the median ___________ (1 point) 22. Compute the mode ___________ (1 point) 23. A sample of construction workers was asked to report the number of times they experienced back pain on the job in the past month. Twenty workers reported their incidents of back pain every day for a month. The data from these 20 workers are found below: 14 16 6 23 27 4 8 15 17 29 15 22 12 19 3 2 16 0 14 5 With the data above, complete the grouped frequency distribution. (6 points) Class Intervals Apparent Limits Real Limits Midpoint Frequency Cum f Relative Percent Cum Relative Percent 0-4 5-9 10-14 15-19 20-24 25-29 23A. What percentage of workers experienced back pain 14.5 or fewer times? (1 point) 17 23B. How many times did the 80th percentile experience back pain? (1 point) 23C. How many workers experienced back pain between 5 and 9 times? (1 point) 23D. What percentile is associated with 14.5 incidents of back pain? (1 point) 23E. Create a frequency histogram of the above data (from the grouped frequency distribution). (2 points) 18Exam 2 Exam 2 will cover Chapters 4-6 in the text, and Lesson 5-8. In chapter 4 we will compute the interquartile range differently than your text pp 79-80. Also, the online lessons and homework packet contain information about conditional probabilities not covered by your text. 19Worksheet Chapter 4 A sample of twenty FSU students were randomly selected and asked, "How many phone calls did you receive last night?" The numbers below are their responses. 0 1 2 0 2 4 2 3 4 5 3 4 5 6 3 10 7 6 7 11 1A. What is the variance? 1B. What is the standard deviation? 1C. What is the interquartile range? _______ 2. How does it affect the standard deviation when you divide a constant into every score? That is, if an instructor divides everyone's score by two, how will the standard deviation change? a. the new standard deviation and the old standard deviation will be the same b. the new standard deviation will be twice as large as the old standard deviation c. the new standard deviation will half the size (twice as small) as the old standard deviation d. not enough information to answer this question . Two samples are as follows Sample A: 7 9 10 8 9 12 Sample B: 13 5 9 1 17 9 3A. Just by looking at these data, which sample has more variability? 3B. What is the standard deviation for sample A? 3C. What is the variance for sample B? 4. A geography exam was given to samples of high school seniors and college students. The lowest possible score on the exam is 0 and the highest possible score is 75. The data showing the test scores is below: high school seniors Sample A: 28 30 33 35 40 40 45 50 50 55 college students Sample B: 35 38 40 40 40 40 40 41 42 45 20 4A. Based on the two means, does it appear that one group is more accurate than the other? 4B. What is the standard deviation for Sample A? 4C. What is the standard deviation for Sample B? 4D. Which group is more consistent (i.e., has less variability)? 5. An instructor gives his class a 10-point quiz. The next day he tells his students that the average score on the quiz was X = 7.5 with a standard deviation of s = 13.5. It should be obvious that the instructor made a mistake in his calculations. Explain why. 21Worksheet Chapter 5 and 6 1. What is the percentage area between a z-score of .43 and a z-score of 1.33? 2. What is the percentage area between a z-score of -1.25 and a z-score of .36? 3. In a normal distribution of test scores with a mean equal to 57 and a standard deviation equal to 6.5, what is the percentile rank is associated with a score of 65? 4. The scores on a personality test are normally distributed with = 250 and = 30. What percentage of people taking the test can be expected to score between 229 and 325? The average man in an industrialized country lives = 70 and = 6.3. Use this information to answer problems 5-8. 5. What percentage of men live 75 years or longer? __________ 6. What percentage of men live between 65 and 75 years? __________ 7. What percentage of men live 65 years or less? _________ 8. What percentage of men live between 55 and 60 years?__________ 229. 95% of the men will live between the ages of ______ and _________ years (i.e. find the raw values that mark the middle 95% of the distribution of ages) 10. In a distribution of scores with a mean of 1500 and a standard deviation of 250, what raw score corresponds with the 67th percentile? Questions 11 - 13 refer to a distribution with = 60 and = 4.3 11. The raw score corresponding to a z-score of 0.00 is . 12. The raw score corresponding to a z-score of -1.51 is . 13. The z-score corresponding to a raw score of 68.7 is . 14. Men in third-world countries have a life expectancy of = 60 and = 4.3. Men in industrialized countries have a life expectancy of = 70 and = 6.3. If a man in a third-world country lives to be 65 and a man in an industrialized country lives 72, who lived longer relative to their age distribution? In a distribution with a mean of 50 and a standard deviation of 5: 15. What raw score corresponds with the 14th percentile? 16. What z-score cuts off the top 10% of this (or any) distribution? 17. What raw score cuts off the top 10% of this distribution? 18. What raw scores mark the middle 60% of this distribution? 23Worksheet: Chapter 6 1. When flipping a coin, heads and tails are mutually exclusive because . a. if the coin comes up heads, it cannot also come up tails. b. if the coin comes up heads on one toss, it has no influence on whether the coin comes up heads or tails on the next toss. c. sampling is with replacement d. sampling is without replacement 2. Jake is having a party for all of his friends in his apartment complex. He knows they all have very different tastes, so he stocks his refrigerator with a large selection. Jake has 12 bottles of Coors beer, 24 bottles of Molson beer, 24 bottles of Heinekin beer, 8 bottles of wine coolers, and 12 bottles of Coke. 2A. Billy wants any beer. What is the probability that the first beverage Jake randomly grabs is a beer? 2B. Allison wants a Coke. Given that the first bottle grabbed was a Coors, what is the probability that the second beverage Jake randomly grabs is a Coke? 3. What is the probability of drawing an ace out of a standard deck of 52 cards? 4. What is the probability of drawing a red card out of a standard deck of 52 cards? 5. What is the probability of drawing a red ace out of a standard deck of 52 cards? 6. What is the probability of drawing three cards out of a standard deck of 52 cards, without replacement, and have all 3 cards turn up red? 24 7. A letter of the English alphabet is chosen at random. Find the probability that the letter selected... 7A. is a vowel (consider y a consonant) 7B. is any letter which follows p in the alphabet 8. If I flip a coin 5 times which set of heads (H) and tails (T) outcomes is more likely: a. HHHHH b. TTTTT c. HTHTH d. all are equally likely 9. There are 105 applicants for a job with a new coffee shop. Some of the applicants have worked at coffee shops before and some have not served coffee before. Some of the applicants can work full-time, and some can only work part-time. The exact breakdown of applicants is as follows... Coffee Shop No Coffee Shop Experience (E) Experience (not E) Available Full-Time (F) 20 12 Available Part-Time (not F) 42 31 Find each of the following probabilities. 9A. P(E): The probability someone has coffee shop experience 9B. P(F): The probability someone is available full-time 9C. P(not E): The probability someone has no coffee shop experience 9D. P(E & F): The probability someone has coffee shop experience and is available full-time. 9E. P(F | E): The probability someone is available full-time given that they have coffee shop experience. 9F. P (not F | not E): The probability someone is available part-time given they have no coffee shop experience. 25Exam 2: Sample Test Multiple Choice (2 points each) 1. If an event can occur once out of 20 times, its probability value is a. .20 b. .80 c. .95 d. .05 2. When the standard deviation has a negative value, then a. most scores were above the mean b. most scores were below the mean c. the distribution is badly skewed d. none of these because the standard deviation can never be negative 3. When the variance is equal to zero... a. the standard deviation is equal to 1 b. the raw scores are negative c. all of the raw scores are the same d. the variance can never be equal to zero 4. How is the standard deviation affected when you divided a constant into every score? That is, if everyone's score is divided by 2, how will the standard deviation change? a. the new standard deviation and the old standard deviation will be the same b. the new standard deviation will be twice as large as the old standard deviation c. the new standard deviation will be half the size (twice as small) as the old standard deviation d. not enough information to answer this question 5. When flipping a coin, heads and tails are independent because . a. if the coin comes up heads, it cannot also come up tails. b. if the coin comes up heads on one toss, it has no influence on whether the coin comes up heads or tails on the next toss. 6. The interquartile range is not the best measure of dispersion because it eliminates 50% of the distribution. The 50% of the distribution that is eliminated is: a. the middle 50% b. the upper 50% c. the lower 50% d. the lower 25% and the upper 25% 7. Which of the following is a conditional probability a. the probability of being struck by lightning b. the probability of it raining c. the probability of having being struck by lightning if it is raining d. the probability of getting heart disease 268. To calculate the probability of the joint occurrence of two independent events, the probabilities for the separate events occurring a. are added together b. are first multiplied together, and then subtracted from 1.0 c. are multiplied together d. are subtracted from each other 9. Which of the following is a conditional probability? a. the probability that the wind will blow tomorrow b. the probability that the wind will blow tomorrow given that it rains c. the probability that it will rain or the wind will blow tomorrow d. the probability that it will rain and the wind will blow tomorrow 10. If there are only 10 red, 5 green, and 10 yellow M & Ms left in the package, what is the probability of drawing a red M & M (which you eat) and then another red one? a. .35 b. .80 c. .16 d. .15 The average score on a test of hand steadiness is 20 ( = 20). The standard deviation is 5 (=5). 11. What proportion of individuals can be expected to score higher than 28? Show your work! (2 points) 12. What proportion can be expected to score between 19 and 21? Show your work! (3 points) 13. Use the following population data set to answer the next problem: (Show Work!) 54 29 35 10 28 36 32 45 48 60 Compute the interquartile range _________ (2 points) 14. The mean of the Stanford Binet IQ is 100 with a standard deviation of 16. 27A. Mensa is an organization that only allows people to join if their IQs are in the top 2% of the population. What is the lowest Stanford-Binet IQ you could have and still be eligible to join Mensa? Show your work. (3 points) B. What percentage of the population has a Stanford-Binet IQ score between 84 and 95? Show your work. (3 points) C. What score falls at the 80th percentile. Show your work (2 points) D. What is the probability of obtaining an IQ score lower than 80? ____________ (2 points) Use the following population data set for the next few problems 5 10 10 12 15 15 18 18 18 20 20 25 15. Compute the variance ___________ Show Work! (3 points) 16. Compute standard deviation ____________ (1 point) 2817. A company hired a psychologist to assist their employees in their personal problems. The psychologist met with 50 employees. The psychologist kept 1 file for each person she helped. That is, she had 50 files. Ten people sought out help for drug related problems. Twenty people needed help for family crisis problems. And the remaining twenty people needed help for miscellaneous reasons. The numbers are summarized below. (3 points each) Problem Frequency Drug 10 Family crisis 20 Other 20 A. If one of the files were selected at random, what is the probability that it would involve a drug case? Leave your answer in decimal form. B. If one of the files were selected at random, what is the probability that it would involves a drug case or a family crisis case? Leave your answer in decimal form. C. If two of the files were randomly selected one at a time, what is the probability that they would involve a drug case and a family crisis case. (Sampling is one at a time with replacement.) Leave your answer in decimal form. D. If two of the files were randomly selected one at a time, what is the probability that they would both involve drug cases. (Sampling is one at a time with replacement.) Leave your answer in decimal form. 29 Exam 3 Exam three will cover Chapter 7 and 8 in the text, and Lesson 9-12 online. 30 Worksheet: Chapter 7 and 8 1. A researcher predicts that someone who exercises regularly should have a different percentage of body fat than people who do not exercise at all. The researcher finds that a person who exercises regularly has a body fat percentage of 13%. Does this percentage differ significantly from the general population of people who do not exercise and have a body fat percentage of 20%? 1A. Was this a one-tailed or a two-tailed test? 1B. What was the null hypothesis in words and symbols? 1B. What was the alternative hypothesis in words and symbols? 2. A study is conducted to determine whether a new drug will improve memory. A person taking the new drug is able to recall 35 words from a list of 50 after studying the list for 10 minutes. Do they recall more words than the general population that can recall only 25 words? 2A. Is this a one-tailed or a two-tailed test? 2B. What is the null hypothesis in words and symbols? 2C. What is the alternative hypothesis in words and symbols? 313. The basketball coach likes to recruit tall students. The height of the students are normally distributed. The mean height of the basketball team is 79 inches high with a standard deviation of 1.76 inches. Someone claims to be a member of the team who is 74 inches tall. What is the probability that someone 74 inches or shorter really is on that basketball team? (Hint: this is mostly a z-score probability problem like we did on Exam 2) 4. What is the critical value for each of the following? 4A. =.05, one-tailed test 4B. =.01, two-tailed test 4C. =.01, one-tailed test 5. One tail-tests: a. predict the direction of the effect and are more likely to result in rejection of Ho b. do not predict the direction of the effect and are more likely to result in rejection of Ho c. predict the direction of the effect, and are less likely to result in rejection of Ho d. do not predict the direction of the effect, and are less likely to result in rejection of Ho 6. If we repeatedly sample from a population and form a distribution of sample means it is: a. a sampling distribution b. a sampling distribution of the mean c. the standard error d. the standard deviation 7. The probability of a Type II error is: a. b. 1 - c. d. 1 - 8. The larger the standard deviation: a. the more variability there is in the set of values b. the less variability there is in the set of values c. standard deviation does not indicate variability d. none of the above 9. The probability of correctly rejecting the null is: a. the probability of a Type II error b. alpha c. power d. none of the above 32 10. What is the probability of committing a Type I error given that the null hypothesis is actually false? 11. What is the probability of committing a Type I error given that the null hypothesis is actually true? 12. Fill in the blanks with correct decision, Type I error, and Type II error. Also include the probability of each cell. Which cell is power? True state of the world Decision Null is true Null is false Reject null Fail to reject null 15. Telling someone that he has a disease when he does not is an example of ... a. Type I error b. Type II error c. Type III error d. Type IV error 16. Telling someone to go home and take an aspirin when in fact he needs immediate treatment is an example of ... a. Type I error b. Type II error c. correct decision 17. Convicting an innocent woman of a crime is an example of ... a. Type I error b. Type II error c. correct decision 18. Letting a guilty woman go free is an example of... a. Type I error b. Type II error c. correct decision 33Worksheet: Chapter 7 and 8 (Part 2) 1. Patients recovering from an appendix operation normally spend an average of 6.3 days in the hospital. The distribution of recovery times is normal with a = 1.2 days. The hospital is trying a new recovery program that is designed to lessen the time patients spend in the hospital. The first 10 appendix patients in this new program were released from the hospital in an average of 5.5 days. On the basis of these data, can the hospital conclude that the new program has a significant reduction of recovery time. Test at the .05 level of significance with a one-tailed test. STEP 1: State your hypotheses (include both H0 and H1). STEP 2: Set up the criteria for making a decision. That is, find the critical value. STEP 3: Summarize the data into the appropriate test-statistic. STEP 4: Evaluate the Null Hypothesis (Reject or Fail to reject?) What is your conclusion? 2. What is the Central Limit Theorem? Why is it so important? 3. From the central limit theorem, we know which of the following characteristics of the sampling distribution... A. its shape B. its mean C. its standard deviation D. all of the above 344. In earlier chapters z = X - . In this chapter the z formula used is z = X - n. What are the differences between the two formulas? Why are the formulas not the same? 5A. From the text, what are some of the factors that affect the likelihood of rejecting H0? 5B. Which of these factors does the experimenter have control over before he/she collects data? 6. Name the factors that affect the z-score, and subsequently your decision about the null. 7. A diligent researchers found that the typical person spends a mean amount of = 25 hours per week using the internet, with a standard deviation of 2.5 hours. The researcher took a random sample of 30 Time Warner cable customers and found that they spend a mean amount of 23.4 hours per week on the internet. Do Time Warner cable customers spend less time on the internet than others? Set = .01. STEP 1: State the null and alternative hypotheses in words. Label H1 and H0 State the null and alternative hypotheses in symbols Label H1 and H0 STEP 2: Set up the criteria for making a decision (find the critical value). (1 point) 35STEP 3: Compute the appropriate test statistic. Show your work. STEP 4: Evaluate the null hypothesis. Reject or Fail to Reject STEP 5:(conclusion) Based on your evaluation of the null hypothesis, what is your conclusion? 7B. Based on your answer above, what type of error might you have made in your decision in Step 4? 36 Exam 3: Sample Test 1. What is the standard error? a. the standard deviation of the sampling distribution of the sample means b. Type I error c. Type II error d. both b and c 2. Professional athletes are now commonly tested for steroid use following competition. It is known that there is some risk of sampling error, but this risk is believed to be minimal. What would constitute a Type II error on the part of the testing agency, if their null hypothesis is that the athlete is drug-free? a. an athlete who is using steroids tests negative (drug-free) b. an athlete who is using steroids tests positive (not drug-free) c. an athlete who is not using steroids tests negative (drug-free) d. an athlete who is not using steroids tests positive (not drug-free). 3. A researcher is very worried about making a Type I error. What is the alpha level she should choose to minimize the risk of a Type I error? a. = .01 b. = .05 c. = .025 d. does not have a direct effect on Type I errors 4. A psychology student was getting ready to propose her thesis, but she was very worried about making a Type I error. She asked her advisor what alpha level she should choose to minimize the risk of Type I error. Which of the following gives the least chance of making a Type I error? a. .01 b. .025 c. .05 d. Alpha does not have a direct effect on Type I errors. 5. When the null hypothesis is rejected, then a. Type II error is committed b. a significant difference has been established c. the sample means are assumed to be equal d. the population means are assumed to be equal 6. According to the Central Limit Theorem, the _________ the size of the samples selected from the population, the __________ likely the sampling distribution of means ___________. a. fewer; more; will approximate the normal curve b. fewer; less; will approximate the standard deviation c. larger; less; will approximate the normal curve d. None of the above is correct 377. Which of the following would constitute a Type II error? a. you test positive for a disease but you really do not have it b. you test negative for a disease and you really do not have it c. you test positive for a disease and you really do have it d. you test negative for a disease but you really do have it 8. A directional test means the same as: a. a test of alpha b. a two-tailed test c. a test of power d. a one-tailed test 9. According to your text, sampling error means the same as: a. the Central Limit Theorem b. the failure to accept the research hypothesis c. a biased sample d. variability due to chance 10. The research (alternative) hypothesis is: a. the hypothesis that states no difference, or no relationship is expected b. the hypothesis that states the error variability is expected to be less than 1 c. the hypothesis that states what the experiment was designed to investigate d. the hypothesis that states the number of subjects to be used in the experiment 11. If all other factors are held constant, decreases in the sample variance will ________ the value of the t-statistic. a. increase b. decrease c. have no effect on d. cant answer: Not enough information 12. What is the critical value for a one tailed-test, Z.05, alpha = .05? a. 1.96 b. 1.64 c. 2.33 d. 2.58 3813. In a large corporation the mean entry level salary is $27,000 with a standard deviation of 6,000. The entry level salaries for a random sample of 15 employees with only high school degrees is $24,100. Do people with only high school degrees earn less than the rest of the company? 13A. Conduct a one-tailed hypothesis test with = .05. STEP 1 State your hypotheses in both words and symbols. Be sure to clearly label your null and alternative hypotheses. (4 points). In words: In symbols: STEP 2: Find the critical value. (1 point) STEP 3: Compute the appropriate test-statistic. (4 points) STEP 4: Evaluate the null hypothesis (based on your answers to the above steps). (1 point) REJECT or FAIL TO REJECT (circle one) What is the best conclusion, according to your decision in STEP 4? (1 point) 3914. Years of population counts have shown African leopards have an average number of spots equal to = 25 with a standard deviation of 7 spots. A biologist claims that Snow leopards have a different number of spots than African leopards. He gets a representative sample of 15 Snow leopards. You notice that these leopards have an average of 30 spots. You want to know, with a 95% level of certainty, whether Snow leopards have a different number of spots compared to those from Africa. Conduct a TWO-TAILED test STEP 1 State your hypotheses in both words and symbols. Be sure to clearly label your null and alternative hypotheses. (4 points). In words: In symbols: STEP 2 Find the critical value. (1 point) STEP 3 Compute the appropriate test-statistic. Show your work (4 points) STEP 4 Evaluate the null hypothesis (based on your answers to this point) REJECT or FAIL TO REJECT (circle one) (1 point) 40 Exam 4 Exam 4 will cover Chapters 9, 10, and 13 in the text, and Lesson 13-16 online. The formulas for ANOVA (Chapter 13) will vary slightly from the text. We will not do the Scheffe test on pp 330-332. Also note that all of our test will now be done as two-tail tests. One-tail tests will not be included in the problems here or on the exams. 41Worksheet: Chapter 9 1. Conduct a t-test to see if a sample of 65 participants with a mean of 83 and a standard deviation of 5.4 is significantly different than a population mean of 80. Set = .05, 2-tail. 2. A psychobiologist hypothesizes that the diastolic blood pressure of Type A persons differs from the average person. In the population, the mean diastolic blood pressure is = 80. The psychobiologist takes the blood pressure of 22 Type A men whose ages range from 21 and 29. The sample mean diastolic pressure is X = 93, with the standard deviation of S = 18.76. Using = .05, two-tailed, conduct a t-test. STEP 1: State your hypotheses (include both H0 and H1).Set = .05, two-tailed. STEP 2: Set up the criteria for making a decision. That is, find the critical value. STEP 3: Summarize the data into the appropriate test-statistic. That is, compute the t statistic. STEP 4: Evaluate the Null Hypothesis (Reject or Fail to reject?) What is your conclusion? Compute 95% confidence limits on . Interpret. 423. A population has = 100 and = 50. Find the t-score for each of the following sample means: a. a sample of n = 25 with X = 220, s = 50 b. a sample of n = 4 with X = 230, s = 50 c. a sample of n = 100 with X = 190, s = 50 4. A particular state knows that its officers can run a mile in = 7 minutes, and they want to improve this overall running performance of the force. You are the chief statistician for the state-attorneys general office, and you have been asked to check to see if new recruits hired under a new standard can run faster than the uniformed officers. You plan to compare the mean-mile run time of ten recruits to the average of 7 minutes to determine if it takes them a different amount of time to run a mile. The run times (in minutes) are: 5.2 5.0 6.8 9.3 11.1 7.0 8.4 8.0 9.9 8.4 (hint: you must compute the mean and standard deviation from the sample) 4A. Should you do a one-tailed or a two-tailed hypothesis test? 4B. Conduct the appropriate hypothesis test. STEP 1: State your hypotheses (include both H0 and H1). Set = .05. STEP 2: Set up the criteria for making a decision. That is, find the critical value. STEP 3: Summarize the data into the appropriate test-statistic. STEP 4: Evaluate the Null Hypothesis (Reject or Fail to reject?) What is your conclusion? 11C. Compute 95% confidence limits. Interpret. 435. A manufacturer of flashlight batteries claims that its batteries will last an average of = 34 hours of continuous use. After receiving several complaints about the batteries, a consumer protection group predicts that the batteries run in a different amount of time than 34 hours. During consumer testing, a sample of n=30 batteries lasted an average of only X = 32.5 hours with a standard deviation of 3. Conduct a two-tailed hypothesis test with = .05. STEP 1: State your hypotheses (include both H0 and H1). STEP 2: Set up the criteria for making a decision. That is, find the critical value. STEP 3: Summarize the data into the appropriate test-statistic. STEP 4: Evaluate the Null Hypothesis (Reject or Fail to reject?) What is your conclusion? 6. In a single-sample t-test, what are the respective critical values for: A. =.05, n=10, two-tailed test B. =.01, n=31, two-tailed test C. =.05, n=40, two-tailed test D. =.01, n=107, two-tailed test 44Worksheet: Chapter 10 1. The standard error of the difference (for the independent measures t-test) is an estimate of a. centrality b. normality c. variability d. none of the above 2. If other factors are held constant, increasing the level of confidence from 95% to 99% will cause the width of the confidence interval to: a. increase b. decrease c. not change d. there is no consistent relation between interval width and level of confidence 3. In an experiment, the experimental group has 13 participants with s2 = 3.24 and the second group has 15 participants with s2 = 2.56. Compute the pooled variance 4. Suppose a teaching methods study was designed to test a hypothesis of equal means on the final examination scores for an experimental teaching method and the traditional lecture method. Subjects were randomly assigned to one of the two methods, classes were taught, and final examination scores were recorded. A summary of the data is as follows Experimental: n = 16 X = 87.5 s2 = 38.13 Traditional: n = 16 X = 82.0 s2 = 42.53 Which type of hypothesis testing should be conducted in order to assess whether there is a difference in the final exam scores of the two teaching techniques? a. single sample t-test b. dependent samples t-test c. independent samples t-test Conduct the appropriate hypothesis test. STEP 1: State your hypotheses (include both H0 and H1). Set = .05, two-tailed. 45 STEP 2: Set up the criteria for making a decision. That is, find the critical value. STEP 3: Summarize the data into the appropriate test-statistic. STEP 4: Evaluate the Null Hypothesis (Reject or Fail to reject?) What is your conclusion? 465. Rapee and Lim (1992) asked 28 persons with social phobias and 33 nonclinical subjects to rate themselves on a public speaking performance that they gave. The participants rated themselves on a 1 to 15 scale with higher numbers indicating worse performance. The sample of phobic patients gave themselves a mean rating of 12.5 with a variance of 9.61, whereas the nonclinical sample had a mean self-rating of 9.4 with a variance of 10.24. Which type of hypothesis test should be conducted in order to assess whether there is a difference in the self report ratings of the two groups? a. single sample t-test b. dependent samples t-test c. independent samples t-test Conduct the appropriate hypothesis test. STEP 1: State your hypotheses (include both H0 and H1). Set = .01, two-tailed. STEP 2: Set up the criteria for making a decision. That is, find the critical value. STEP 3: Summarize the data into the appropriate test-statistic. That is, compute the t statistic. STEP 4: Evaluate the Null Hypothesis (Reject or Fail to reject?) What is your conclusion? 476. A researcher is studying whether diet pills really work. The researcher gets two groups of people. The first group of 20 people is given the diet pill to help suppress their appetite. The second group of 15 people is given a placebo. Both groups are then instructed to try to lose weight. The researcher hypothesizes that the people who were given the diet pill will lose more weight. The diet pill group lost a mean of 4.78 pounds (with a variance of 10.63) during the one month experiment. The members of the placebo group, on the other hand, lost a mean of 3.61 pounds (with a variance of 12.04). Which type of hypothesis test should be conducted in order to assess whether people using the diet pills lost more weight? a. single sample t-test b. dependent samples t-test c. independent samples t-test Conduct the appropriate hypothesis test. STEP 1: State your hypotheses (include both H0 and H1). Set = .05, two-tailed. STEP 2: Set up the criteria for making a decision. That is, find the critical value. STEP 3: Summarize the data into the appropriate test-statistic. That is, compute the t statistic. STEP 4: Evaluate the Null Hypothesis (Reject or Fail to reject?) What is your conclusion? 48 Worksheet: Chapters 13 1. What is the abbreviation for analysis of variance? 2. When does one conduct an ANOVA? 3. If you obtain a significant F statistic you know that: a. at least one mean is statistically different from one other mean b. all the means are different from each other c. all the means come from the same population d. the null hypothesis is probably correct 4. When the null hypothesis is true, then F = MSbetween / MSwithin will be equal to: a. 0 b. 1 c. greater than 1 d. not enough information given 5. To test the truth or falsity of H0, we calculate two estimates of the population variance. Which estimate of the population variance is independent of the truth or falsity of H0? 6. In an ANOVA summary table, what are the sources of variability? 7. Between variability can also be thought of as A) between groups variability B) within groups variability C) total variability D) both A and B 8. Within variability can also be thought of as A) between groups variability B) within groups variability C) total variability D) both A and B 499. The total variability can also be thought of as A) between variability + within variability B) error variability C) within variability D) between variability Use the following example for questions 10 - 12. Suppose I was conducting a study to see which network can make people laugh more on Thursday nights. I have three groups: One group watches NBC, the second group watches ABC, and the third group watches CBS. All participants watch television from 8:00 to 10:00 with a tape recorder. The experimenter listens to the tape to record laughter. 10. What is the appropriate statistical test? A) Pearson's r B) single sample t-test C) ANOVA D) related measures t-test 11. In this experiment, what are some of the reasons for between groups variability. That is, what are some of the reasons that the groups in an experiment may have different values? (In other words, what are some of the reasons that people in the NBC group have higher laughter scores than people in the CBS group?) 12. In this experiment, what are some of the reasons for within group variability. That is, what are some of the reasons that the subjects within each group may have different scores? (In other words, how come everyone in the NBC group does not have the same laughter score?) 13. What is a multiple comparison procedure (post-test) and why does one need to conduct one when conducting ANOVA? 14. What is Tukeys HSD? When does one compute Tukeys HSD? What does HSD stand for? 50 15. What do eta-squared and omega-square measure? Which one is more accurate? 16. What are two measures of magnitude of effect? Which measure is less biased? 17. A pool of subjects was randomly divided into five treatment groups. The groups were administered daily doses of vitamin C over a 12-month period. The data in the table represent the number of cold and flu viruses reported by the participants as a function of their vitamin C dosage. Using the .05 level of significance, carry out a complete ANOVA on these data. 0mg 250mg 500mg 1000mg 2000mg 6 3 3 4 1 5 4 3 1 0 3 5 4 0 2 2 4 2 3 1 STEP 1: State your hypotheses. STEP 2: Set up the criteria for making a decision STEP 3: Summarize the data into the appropriate test-statistic. 51 STEP 4: Evaluate H0. (Reject or Fail to reject) Conclusion: 18. If appropriate, use Tukey's HSD test to perform pairwise comparisons on the means of the data in the above question. 19. Calculate and interpret 2 (eta squared) on the data in question 17. 20. Use Tables D.3 and D.4 to determine the critical value for F (Fcrit) for each of the following situations: 20A. = .01, dfgroup = 7, dferror = 60 20B. = .01, dfgroup = 4, dferror = 30 20C. = .05, dfgroup = 5, dferror = 120 20D. = .05, dfgroup = 3, dferror = 24 21. Complete the ANOVA summary table. You do not need the raw data to complete this table. Source SS df MS F Group (Between)80 40 Error (Within) Total 100 14 52Exam 4: Sample Test 1. If other factors are held constant, increasing the level of confidence from 95% to 99% will cause the width of the confidence interval to: a. increase b. decrease c. not change d. there is no consistent relation between interval width and level of confidence 2. In an Analysis of Variance test (ANOVA), what term is used to signify (or is equivalent to) variance? a. F-ratio b. sum of squares c. mean square d. degrees of freedom 3. In ANOVA, MS group is best described as the a. variance due to between group differences b. variability due to individual differences c. proportion of total variance due to between group differences d. proportion of total variance due to individual differences 4. When conducting an independent measures t-test , if the null hypothesis is rejected: a. the samples were drawn from populations that were actually dependent rather than independent. b. the mean of one sample is so far from the mean of the other sample that the decision is that the samples come from populations that have different mean values. c. the mean of one sample is statistically the same as the mean of the other sample so the decision is that they come from populations that have the same mean value. d. both a and c 5. Each of the following is part of conducting a independent measures t-test, EXCEPT a. difference scores are found for each subject b. the population variances are estimated c. the comparison is made against a t-distribution d. the variance of the distribution of differences between means is computed 6. When conducting an independent measures t-test: a. the medians of the two populations are assumed to be equal b. the null hypothesis is rejected if the calculated t-statistic you compute is more extreme than the critical-t c. only the .01 significance level should be used to increase power d. all of the above 537. When conducting an ANOVA, you decide to reject the null hypothesis. Which of the following must be true? a. between variability > within variability b. between variability = within variability c. between variability < within variability d. between variability > total variability 8. When do you normally use analysis of variance rather than the independent measures t-test? a. when the population means are unknown b. when the population variances are unknown c. when there are more than two means to compare d. when the data is badly skewed 9. The assumption that the population variances are the same is called a. the normality assumption b. a one-tailed test c. homogeneity of variance d. the repeated measures assumption 10. If there is no treatment effect, the F ratio is near a. zero b. ten c. infinity d. one 11. Keeping everything else constant, if we changed from a one-tailed to a two-tailed test, we would expect power to a. remain unchanged b. decrease c. increase 12. If you obtain a significant F-statistic then you know that: a. at least two means are significantly different from one another b. all of the means are significantly different from one another c. all of the means belong to the same population d. then null hypothesis is probably correct 13. An independent measures experiment uses two samples with n = 8 in each group to compare two experimental treatments. The t-statistic from this experiment will have degrees of freedom equal to a. 7 b. 14 c. 15 d. 16 5414. When doing an independent samples t-test, when MUST you pool the variance? a. when the sample size is less than 30 b. when the samples are of unequal sizes c. when you are performing a one-tailed test d. when you are using an alpha level less than .05 15. A researcher is interested in whether a certain hour-long film that portrays the insidious effects of racial prejudice will affect attitudes toward a minority group. One group of participants (n = 10) watched the movie, and a control group (n= 10) spent the hour playing cards. Both groups were then given a racial attitude test, wherein high scores represented a higher level of prejudice. Summary data were as follows: Movie Control X = 9.6 94.821 =s X = 11.75 86.922 =s Conduct a two-tailed test with = .05. Step 1: State the null and research hypotheses in symbols: (2 points) Step 2: Set up the criteria for making a decision. (1 point) Step 3: Conduct the appropriate statistical test. (3 points) Step 4: Based on your answers above, state your decision about the null (1 point) REJECT FAIL TO REJECT (circle one) What does your decision lead you to conclude about the research question? In other words, state the results of the experiment. (1 point) 16. Which of the following is the least biased measure of magnitude of effect? a. eta-squared b. omega-squared c. beta d. delta 17. A pool of subjects was randomly divided into 4 treatment groups. The groups were administered daily doses of Vitamin C over a 12-month period. The data in the table 55represents the number of cold and flu viruses reported by the participants as a function of their vitamin C dosage. Using the .05 level of significance, complete the ANOVA. 0mg 500mg 1000mg 2000mg 1x = 16 2x = 12 3x = 8 4x = 4 21x = 74 22x =38 23x = 26 24x = 6 n 1 = 4 n 2 = 4 n 3 = 4 n 4 = 4 Step 1: State the null hypotheses in words or symbols. (1 point) Step 2: Set up the criteria for making a decision (1 point) Step 3: Conduct the appropriate statistical test. (8 points) Source SS df MS F Group Error 2 Total 44 Step 4: Based on your answers above, state your decision about the null (1 point) REJECT FAIL TO REJECT (circle one) Based on your decision about the null, is it appropriate to conduct a post-hoc test? (1 point) YES NO (circle one) Just by looking at the data you used to conduct the test, which group reported the least number of colds and viruses? (1 point) Conduct a test of Magnitude of Effect using the least biased estimator (2 points) 56 18B. Interpret the effect size you computed above. (2 points) 19. A researcher conducts an ANOVA test to determine which of 3 treatments (using 33 total subjects) will extend terminal cancer patients lives the longest. The omnibus ANOVA was significant with a MSwithin = 36.89. The mean number of months patients survived for each of the groups is printed below. Conduct a Tukey's post-hoc test to determine which of the groups differed from one another. Set = .05. (5 points) 1X = 28.26 2X = 18.39 3X = 17.15 57Final Exam The non-comprehensive part final exam will be worth 50 points (the same as the other exams) and will cover Chapters 15 and 16 in the text, and Lesson 17-20 online. There will also be a comprehensive section on the final exam worth 15 points. These points will be taken from material on the previous three exams. In chapter 15 we will not cover the Spearman correlation pp 404-409. Also, the formulas the homework packet and online lecture notes contain for this chapter differ from those the text uses. 58Worksheet: Chapter 15 1. A previous student of this class was curious about the relationship between number of hours a person slept before an exam and the number of correct answers on the exam. She asked a sample of 5 people from her residence hall the number of hours they slept before and the number of correct answers they got on their first exam. The data are as follows... Number of hours Number of correct slept before exam answers on exam X Y 10 5 12 11 3 0 8 13 5 9 1A. Compute the correlation coefficient and conduct a hypothesis test using the following steps. STEP 1: State your hypotheses (include both H0 and H1). Set = .01, two-tailed. STEP 2: Set up the criteria for making a decision. That is, find the critical value. STEP 3: Summarize the data into the appropriate test-statistic. That is, compute the correlation. STEP 4: Evaluate the Null Hypothesis (Reject or Fail to reject?) What is your conclusion? 1B. According to the data, how many correct answers should they get if they sleep 9 hours (Hint: Compute the regression equation). 592. Use the regression equation below to predict the yearly salary (in thousands) from the number of years of higher education. Y = 2X + 12.98 2a) Samantha has had 0 years of higher education. Estimate her annual salary. 2b) Tabatha has had 11 years of higher education. Estimate her annual salary. 2c) What is the slope of this regression equation? 2d) What is the intercept of this regression equation? 2e) What is the regression coefficient and y-intercept of this regression equation? 3. Which type of correlation coefficient should be computed when both the X variable and the Y variable are dichotomous? a. Pearson c. Phi b. Point biserial d. Spearman 4. What is the difference between the predictor variable and the criterion variable? 5. Listed are 4 correlations. Put them in order showing the highest to lowest degree of relationship: -0.05 +0.26 -0.97 +0.84 6. For the test for significance of a correlation, the null hypothesis states a. the population correlation is zero c. the sample correlation is zero b. the population correlation is not zero d. the sample correlation is not zero 7. Suppose the correlation between hot chocolate sales and weather temperature is -0.80. What proportion (or percent) of the variability is predicted by the relationship with weather? a. 80% b. 40% c. 20% d. 64% e. not enough information to answer this question 8. What is the "best" fitting line? 9. What is predicted (or predictable) variability (r2)? 60 10. Use the following data for the next 2 problems. X Y 0 9 1 7 2 11 10A. Find the regression equation for predicting Y from X from the above data. 10B. What is the standard error of estimate for the above data. Interpret. 11. A sample of n = 27 pairs of scores (X and Y values) produces a correlation of r = +0.50. Are these sample data sufficient to conclude that there is a non zero correlation between X and Y in the population? Test at the .05 level of significance, two-tailed. 61Worksheet: Chapter 16 1. What type of data does one need to have in order to conduct a chi-square test? 2. What is the goodness-of-fit test? 3. What are observed frequencies? What are expected frequencies? 4. Degrees of freedom for the goodness-of-fit test are defined as df = k - 1. What is k? 5. Nonparametric tests are referred to as ____ free tests. a. distribution c. definition b. measurement d. parameter 6. Degrees of freedom for the test of independence is defined as df = (R - 1) (C - 1). What is R? What is C? 7. A chi-square test on two categorical variables is called a a. parametric test c. contingency test b. goodness-of-fit test d. test of independence 8. Which one of the following statements about chi-square is not true? a. chi-square is used primarily with nominal data b. the observations must be dependent c. no expected frequencies should be less than 5 629. The table below shows the frequencies of new admissions to a metropolitan psychiatric clinic as a function of season. Test the hypothesis that the incidence of depression, as measured in this way, is independent of season. Use = .01. Be sure to state your hypotheses, find your critical value, calculate your test-statistic, and evaluate the null hypothesis. Also state a conclusion. Spring Summer Fall Winter Depression 20 10 12 25 Other diagnosis 15 15 25 20 10. A potential sponsor would like to know whether local viewers prefer some evening news programs over others. The sponsor conducts a viewer preference survey based on a simple random sample of 1000 households. The results are given in the table. Perform a goodness-of-fit test on these data, using = .05. KTVO KMDT KLPF KZTV 220 200 300 280 STEP 1: State your hypotheses. STEP 2: Set up the criteria for making a decision STEP 3: Summarize the data into the appropriate test-statistic. STEP 4: Evaluate H0. (Reject or Fail to reject) Conclusion: 6311. The data in the table were gathered in an investigation of possible gender differences in book-carrying behavior among college students. The researcher wants to know if men, compared with women, tend to carry books down at their side rather than in front of them. Using = .05, test this hypothesis. Be sure to state your hypotheses, find your critical value, calculate your test-statistic, and evaluate the null hypothesis. Also state a conclusion. Book-Carrying Styles Down at the Side In Front Other Women 24 70 6 Men 100 46 4 12. How does the Chi-square test of independence differ from the chi-square goodness of fit test? 64 Final Exam: Sample Test 1. If two variables are related so that as values of one variable increase the values of the other also increase, then the relationship is said to be... a. positive b. negative c. non-existent d. neutral 2. The amount of change in a Y variable that accompanies a given amount of change in X is: a. slope of a straight line b. Y-intercept of a straight line c. correlation between X and Y d. length of the prediction line 3. The Y-intercept is the value of when the value of is equal to zero. a. X; X b. X; Y c. Y; X d. Y; Y 4. The direction of a linear relationship between two variables is given by of r. a. the numerical value b. the plus or minus sign c. both the sign and the numerical value d. the numerical value of the denominator 5. In regression analysis, when Y increases by two units for each equal single-unit increase in X, then a. the slope equals +2.00 b. the slope equals +0.50 c. the intercept equals +0.50 d. the intercept equals +2.00 6. In a survey of 20 individuals, one of the survey questions provided 7 response alternatives. If the responses were evaluated using a 2 test for goodness of fit at the = .05 level of significance, the critical value for the test-statistic would be a. 10.11 b. 1.63 c. 12.59 d. 30.11 7. A perfect linear relationship of variables X and Y would result in a value of r equal to... a. zero b. a large value but not +1.00 or -1.00 c. a small value but not zero d. either +1.00 or -1.00 658. Which of the following values of r allows perfect prediction of the Y score from knowledge of the X score? a. +2.00 b. -.50 c. zero d. -1.00 9. Which correlation coefficient represents the weakest association between the X and Y variables? a. r = +0.20 b. r = +0.60 c. r = -0.50 d. r = -0.90 10. A study has found a negative correlation between a person's income and his or her blood pressure. This study indicates that . a. income and blood pressure are not related b. higher income is associated with higher blood pressure c. as income increases, blood pressure tends to increase also d. as income increases, blood pressure tends to decrease 11. The population correlation coefficient is represented by... a. b. c. d. 12. A psychologist has found a correlation of +0.54 between measures of need for achievement and college grade point average. Given this knowledge, you would expect that . a. if you knew a student's need for achievement score, you could predict the student's grade point average perfectly b. as need for achievement scores decrease, there is a tendency for college grade point to decrease c. as need for achievement scores increase, there is a tendency for college grade point to decrease d. there is no relationship between need for achievement and college grade point average 13. The equation of a regression line is Y = -1.4X + 5.0. From this equation we know that a. the line has a negative slope and intersects the X axis at +5.0 b. the line has a slope of +5.0 and intersects the Y axis at -1.4 c. the line has a slope of -1.4 and intersects the Y axis at +5.0 d. X and Y are not linearly related 14. In linear regression the difference between a value of Y and Y is known as the ... a. error of measurement b. standard error of estimate c. standard deviation d. residual 6615. The standard error of estimate in linear regression will be zero when a. r = zero b. r = -1.00 or +1.00 c. the slope of the regression line is 0.00 d. the slope of the regression line is 10.00 16. When computing a chi-square test of independence one compares to . a. sample means; population means b. sample variances; population variances c. observed frequencies; expected frequencies d. sample statistics; population parameters 17. If you fail to reject the null hypothesis in a chi-square test for goodness of fit, then the expected and observed a. variances should be about equal b. variances should be unequal c. frequencies for the cells should be unequal d. frequencies for the cells should be equal (1 points each) Below are three scattergrams. (Note: A scattergram may be the correct answer to more than one question.) A B C _____ 18. If you were to compute a correlation between the X and Y variables for each of the three sets of data, which set of data would yield a correlation closest to zero? _____ 19. If you were to construct a regression equation using the X variable to predict the Y variable for each of the three sets of data, for which set of data would the regression equation have the largest, positive slope? _____ 20. If you were to construct a regression equation using the X variable to predict the Y variable for each of the three sets of data, for which set of data would the regression equation have the most negative slope? 6721. (Runyon & Haber, 1991) In a recent study, Thornton (1977) explored the relationship of marital happiness to the frequency of sexual intercourse and to the frequency of arguments. Twenty-eight married couples volunteered to monitor their daily frequency of sexual intercourse and arguments for 35 consecutive days, and then they indicated their perceived marital happiness using an 11-point scale ranging from very unhappy (1) to perfectly happy (11). Thornton (1977) reported that the Pearson correlation between ratings of marital happiness and number of arguments was -0.74. Do the appropriate statistical test to determine whether there is a significant linear relationship between happiness and arguments. Set = .05, two-tailed. STEP 1: State your hypotheses in either words or symbols (2 points) STEP 2: Set the criteria for making a decision. That is, find the critical value (2 points) STEP 3: Summarize the data into the appropriate test-statistic. I have already done this for you: r = -0.74 STEP 4: Evaluate the null hypothesis. (1 point) Based on your results, is there a relationship in the population between happiness and arguments? YES NO (circle one) (1 point) What proportion of the variability in happiness can be explained by the number of arguments? (1 point) 6822. Soldiers at Fort Gordon, Georgia and Fort Campbell, Kentucky completed a questionnaire, which included items about cigarette use, alcohol consumption, and coffee consumption (Zvela, Barnett, Smedi, Istvan, & Matarazzo, 1990). One of the questions the researchers wanted to answer was the following: Is there a relationship between smoking and gender in the military? The data are below. Gender Smoking Status Male Female Total Current smokers 252 46 298 Ex-smokers 62 29 91 Nonsmokers 170 51 221 Total 484 126 610 Perform a chi-square test of independence on these data. Set = .05 STEP 1: State your hypotheses. (I have already done this for you). H0: Gender and smoking status are independent H1: Gender and smoking status are not independent STEP 2: Set up the criteria for making a decision. That is, find the critical value. (2 points) STEP 3: Summarize the data into the appropriate test-statistic (3 points) STEP 4: Evaluate the null hypothesis (1 point) What is your conclusion? (1 point) 23. (Birkes & Dodge, 1993) Below is the weight (in kilograms) and the time to run 1.5 miles (in minutes) for a sample of 5 individuals. Person Weight (X) Time (Y) X2 Y2 XY 1 89 11.4 7,921 129.96 1,014.6 2 75 10.1 5,625 102.01 757.5 3 66 11.1 4,356 123.21 732.6 4 92 12.3 8,464 151.29 1,131.6 5 83 10.5 6,889 110.25 871.5 405 55.4 33,255 616.72 4,507.8 23a. (3 points) Compute the correlation between weight and running time. (Set up the appropriate formula to receive credit for your answer.) 69 23b. (5 points) Write the regression equation for predicting running time from weight. (Set up the appropriate formulas to receive credit for your answer.) 23c. (1 point) What is the value for the slope of the regression line in 27b. 23d. (1 point) Predict the running time for a child who weighs 77 kilograms. 23e. (1 point) is the predictor variable and is the criterion variable in the regression equation. (circle one) a. weight; time b. time; weight 7024. A discount store has prepared a customer survey to determine which factors influence people to shop in the store. A sample of 90 people is obtained and each person is asked to identify from a list of alternatives the most important factor influencing their choice to shop in the store. The data are as follows: Convenient Low Good Location Prices Selection 30 40 20 On the basis of these data can you conclude that there is any specific factor (or factors) that is most often cited as being important? Test at the .05 level of significance with the goodness of fit chi-square test. Determine the critical region (1 point) Summarize the data into the appropriate test-statistic (3 points) Evaluate H0. (Reject or Retain) (1 point) Part 2 25. Compute the median and the mode of the following data set. 9 7 4 5 7 2 median (1 point) mode (1 point) 71 26. A national test has a mean of 192 and a standard deviation of 10. The author of the exam wants the test to have a mean of 500. What specifically does the author have to do so that her test has a mean of 500 (and the standard deviation remains 10)? (1 point) 27. Which measure of central tendency is used with nominal data? (circle one)(1 point) a. mean b. median c. mode 28. In October of 1981 the mean and the standard deviation on the Graduate Record Exam (GRE) for all people taking the exam were 489 and 126, respectively. Scores on the GRE are normally distributed. 28a. What percentage of students would you expect to have a score between 400 and 500? (1.5 points) 28b. What is the median of this distribution? (1.5 points) 29. A psychologist would like to know how much difference there is between the problem-solving ability of 8-year-old children versus 10-year-old children. A random sample of 10 children is selected from each age group. The children are given a problem-solving test, and the results are summarized as follows: 8-year-olds 10-year-olds n = 10 n = 10 x = 36 x = 39 s = 3.50 s = 5.27 30. Perform the appropriate analysis on this data. Set = .05, two-tailed. STEP 1: State your hypotheses in symbols (1 point) 72STEP 2: Set the criteria for making a decision. That is, what is your critical value? (1 point) STEP 3: Summarize the data into the appropriate test-statistic. (2 points) STEP 4: Evaluate the null hypothesis (1 point) In a controlled study, more than 70 Dartmouth College students were instructed to use orange-flavored lozenges at the first sign of an incipient cold, sucking on one as often as every two hours. Half the students got zinc lozenges; half the students were given candies that looked and tasted the same, so that none knew who was really taking the zinc. The participants who were given the zinc had a cold for 4.3 days, as against 9.2 days for those who got the look-alike candies 31A. What was the dependent variable in this study? (circle one) (1 point) a. type of cold treatment b. 70 students c. number of days cold continues d. Dartmouth College 31B. What was the independent variable in this study? (circle one)(1 point) a. type of cold treatment b. 70 students c. number of days cold continues d. Dartmouth College 31C. What is the correct analysis for this experiment? (circle one) (1 point) a. independent measures t-test c. chi-square test of goodness of fit b. related measures t-test d. single sample t-test 73Answers: Chapters 1 and 2 1. A 2A. nominal 2B. ordinal 2C. nominal 2D. ratio 2E. ratio 3A. continuous 3B. discrete 4A. exercise regimen 4B. body fat 5A. music 5B. words recalled 5C. discrete 6A. physical fitness 6B. amount of sleep 6C. continuous 6D. measurement 6E. ratio 7A. 14 7B. 14 7C. 55 7D. 54 7E. 144 7F. 168 7G. 2 8. 9. Real Mid- Fre- Cumulative Relative Cumulative Interval Limits point quency Frequency Percentage Percentage 0-1 -0.5-1.5 .5 3 3 15 15 2-3 1.5-3.5 2.5 6 9 30 45 4-5 3.5-5.5 4.5 5 14 25 70 6-7 5.5-7.5 6.5 4 18 20 90 8-9 7.5-9.5 8.5 0 18 00 90 10-11 9.5-11.5 10.5 2 20 10 100 10. 30% 11. 18 12. 5.5. 70% of the scores fall at or below 5.5. 13. 45th percentile. 45% of the scores fall at or below 3.5. 74Answers: Chapters 2 and 3 1A. 8.00 1B. 8.00 1C. 8.00 2. 01234564.58.512.516.520.524.528.532.536.5No. of WeeksFrequency 3) A 4A) B 4B) C 4C) C 5) B 6) C 75Answers: Chapters 3 1A. 2,3,4 1B. 4 1C. 4.25 2 c 3 a 4A. c 4B. Papa John's 4C. Papa John's 5. b 6 170 7 9 8. 9 9. 9.17 10. 9 11 b 12. 40.6 12. 40.1 12. No, on average both groups are fairly accurate 76 Exam 1: Sample Test Answers 1) D 2) A 3) A 4) B 5) D 6) D 7) D 8) C 9) A 10) B 11) C 12) A 13) D 14) D 15) B 16) B 17) 2 18) -18 19) -6 20) 15.5 21) 16.5 22) 18 23A) 50% 23B) 19.5 23C) 3 23D) 50th Class Intervals Apparent Limits Real Limits Midpoint Frequency Cum f Relative Percent Cum Relative Percent 0-4 -.5-4.5 2 4 4 20 20 5-9 4.5-9.5 7 3 7 15 35 10-14 9.5-14.5 12 3 10 15 50 15-19 14.5-19.5 17 6 16 30 80 20-24 19.5-24.5 22 2 18 10 90 25-29 24.5-29.5 27 2 20 10 100 23E) 01234567# of Back PainsFrequency 77Answers: Chapter 4 1A. 8.83 1B. 2.97 1C. 4 2. C 3A. Sample B 3B. 1.72 3C. 32 4A. No, on average both groups are fairly accurate 4B. 9.22 4C. 2.56 4D. Sample B 5. Scores, on the average, cannot be 13.5 points away from the mean on a 10-point scale 78Answers: Chapters 5 and 6 1. 24.18% 2. 53.50% 3. 89th 4. 75.18 5. 21.48% 6. 57.04% 7. 21.48% 8. 4.84% 9. 57.65 to 82.35 10. 1610 11. 60 12. 53.51 13. 2.02 14. Third-world man lived longer for his distribution (z=1.16) than the other man (z=.31) 15. 44.6 16. 1.28 17. 56.4 18. 45.8, 54.2 79 Answers: Chapter 6 1. a 2A. .75 2B. .15 3. .0769 4. .5 5. .0385 6. .1176 7A. .19 7B. .38 8. D 9A. .5905 9B. .3048 9C. .4095 9D. .1905 9E. .3226 9F. .7209 80Exam 2: Sample Test Answers 1) D 2) D 3) C 4) C 5) B 6) D 7) C 8) C 9) B 10) D 11) .054 12) .1586 13) 19 14A) 132.8 14B) 21.96% 14C) 113.44 14D) Z= -1.25, prob = .1056 15) 27.75 16) 5.26 17a) 0.2 b) 0.6 c) .08 d) 0.04 81Answers: Chapters 7 and 8 1A. two-tail 1B. The percentage of body fat for exercisers will not differ from those who do no exercise exercise = 20 1C. The percentage of body fat for exercisers will be different from those who do no exercise exercise 20 2A. one-tail 2B. Participants taking the new drug will recall less than or the same amount of words as the untreated population. new drug 25 2C. Participants taking the new drug will recall more words than the untreated population. new drug > 25 3. .0023 4. 1.64, + 2.58, 2.33 5. A 6. B 7. A 8. A 9. C 10. p(Type II error | null is false) = 0 11. p(Type II error | null is true) = 12. see text 15. a 16. b 17. a 18. b 82Answers: Chapter 7 & 8 (Part 2) 1. H0: > 6.3 H1: < 6.3 critical z = z.05 = -1.64 zobtained = -2.11 Reject H0 Patients in the new program are released from the hospital in less time. 2. see sampling distributions on web page 3. D 4. The first is used for finding the probability of an individual value, the second for finding the probability of a sample of values. In the same way estimates the average difference between and X, / n estimates the average difference btw/ and X 5A. alpha, N, distance between means, sigma, one-tail vs. two-tail test. 5B. Sample size, alpha level, one- or two-tailed test 6. Sample size, distance between means, sigma 7. H1: Time Warner cable customers spend less time on the internet than the general population H1 tw < 25 H0: Time Warner cable customers spend the same or more time on the internet than the general population H0: tw > 25 Critical value = -2.33 Z= -3.51 Reject H0 Time Warner cable customers spend less time on the internet than the general population. 7B. Type I (because you rejected the null it is the only type of error possible) 83 Exam 3: Sample Test Answers 1) A 2) A 3) A 4) A 5) B 6) D 7) D 8) D 9) D 10) C 11) A 12) B 13) Step 1: H1: People with a HS degree earn less than other company employees HS < 27,000 H0: People with a HS degree earn the same or more than other company employees HS > 27000 Step 2: -1.64 Step 3: t = 24,1000 27,0006, 000 / 15= 29006000 / 3.873= 29001549.19= 1.87 Step 4: Reject People with a HS degree earn less than other company employees 14) Step 1: H1: Snow leopards have a different number of spots than African leopards H0: Snow leopards have the same number of spots as African leopards H1: Snow 25 H0: Snow = 25 Step 2: +1.96 Step 3: 76.281.15873.3/7515/72530 ====z Step 4: Reject the null conclusion: there is a different number of spots for Snow leopards 84Answers: Chapter 9 1. H0: = 80 H1: 80 critical t = +2.00 tobtained = 4.48 Reject H0 2. H0: = 80 H1: 80 critical t = t.05 = 2.08 tobtained = 3.25 Reject H0 Type A persons have significantly higher blood pressure than the average person. (CI.95 = 84.68 101.32), 95% sure that the population of Type A men have a mean blood pressure in this range. 3. a. 12 b. 5.2 c. 18 4A. two-tailed 4B. H0: = 7 minutes H1: 7 minutes critical t = t.05 = +2.262 tobtained = 1.47 Retain H0 The sample does not run the mile in less time than the pop. 4C. CI.95 = 6.51 9.31, 95% sure that the population the sample of troopers comes from has a mean running time in this range. 5. H0: = 34 H1: 34 critical t = t.05 = +2.045 tobtained = -2.73 Reject H0 The batteries last significantly less time than claimed by the manufacturer. 6A. 2.262 6B. +2.750 6C. +2.021 6D. 2.62 85Answers: Chapter 10 1. c 2. a 3. 2.87 4. c H0: em = tm H1: em tm critical t = t.05 = 2.042 tobt = 2.46 Reject H0 The students in the experimental teaching class performed significantly better on the final exam than students in the traditional class. 5. c H0: sp = nc H1: sp nc critical t = t.05 =2.678 tobt = 3.83 Reject H0 Social phobic patients rated themselves significantly worse on public speaking performance than did nonclinicals. 6. c H0: diet = placebo H1: diet placebo critical t = t.05 = +2.042 tobt = 1.02 Fail to reject H0 Diet pills do not work. Diet pills are not significantly more effective than placebos in losing weight. 86Answers: Chapters 13 1. ANOVA 2. When you wish to compare more than two sample means. 3. A 4. B 5. within group variability (variance) 6. Between/ Within/Total . 7. A 8. B 9. A 10. C 11. Individual differences (e.g., Some people laugh more than others.) Error (e.g., The tape recorder picked up other noise which made it difficult to hear the laughter.) Treatment (e.g., Some networks are funnier than others.) 12. Individual differences Error 13. ANOVA only tells us that at least 2 means differ, but not which onesmust do Tukeys post-hoc test to compare multiple groups and determine which means differ. 14. Tukeys HSD is a post-test (multiple comparison procedure). One computes a Tukeys HSD when the null hypothesis has been rejected to determine which of the groups are significantly different from each other. HSD stands for honestly significant difference. 15. both measure magnitude of the effect. Omega-square is more accurate. 16. eta-square and omega-square. Omega-square is less biased. 17. H0: 1=2=3=4=5 H1: At least one mean is different from the others F.05(4,15) = 3.06 Fobt = 3.93 Reject H0 At least one group reported more cold and flue viruses than at least one other group. After conducting the Tukey HSD, we can conclude, Subjects taking no Vitamin C and subjects taking 250 mg. of Vitamin C reported significantly more cold and flu viruses than persons taking 2000 mg. of Vitamin C. 18. Tukeys HSD = 2.88 19. 2=.51 20A. 2.95 20B. 4.02 20C. 2.29 20D. 3.01 21. Source SS df MS F Group 80 2 40 23.95 Error 20 12 1.67 Total 100 14 87Exam 4: Sample Test Answers 1)A 2) C 3) A 4) B 5) A 6) B 7)A 8) C 9) C 10) D 11) B 12) A 13) B 14) B 15A) Step1: H0: film=nofilm H1: film nofilm Step2: + 2.1009 Step 3: [ ]56.137.115.22.18736.8846.8015.210110121010)14.3(9)99.2(975.116.922==+= +++=t Step 4: Fail to Reject......so, no differences in attitudes between the film and no film group 16) B 17) Step 1: 1=2=3=4 Step 2: 3.49 Step 3: Source SS df MS F Group 20 3 6.67 3.33 Error 24 12 2 Total 44 15 Step 4: Fail to reject......no......200mg group 18A) =2 .30 18B) 30% of the variability in number of cold virus reported is due to amount of vitamin C consumed. 19. HSD = 6.39 Groups 1 and 2 differ (difference=9.87), Groups 1 and 3 significantly differ (difference = 11.11), Groups 2 and 3 do not differ (difference = 1.24). 88Answers: Chapter 15 1A. H0: = 0 H1: 0 rcrit = .959 r = .56 Fail to reject H0 There is insufficient evidence to conclude that there is a significant linear relationship. 1B. The regression equation is Y =.79X +1.57. The answer is 8.68 or 8 answers. 2A. 12,980 2B. 34,980 2C. 2 2D. 12.98 2E. 2 and 12.98 3. c 4. See page 414 of your text 5. -0.97, +0.84, +0.26, -0.05 6. a 7. d 8. See page 414 of your text 9. variability in Y that is explained by differences in X 10A. Y = X +8 10B. 2.45 The standard deviation of points about the regression line (standard error) is 2.45. 11. Yes 89Answers: Chapter 16 1. Categorical or frequency data 2. See p. 428 of your text 3. See pp. 230-431 of your text 4. k stands for k-k-categories (number of groups) 5. A 6. See p. 442 of your text 7. D 8. B 9. H0: The incidence of depression is independent of season. H1: The incidence of depression is not independent of season. 2crit = 11.35 2 = 7.22 Retain H0 The incidence of depression is not independent of season. 10. H0: observed frequencies are equal to the expected frequencies H1: observed frequencies are not equal to the expected frequencies 2crit = 7.82 2 = 27.2 Reject H0 Local viewers prefer some evening news programs over others. 11. H0: Book-carrying styles are independent of gender H1: Book-carrying styles are not independent of gender 2crit = 5.99 2 = 43.69 Reject H0 Men compared with women tend to carry books down at their side rather than in front of them. 12. Chi-square test of independence: consider 2 variables at once to determine if they are independent (related). Chi-square goodness of fit test: consider 1 variable at a time. Compares actual data to what we expect by chance. 90Final Exam: Sample Test Answers 1) A 2) A 3) C 4) B 5) A 6) C 7) D 8) D 9) A 10) D 11) D 12) B 13) C 14) D 15) B 16) C 17) D 18) B 19) B 20) C 21) STEP 1: H0: = 0 H1: 0 Null: The correlation does not exist in the population Alternative: The correlation does exist in the population STEP 2: df = n - 2 = 28 - 2 = 26 rcrit = 0.374 STEP 4: Reject the null.......so, yes there is a relationship What proportion of the variability....? .5476 22) STEP 2: df = (3-1)(2-1) = 2 2crit = 5.99 STEP 3: 2obtained = 1.02 + 3.93 + 1.44 + 5.53 + .16 + .63 = 12.71 STEP 4: Reject null conclusion? Gender and smoking status are not independent. There is a relationship between gender and smoking status 23a) r = +0.5658 = +.57 23b) Y = .0453X + 7.41 23c) .0453 23d) 10.898 rounds to 10.90 23e) a. weight; time 24) critical = 5.99; 2obtained = 6.67; Reject 25) median 6 mode 7 26) Adding a constant to each score will change the mean without having an effect on the standard deviation. Add 308 to each score. 27) c. mode 28a) 29.7% z = +0.09 z = -0.71 area = .0359 area = .2611 .0359 + .2611 = .2970 28b). 489 29)STEP 1: H0: 1 = 2 H1: 1 2 STEP 2: df = 18 tcrit = 2.101 STEP 3: tobtain = -1.5 STEP 4: Fail to reject null 30a) C 30b) A 30c) A Lesson 1 Introduction.pdfLesson 2 Scales of Measure.pdfLesson 3 Data Displays.pdfLesson 4 Measures of Central Tendency.pdfLesson 5 Measures of Dispersion.pdfLesson 6 Z - Scores.pdfLesson 7 Z - Scores and Probability.pdfLesson 8 Probability.pdfLesson 9 Hypothesis Testing.pdfLesson 10 Steps in Hypothesis Testing.pdfLesson 11 Hypothesis Testing with a Sample of Values.pdfLesson 12 Errors in Hypothesis Testing.pdfLesson 13 Hypothesis Testing with the t - test statistic.pdfLesson 14 Independent Samples t - test.pdfLesson 15 ANOVA ( Analysis of Variance ).pdfLesson 16 Post - hoc Tests.pdfLesson 17 Pearson's Correlation Coefficient.pdfLesson 18 Regression.pdfLesson 19 Chi - Square.pdfLesson 20 Chi - Square ( Test of Independence ).pdfHomework Packet.pdf

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