gm 533 final_prep_2012

Preparing for theGM 533 Final Exam

Professor Brent Heard

GM 533 Final Exam Prep• Summary of topics covered

– Week 1 - Descriptive Statistics: includes central tendency, dispersion, and the shape of the distribution, in numbers, pictures, and tables.

– Week 2 – Probability: includes 3 major problem types, and their most important variations: contingency tables, expected value, and the binomial distribution.

– Week 3 – Probability continued: includes the normal distribution, its application to sampling distributions, and its most important variations.

– Week 4 – Confidence intervals and sample size determinations, and their most important variations.

– Week 5 – Hypothesis testing: includes the 5-step hypothesis testing procedure, applied to means and proportions, and its most important variations.

– Week 6 – Simple linear regression: includes interpreting Minitab output for point estimates, hypothesis tests, and confidence intervals.

– Week 7 – Multiple regression: includes the same elements as simple regression, but also includes the application to multiple independent variables.

• Examples and topic areas follow

GM 533 Final Exam Prep• Sample Question on Binomial Distribution

– Assume that a study was done finding that 70 percent of males in Georgia are football fans. If a researcher asks 8 Georgia Males if they are fans, the following binomial distribution would be applicable. What is the probability that at least 5 will be football fans?

n p8 0.7

x P( x) Cumulative0 0.0001 0.00011 0.0012 0.00132 0.0100 0.01133 0.0467 0.05804 0.1361 0.19415 0.2541 0.44826 0.2965 0.74477 0.1977 0.94248 0.0576 1.0000

GM 533 Final Exam Prep• Sample Question on Binomial Distribution

– Assume that a study was done finding that 70 percent of males in Georgia are football fans. If a researcher asks 8 Georgia Males if they are fans, the following binomial distribution would be applicable. What is the probability that at least 5 will be football fans?

n p8 0.7

x P( x) Cumulative0 0.0001 0.00011 0.0012 0.00132 0.0100 0.01133 0.0467 0.05804 0.1361 0.19415 0.2541 0.44826 0.2965 0.74477 0.1977 0.94248 0.0576 1.0000

“At least 5” is the probability that 5, 6, 7 or 8 will be fans. Simply add those probabilities.

0.25410.29650.19770.0576

0.8059

My total is 0.8059 or about 81% which is the probability of at least 5 being football fans.

SUM

GM 533 Final Exam Prep• Analysis Example

– 9 members of the local college baseball team had the following number for extra base hits for the year. Using the Minitab output given, determine: A. MeanB. Standard DeviationC. Range D. MedianE. The range of the data that would contain 68% of the results.

Data79424151715629

– Minitab Follows

GM 533 Final Exam PrepDescriptive Statistics: Extra Base Hits Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3Extra Base Hits 9 0 14.00 2.82 8.47 4.00 6.50 15.00 20.50 Variable MaximumExtra Base Hits 29.00 Stem-and-Leaf Display: Extra Base Hits Stem-and-leaf of Extra Base Hits N = 9Leaf Unit = 1.0 4 0 4679(3) 1 557 2 2 49

A. MeanB. Standard DeviationC. Range (Max – Min = 29 – 4 = 25)D. Median (I would enter data into Excel to

find the Median is 15 or do by ordering data and identifying)

E. The range of the data that would contain 68% of the results. (Mean – Std Dev, Mean + Std Dev) which is (14 – 8.47, 14 + 8.47) or (5.53,22.47)

This is technically NOTthe correct Median.

GM 533 Final Exam Prep

Here I used

=MEDIAN(A1:A9)

It returned the median of 15.


• Sample Question on Hypothesis Testing– Pepito’s Pizza Works is putting pizzas out by delivery as fast as they

can. Pepito’s claims they can deliver pizzas within their delivery area in less than 29 minutes. You are given the following data from a sample.Sample size: 120 DeliveriesPopulation standard deviation: 1.4Sample mean: 28.3Formulate a hypothesis test to evaluate the claim.


• Sample Question on Hypothesis Testing– Pepito’s Pizza Work is putting pizzas out by delivery as fast as they can.

Pepito’s claims they can deliver pizzas within their delivery area in less than 29 minutes. You are given the following data from a sample.Sample size: 120 DeliveriesPopulation standard deviation: 1.4Sample mean: 28.3Formulate a hypothesis test to evaluate the claim.

– Answer: Ho: µ = 29, Ha : µ < 29– (In this case, the claim was Ha)– Remember Ho always contains equality (It will either be =, ≤ or ≥)– Ha will be either ≠, < or >


• Confidence interval Example– Acme computers needs to find a new vendor for

their hard drives. They are considering using Howie’s Hard Drives as a vendor. Acme’s requirement is that 95% of the hard drives last 24000 hours ± 2000 hours. The following data is from an independent source who evaluated Howie’s. Should Acme buy from Howie’s? Explain your answer. (Follows on next page)


• Mean = 24500• Sample Standard Deviation 2250• Min 21402• Max 29463• Margin of Error 4500

• Answer Follows


• No, Acme shouldn’t buy from Howie’s looking at their requirements (24000 – 2000, 24000 + 2000) which is (22000, 26,000). Based on the results given, Howie’s would yield a tolerance of (24500 – 2*2250, 24500+2*2250) which is (20000, 29000). This does not meet Acme’s requirement. You could also see this by looking at the margin of error.


• Example on Pivot/Contingency Tables• The table below gives the number of cars of

various colors and the state tag on the car for a parking lot in a mall close to DC.

VA MD DCOther State Total

Blue 4 8 9 3 24Black 9 7 11 8 35White 12 14 21 15 62Other 37 10 29 35 111Total 62 39 70 61 232


• Based on the table, find the probability that a car is from VA or MD.

• Based on the table, given that a car is from DC, find the probability it is black.




Find the probability that a car is from VA or MD. Add 62 + 39 to getSo the answer would be 101/232 or

it’s decimal form.

62 + 39 = 101




Given that a car is from DC, find the probability it is black.Given that it is from DC means we are onlydealing with the 70 cars from DC.There are 11 of those that are black, so theprobability is 11/70


• Normal Distribution Example– The number of students who use the dining hall at

an urban college on a given day is normally distributed with a mean of 1578 students and a standard deviation of 274 students.


• I’m suggesting Excel to work these types of problems even if you are given partial Minitab results.

• Go tohttp://highered.mcgraw-hill.com/sites/0070620164/student_view0/excel_templates.htmlAnd download the template titled Normal

Distribution.

http://highered.mcgraw-hill.com/sites/0070620164/student_view0/excel_templates.html

http://highered.mcgraw-hill.com/sites/0070620164/student_view0/excel_templates.html


• It will look something like this when you open it


• Before doing anything else, click the “Review” tab at the top of Excel (Between Data and View), then click “Unprotect Sheet”.

After clicking “Unprotect Sheet” it will say “Protect Sheet.” Leave it that way and save to your computer. You now have a cool Normal Distribution Calculator.


• Back to our problem….• Questions• What is the probability that less than 1400 students

will use the dining hall?• What is the probability that more than 1700 will use

the dining hall?• What is the probability that between 1400 and 1600

students will use the dining hall?• Get that Normal Distribution Excel Calculator ready

and be amazed!


Just enter your meanand standard deviation.

Don’t do anything else yet.


Clear out the otherGreen cells by using

the backspace key, then“Enter”. It should looklike the one on the left.

NEVER ENTER ANYTHINGinto the white cells. They

give you the results.


This gives you area to the left based on your mean and standard deviation.

This gives you area to the right based on your mean and standard deviation.

This gives you area between two values based on your mean and standard deviation.


I entered 1400 in the green cell which gives me the probability of less than 1400 students using the dining hall. The answer is 0.2580

I entered 1700 in the green cell which gives me the probability of more than 1700 students using the dining hall. The answer is 0.3281

I entered 1400 in the left cell and 1600 in the right green cell which gives me the probability of between 1400 and 1600 students using the dining hall. The answer is 0.2740


• Another Confidence Interval Example– I randomly sampled 18 engineers where I work and asked them

how many projects they have worked on in the last five years. The sample mean was 21, with a standard deviation of 5. What is the mean number of projects of all engineers at my research center? Why? What is the 95% confidence interval for the population mean? You are given the information below from Minitab.

One-Sample T N Mean StDev SE Mean 95% CI18 21.00 5.00 1.18 (18.51, 23.49)


• Another Confidence Interval Example– I randomly sampled 18 engineers where I work and asked them

how many projects they have worked on in the last five years. The sample mean was 21, with a standard deviation of 5. What is the mean number of projects of all engineers at my research center? Why? What is the 95% confidence interval for the population mean? You are given the information below from Minitab.

Answer:21 projects would be the best estimate for the mean. I would

expect 95% of the population mean to fall between 18.51 and 23.49 projects. The t is used because of the sample size.


• Regression Example– I did an analysis to determine if the number of

hours studied for a final exam related to the Final Exam grade for students. On the sheets that follow you will see what my Minitab results were.

General Regression Analysis: Final Grade versus Hours of Study

Regression Equation

Final Grade = 34.2845 + 1.45508 Hours of Study

Coefficients

Term Coef SE Coef T PConstant 34.2845 3.38091 10.1406 0.000Hours of Study 1.4551 0.11118 13.0872 0.000

Summary of Model

S = 6.87303 R-Sq = 88.62% R-Sq(adj) = 88.10%PRESS = 1426.21 R-Sq(pred) = 84.38%

Analysis of Variance

Source DF Seq SS Adj SS Adj MS F PRegression 1 8090.71 8090.71 8090.71 171.274 0.000000 Hours of Study 1 8090.71 8090.71 8090.71 171.274 0.000000Error 22 1039.25 1039.25 47.24 Lack-of-Fit 18 936.58 936.58 52.03 2.027 0.259215 Pure Error 4 102.67 102.67 25.67Total 23 9129.96

Fits and Diagnostics for Unusual Observations

FinalObs Grade Fit SE Fit Residual St Resid 24 19 37.1947 3.17993 -18.1947 -2.98609 R

R denotes an observation with a large standardized residual.


Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 48.8353 2.41382 (43.8294, 53.8413) (33.7280, 63.9426)

Values of Predictors for New Observations

Hours ofNew Obs Study 1 10

GM 533 Final Exam PrepI did an analysis to determine if the number of hours studied for a final exam related to the Final

Exam grade for students. On the sheets that follow you will see what my Minitab results were.Answer the following questions.

Determine the regression equation.

What conclusions are possible using the meaning of bo (intercept) and b1 (regression coefficient) in this problem?

What does the coefficient of determination (r-squared) mean?

Calculate the coefficient of correlation and explain what it means.

Does this data provide significant evidence (a=0.05) that the final exam grade is associated with the hours studied? Find the p-value and interpret.

Determine the predicted grade for someone who spends 10 hours studying for the final exam.

What is the 95% confidence interval for the score for spending 10 hours studying on the test? What conclusion is possible using this interval?

GM 533 Final Exam PrepI did an analysis to determine if the number of hours studied for a final

exam related to the Final Exam grade for students. On the sheets that follow you will see what my Minitab results were.

Answer the following questions.

Determine the regression equation. y= 34.2845 + 1.45508x

What conclusions are possible using the meaning of bo (intercept) and b1 (regression coefficient) in this problem? For each hour of study the final grade is increased by about 1.5 points (1.45508). bo represents the y intercept or 34.2845 in our case. It is the score that a student could expect to get without studying.

What does the coefficient of determination (r-squared) mean? The .886 means that 88.6 percent of the variability of the final grade can be explained by the number of study hours. The other 11.4% would be due to something else or be unexplained.


Calculate the coefficient of correlation and explain what it means. Square Root of (0.886) is 0.942 which is r, the correlation coefficient. With a value this close to one, we could say there is strong positive correlation.

Does this data provide significant evidence (a=0.05) that the final exam grade is associated with the hours studied? Find the p-value and interpret. Yes, the p value was 0. If it were above 0.05, I would have said “no.”

Determine the predicted grade for someone who spends 10 hours studying for the final exam. 48.8353

What is the 95% confidence interval for the score for spending 10 hours studying on the test? What conclusion is possible using this interval? (43.8294, 53.8413) We would be 95% confident that if someone studied for 10 hours they would score on average between those two values.


• Multiple Regression– Be able to identify the multiple regression output

from a Minitab analysis• Equation• F value• p value• Confidence intervals• (Basically reading from Minitab output)


• There is no live lecture this week since the University is transitioning to a new live lecture format.– www.facebook.com/statcave

– Good Luck!

http://www.facebook.com/statcave

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