Question

analve the mework on confidence intervals, you will use JMP to analyze the Deflategate data set. This time you will As a reminder, each student is randomly assigned one of 10 versions of this dataset. This time, your version is 8 document on ednvas under adtiPaemnt Butiing what is in this dataset before you begin. You can find this answere numbers exactly as they appear in JMP. WebAssign is set to use three decimal place precision to determine if the denticaladestion 1 from the last homework, however your data version may be different PSI Prega PSI Halftim Team me 1 P 12.5 11.54 the following 2 P Firerons thees 12.5 10.88 3 P 12.5 11.12 a. Mean halftime PSI for Colts 12.628 b. Standard deviation of mean halftime PSI for Colts 141 10.58 4 P 12.5 c. Standard error of mean halftime PSI for Colts = 070 d. 95% CI for uColts PSI Halftime e. Mean halftime PSI for Patriots 11.017 5 P 12.5 11.1 to 6 P 12.5 11.68 for atriots 413 q. Standard error of mean halftime PSI for Patriots 414 7 P 12.5 11.78 h. 95% CI for uPatriots PSI Halftime to 8 P 12.5 10.99 Pregae Wed ike tok0tpalls was.13 ony "Analvze Distribution". Next, click tho ro dron dau me ng 9 P 12.5 11.04 10 P 10.5 12.5 a. Perform a two sided, one sample test of hypothesis pair: Ho: HColts PSI Halftime HA: HColts PSI Halftime 3 11 P 12.5 10.97 Analvze / 12 C 13 12.76 altimeand Select TestMeansOr you n dBenanesquestons ineart dethen click che red arow bY 13 C 13 12.68 Test statistic: -value: 14 C 13 12.64 b. What would the p-value for the previous test be if the alternative had been HA: HColts PSI Halftime < 137 15 C 13 12.43 fymer toides henumbersHed to compute the test statistic youreported above. Calculate the values for the below. Numerator of test statistic: t statistic: Denominator f the hy "agree" withthe s57 reported in art 5% CL did not contain zero, 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 ot agree and the hypothesis test produced a p-value less than 0.05. The 959% CL contained the The 95% CI did not contain the pregame PSI, and the hypothesis test produced a p-value less 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 greater than 0.05. PSI between the two teams. 3. As with the previous homework, we want compare the average decrease Greate a new column for decrease in PSdfined as PSPregame minys SHalftime se Apalyze FitY by Xelect mend on the next coes (see the netes u anored in the previous homewdrk are the resul nrse rp down ework a. What is the value of the test statistic testing Ho: uPatriots HColts 07 b. What is the value of the numerator of this test statistic? c. What is the value of the denominator of this test statistic? PWna e significant. he uandoncude that the dilierence in mean PSI decrease is statistically significant'ail to reject the null and conclude that the difference in mean PSI decrease is statistically p>0.05, so reject the null and conclude that the difference in mean PSI decrease is not statistically 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 not statistically significant ps0.05, so reject the null and conclude that the difference in mean PSI decrease is not statistically PSO.05, So fail to reject the null and conclude that the difference in mean PSI decrease is not statistically o reject the null and conclude that the difference in mean PSI decrease is statistically significant. p>0.05,

Answer 1

1.

(d)

α=0.05

Critical Z​=Z_1−α/2​=1.96

X CI 0.141 12.628 1.96 0.141 12.628 1.96 (12.49, 12.765)

(e)

Mean = (11.54 + 10.88 + 11.12 + 10.58 + 11.1 + 11.68 + 11.78 + 10.99 + 11.04 + 10.5 + 10.97)/11
= 122.18/11
Mean = 11.107

(f)

Standard Deviation σ = √(1/11 - 1) x ((11.54 - 11.1073)² + (10.88 - 11.1073)² + (11.12 - 11.1073)² + (10.58 - 11.1073)² + (11.1 - 11.1073)² + (11.68 - 11.1073)² + (11.78 - 11.1073)² + (10.99 - 11.1073)² + (11.04 - 11.1073)² + (10.5 - 11.1073)² + (10.97 - 11.1073)²)
= √(1/10) x ((0.4327)² + (-0.2273)² + (0.012699999999999)² + (-0.5273)² + (-0.0073000000000008)² + (0.5727)² + (0.6727)² + (-0.1173)² + (-0.067300000000001)² + (-0.6073)² + (-0.1373)²)
= √(0.1) x ((0.1872) + (0.0517) + (0.0002) + (0.278) + (0.0001) + (0.328) + (0.4525) + (0.0138) + (0.0045) + (0.3688) + (0.0189))
= √(0.1) x 1.7036
= √0.1704
Standard Deviation σ = 0.413

(g)

Standard Error = σ√n
= 0.4127√11
= 0.41273.3166
Standard Error = 0.124

(h)

α=0.05

Critical Z​=Z_1−α/2​=1.96

X — 2с X =, X + ze X CI = 0.413 \1.107+1.96- ,0.413 11.107 1.96- V11 (10.863, 11.351)