The computer program JMP will need to be used to compute the following data set, the "Wells Report", outlining the finding from an investigation into whether there was strong evidence that the P football team intentionally let air out of footballs during a playoff game.

Each student is randomly assigned one of 10 versions of this dataset. Your version is3.

Team PSI PreGame PSI Halftime

1 P 12.5 11.54

2 P 12.5 10.78

3 P 12.5 11.06

4 P 12.5 10.71

5 P 12.5 11.04

6 P 12.5 11.54

7 P 12.5 11.83

8 P 12.5 11.14

9 P 12.5 11.02

10 P 12.5 10.41

11 P 12.5 11.03

12 C 13 12.52

13 C 13 12.74

14 C 13 12.5

15 C 13 12.47

Enter the numbers exactly as they appear in JMP. WebAssign is set to use three decimal place precision to determine if the answer is correct. 1. Report the values of the following: a. Mean halftime PSI for Colts = b. Standard deviation of mean halftime PSI for Colts = c. Standard error of mean halftime PSI for Colts = d. 95% CI for_{Colts PSI Halftime}= to e. Mean halftime PSI for Patriots = f. Standard deviation of mean halftime PSI for Patriots = g. Standard error of mean halftime PSI for Patriots = h. 95% CI for_{Patriots PSI Halftime}= to 2. Suppose we'd like to know if the mean PSI for the Colts' footballs changed significantly from pregame to halftime. The pregame PSI for all Colts footballs was 13. a. Get JMP to perform a test against: H₀:_{Colts PSI Halftime}= 13 To do this in JMP, start from the Analyze / Distribution output you used to answer the questions in part 1, then click the red arrow by "PSI_Halftime" and select "Test Means". Test statistic: p-value: b. JMP provides all the numbers used to compute the test statistic you reported above. Calculate the values for the numerator and denominator of this test statistic separately, and report them below. Numerator of test statistic: Denominator of test statistic: c. In what way to the results of the hypothesis test in part a. "agree" with the 95% CI forreported in part 1? (You have two attempts) The 95% CI contained the pregame PSI, and the hypothesis test produced a p-value greater than 0.05.The 95% CI did not contain the pregame PSI, and the hypothesis test produced a p-value greater than 0.05. The 95% CI contained the pregame PSI, and the hypothesis test produced a p-value less than 0.05.The 95% CI did not contain zero, and the hypothesis test produced a p-value less than 0.05.Actually, the results did not agree with each other.The 95% CI did not contain the pregame PSI, and the hypothesis test produced a p-value less than 0.05.The 95% CI contained zero, and the hypothesis test produced a p-value less than 0.05. 3. Now we want to compare the average decrease in PSI between the two teams. Create a new column for decrease in PSI, defined as PSI_Pregame minus PSI_Halftime. Use "Analyze / Fit Y by X", select the appropriate variables (see the notes if you have trouble with this), hit "OK", then select "t Test" from the drop down menu on the next screen. a. What is the value of the test statistic testing H₀:_Patriots-_Colts= 0? b. What is the value of the numerator of this test statistic? c. What is the value of the denominator of this test statistic? d. What is the statistical decision resulting from this hypothesis test? (You have two attempts)

p>0.05, so reject the null and conclude that the difference in mean PSI decrease is not statistically significant.

p>0.05, so reject the null and conclude that the difference in mean PSI decrease is statistically significant. p<0.05, so fail to reject the null and conclude that the difference in mean PSI decrease is statistically significant.

p<0.05, so fail to reject the null and conclude that the difference in mean PSI decrease is not statistically significant.

p>0.05, so fail to reject the null and conclude that the difference in mean PSI decrease is not statistically significant.

p>0.05, so fail to reject the null and conclude that the difference in mean PSI decrease is statistically significant.

p<0.05, so reject the null and conclude that the difference in mean PSI decrease is not statistically significant.

p<0.05, so reject the null and conclude that the difference in mean PSI decrease is statistically significant. 4. Some people criticized the Wells report on the grounds that the data were biased by the fact that the footballs had been moved indoors at halftime, and the Patriots' footballs were measured first, giving the Colts' footballs more time to warm up and gain air pressure naturally. Can we use the hypothesis test from part 3. to assess this claim? (You have two attempts)

No, hypothesis tests take account of bias, but not sampling variability.

Yes, hypothesis tests take account of both sampling variability and bias.No, hypothesis tests take account of sampling variability or bias.

Yes, hypothesis tests produce numbers, and once you have a number you are basically holding truth in your hands.

No, hypothesis tests take account of sampling variability, but not bias.

Yes, hypothesis tests take account of bias, but not sampling variability, and in this case we're only interested in accounting for bias.