chapter 7 sampling distributions & confidence intervals
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Doing Statistics for Business Data, Inference, and Decision Making Marilyn K. Pelosi Theresa M. Sandifer. Chapter 7 Sampling Distributions & Confidence Intervals. Doing Statistics for Business. Chapter 7 Objectives Motivation for Point Estimators Common Point Estimators - PowerPoint PPT PresentationTRANSCRIPT
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DoingDoing Statistics for BusinessStatistics for Business Data, Inference, and Decision Making
Marilyn K. PelosiTheresa M. Sandifer
Chapter 7Chapter 7Sampling Sampling
Distributions &Distributions &Confidence Confidence IntervalsIntervals
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Chapter 7 Objectives
Motivation for Point EstimatorsCommon Point EstimatorsDesirable Properties of Point EstimatorsDistribution of the Sample Mean:
Large Sample orKnown
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Chapter 7 Objectives (con’t)
The Central Limit Theorem - A More Detailed Look
Drawing Inferences by Using the Central Limit Theorem
Large Sample Confidence Intervals for the Mean
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Chapter 7 Objectives (con’t)
Distribution of the Sample Mean:Small Sample andUnknown
Small Sample Confidence Intervals for the Mean
Confidence Intervals for Qualitative DataSample Size Calculations
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Figure 7.1 Relationship between probability and inferential statistics
Probability
InferentialStatistics
Population Sample
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A Point Estimate is a single numbercalculated from sample data. It is used toestimate a parameter of the population.A Point Estimator is the formula or rulethat is used to calculate the point estimatefor a particular set of data.
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TRY IT NOW!Sales of CD’sComparing Point Estimators CalculatedFrom Samples Selected From Different Populations
Use a software package such as Excel or Minitab to simulate picking asample of size n=10 from two different populations or use the samples given on the following slide.
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TRY IT NOW!Sales of GizmosComparing Point Estimators CalculatedFrom Samples Selected From Different Populations (con’t)
Store 1 Store 2. 95 99 97 97
95 99 103 103101 102 101 98102 102 98 106 92 99 100 96
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TRY IT NOW!Sales of GizmosComparing Point Estimators CalculatedFrom Samples Selected From Different Populations (con’t)Suppose the first variable is daily sales of a new CD at Store 1. Thesecond variable is daily sales of the new CD at Store 2. Select a sample of size n=10 days from both stores. Assume that the days sales at both stores are normally distributed with a mean of 100 and a standard deviation of 3.Find andX1 X 2 X1-X2
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TRY IT NOW!Sales of CDsComparing Point Estimators CalculatedFrom Samples Selected From Different Populations (con’t)What do you notice about the difference in sample means even though thepopulation means are the same?
Find (Store 1 standard deviation squared), and
What do you notice about the ratio of the two sample variances?
s12 s2
2 s12/s22
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An Unbiased Estimator yields an estimatethat is fair. It neither systematically over-estimates the parameter nor systematicallyunderestimates the parameter.
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TRY IT NOW!The Diaper CompanyComparing the Variability of TwoPoint EstimatorsWeights in grams for the next 5 hourly samples taken at the diapercompany are shown below:
Hour 6: 54.89 55.06 55.45 55.23 55.75Hour 7: 54.32 55.72 54.91 54.40 55.78Hour 8: 54.14 55.18 55.78 55.37 55.69Hour 9: 54.11 54.05 53.60 55.97 55.86Hour 10:55.21 55.40 53.87 55.09 55.70
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TRY IT NOW!The Diaper CompanyComparing the Variability of TwoPoint Estimators (con’t)For each sample, calculate the sample mean and the sample median.Find the average of the sample means and the average of the samplemedians.Find the standard deviation of the sample means and the standarddeviation of the sample medians.
Which point estimator has less variability?
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The Standard Error is the standard deviation of the sampling distribution of a point estimator. It measures how muchthe point estimator or sample statistic variesfrom sample to sample.
The probability distribution of a point estimator or a sample statistic is called aSampling Distribution.
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Central Limit Theorem (CLT)In random sampling from a population, withmean and standard deviation , when n islarge enough, the distribution of is approximately normal with a mean equal to and a standard deviation equal to n/
X
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Histogram of Individual DiaperWeights
01020304050
53.5 54 54.5 55 55.5 56 56.5
Weight (grams)
Frequency
Figure 7.2 The Diaper Company
Histogram of Individual Diaper Weights
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Figure 7.3 The Diaper Company
Histogram of 52 Sample MeansHistogram of Average DiaperWeight
05101520
53 53.5 54 54.5 55 55.5 56 56.5Average Weight (grams)
Frequency
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Figure 7.4 Graphs of two populations with the same mean but different standard deviations
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. .. ..:. : -+---------+---------+---------+---------+---------+-----St Dev = 1
. . . . . . : . . -+---------+---------+---------+---------+---------+-----St Dev = 5 23.50 24.00 24.50 25.00 25.50 26.00
Figure 7.5 Dotplots of Sample Means
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TRY IT NOW!The Central Limit TheoremExploring the Third PointUse a software package such as Excel or Minitab to simulatepicking 10 samples each of size n = 35 from two different populations:
Population 1: Monthly sales of a leading on-line bookstore- normally distributed with a mean of = 25 ($1,000) and a standard deviation of = 1($1,000)Population 2: Monthly sales of a leading on-line bookstore- normally distributed with a mean of = 25 ($1,000) and a standard deviation of = 3 ($1,000)
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TRY IT NOW!The Central Limit TheoremExploring the Third Point (con’t)
For each sample, calculate a sample mean, .Find the average and standard deviation of the 10 ‘s from population1 samples.
Find the average and standard deviation of the 10 ‘s from population 2samples.
X
X
X
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TRY IT NOW!The Central Limit TheoremImpact of Sample Size on Standard ErrorUsing the random number table in Appendix A and the 350 values shown in the previous Try It Now! exercise in your textbook,select 10 samples of size 5 from population 1. Review Section 2.4 if you need a refresher on how to use the random number table.For each sample, calculate a sample mean, Find the average and standard deviation of the 10 ‘s. Compare thesevalues to the corresponding values that you found in the previous Try ItNow! for population 1. In that case the sample size was n = 35.
XX
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68%
95%
99.7%
Figure 7.6A Sampling Distribution of X
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Discovery Exercise 1.1The Central Limit Theorem in Action
Part I. Draw a picture of a normal distributionwith mean of 80 and standard deviation of 5. This is the population we will sample from.
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Discovery Exercise 1.1The Central Limit Theorem in Action(con’t)
Part II. Generate and examine 100 random samples.For this exercise you will need to generate 100 samples each consisting of30 value selected from a normal distribution with a mean of 80 and a standard deviation of 5.
Part III. Create a distribution of for samples ofsize n = 30.
X
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Figure 7.6B Sampling Distribution for when = 55.00
Figure 7.6C Sampling Distribution for when = 54.50
68%
95%
99.7%
68%
95%
99.7%
X X
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TRY IT NOW!Cost of BooksComparing The Sample Mean to theClaimed Population Mean
A university states the average student spends $225 per semester on books. Based on your own experience you feel that this is anunderestimate of the true expenditure. You ask 30 of your friends howmuch they spent on textbooks this past semester and you obtain thefollowing data:
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TRY IT NOW!Cost of BooksComparing The Sample Mean to theClaimed Population Mean (con’t)
214 241 248 258 269233 244 249 260 274234 245 250 262 276236 247 253 262 277239 248 254 263 277241 248 254 265 281
Based on these data, do you have reason to tell the university that itsstatement is inaccurate?
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A Confidence Interval or an Interval Estimate is a range of values with an associatedprobability or Confidence Level, 1 – . Theprobability quantifies the chance that you have an interval that contains the true population parameter.
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Figure 7.7. NormalDistribution with 0.05 inthe tails.
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TRY IT NOW!The Bottle-Filling Problem
A sample of 36 bottles had a sample mean of = 32.10 oz.The population standard deviation, , was assumed to be 0.1 oz. Find a 95% confidence interval for . How wide is the interval?
Now find a 98% confidence interval for .
Which interval is wider?
x
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Figure 7.8 Comparison of Confidence Intervals and µ
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Discovery Exercise 7.2Exploring Confidence Intervals for From a population of college students across theUnited States, a sample was selected to find our howmany hours per week a typical student spends playing sports.Part I. A random sample of 2500 students wasselected. The sample mean, was found to be 12.5 hours. The population standard deviation, , is known to be 1.05 hours. Given this information,find:
x
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Discovery Exercise 7.2 (con’t)Exploring Confidence Intervals for (a) a 90% confidence interval for .
(b) a 92% confidence interval for .
(c) a 94%confidence interval for .
(d) a 96% confidence interval for .
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Discovery Exercise 7.2 (con’t)Exploring Confidence Intervals for (e) a 98% confidence interval for
(f) Discuss what happens to the size of the interval as the level of the confidence increases.
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Discovery Exercise 7.2 (con’t)Exploring Confidence Intervals for
Part II. A random sample of 2500 students wasselected. The sample mean, was found to be 10.5 hours. The population standard deviation, , is known to be 1.05 hours. Given this information,find:
(a) a 90% confidence interval for
x
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Discovery Exercise 7.2 (con’t)Exploring Confidence Intervals for
(b) a 92% confidence interval for .
(c)a 94%confidence interval for .
(d) a 96% confidence interval for
(e) a 98% confidence interval for
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Discovery Exercise 7.2 (con’t)Exploring Confidence Intervals for Compare the intervals found in Part I with thosefound in Part II. Discuss what happened to the confidenceinterval due to the change in the value of the sample mean x.
Part III. A random sample of 2500 students wasselected. The sample mean, was found to be 12.5 hours. Suppose you that the populationstandard deviation, , is actually 2.05 hours.Given this information, find:
x
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Discovery Exercise 7.2 (con’t)Exploring Confidence Intervals for (a) a 90% confidence interval for
(b) a 92% confidence interval for .
(c)a 94%confidence interval for .
(d) a 96% confidence interval for
(e) a 98% confidence interval for
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Discovery Exercise 7.2 (con’t)Exploring Confidence Intervals for Compare the intervals found in Part I with thosefound in Part III. Discuss what happened to the confidence intervals due to the change in the value of the population standard deviation, .Part IV. A random sample of 2000 students wasselected. The sample mean, was found to be 12.5 hours. The population standard deviation, , is known to be 1.05 hours. Given this information,find:
x
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Discovery Exercise 7.2 (con’t)Exploring Confidence Intervals for (a) a 90% confidence interval for
(b) a 92% confidence interval for .
(c)a 94%confidence interval for .
(d) a 96% confidence interval for
(e) a 98% confidence interval for
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Discovery Exercise 7.2 (con’t)Exploring Confidence Intervals for
Compare the intervals found in Part I with thosefound in Part IV. Discuss what happened to the confidenceinterval due to the change in the value of the sample size, n.
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Figure 7.9 t-distribution with 5 Degrees of Freedom
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Figure 7.10 t-distribution with 25 Degrees of Freedom
Figure 7.11t-distribution with 1 and50 Degrees of Freedom
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Upper Tail Areas
Degrees of Freedom 0.25 0.1 0.05 0.025 0.01 0.005
20 0.6870 1.3253 1.7247 2.0860 2.5280 2.845321 0.6864 1.3232 1.7207 2.0796 2.5176 2.831422 0.6858 1.3212 1.7171 2.0739 2.5083 2.818823 0.6853 1.3195 1.7139 2.0687 2.4999 2.807324 0.6848 1.3178 1.7109 2.0639 2.4922 2.797025 0.6844 1.3163 1.7081 2.0595 2.4851 2.787426 0.6840 1.3150 1.7056 2.0555 2.4786 2.7787
Figure 7.12 A portion of the t table
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TRY IT NOW!Retirement YearsConfidence Interval forA survey shows that a growing number of Americans are willing to make sacrifices to become home owners despite increasing job andfinancial worries. The Federal National Mortgage Association surveyed1857 Americans and found that 67% would put off retirement for 10 yearsto own a home.
Find a 90% confidence interval for the proportion of all Americans who would put off retirement for 10 years to own a home.
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TRY IT NOW!Bottle FillingFinding the Sample Size
How many bottles does the bottle manufacturer need to sample tobe 98% confident that the error is at most 0.002 oz? Remember that the population standard deviation is 0.1 oz.
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TRY IT NOW!Retirement YearsSample Size Calculation for
How many Americans must be sampled to determine the percentage whowould put off retirement for 10 years to own a home? The estimateshould not differ from the actual population proportion by more than 3%with a confidence of 90%.
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Finding Confidence Intervals Using KaddStat Instructions for small sample confidence interval for mean- all others are done similarly
From the Kadd menu select Confidence Intervals>One Sample>Population Mean using t
The dialog box opens.
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Finding Confidence Intervals Using Kadd(con’t)
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Finding Confidence Intervals Using Excel (con’t)
1. First, indicate the level of confidence as a percent.
2. Select User Input if you already have the summary statistics or Input Range if you have raw data.
3. Indicate how the sampling was done4. Indicate where you want the output to appear5. Click OK
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Output for Confidence Interval for Small Sample usiing t
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Chapter 7 SummaryIn this chapter you have learned: The basics of estimating population parameters, in
particular how to estimate the average of a numeric characteristic of a population, , and the proportion of a population that has a certain characteristic, .
The estimates are calculated from a sample selected from the population.
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Chapter 7 Summary
Each sample yields a slightly different estimate of the population parameter.
Thus, estimators are themselves random variables. When the random variable is an estimator, the
distribution is called a sampling distribution. The sampling distribution as a mean and a
standard deviation, called the standard error.
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Chapter 7 SummaryIn this chapter you have learned:
How to use the sampling distribution of to calculate probabilities and make inferences about and for .
How to create confidence intervals for and for . How to calculate the required sample size to
achieve a certain level of precision with a specified confidence.
X