### Hypothesis Testing: America's Cup Yacht Race AnalysisIn the America's Cup, two teams, Team Oracle U.S.A. and Land Rover BAR, compete by racing their yachts over a course multiple times. We're analyzing sample times in minutes for each yacht, assuming that the population standard deviations are equal.#### ObjectiveDetermine if there is a statistically significant difference in their mean times at a 0.05 significance level.#### MethodologyApply the 6-step hypothesis test procedure, and express the decision rule in terms of:a) Critical value termsb) P-value terms#### Data TableThe table below shows the times recorded for each team:| America's Cup | | ||---------------------|------------|----------|| **Team Oracle** | **Land Rover BAR** || 13.1 | 11.9 || 13.1 | 11.5 || 13.2 | 10.0 || 6.4 | 12.3 || 14.8 | 10.2 || 15.7 | 10.4 || 15.1 | 10.6 || 14.3 | 12.3 || 12.4 | 14.2 || 12.6 | 13.3 || 9.8 | || 18.0 | |*You can copy and paste the numbers above directly into Microsoft Excel® for further analysis.*This data will be used to perform t-tests to compare the mean times of the two teams and determine if there's a significant difference.

Solution: Given information: n1=12 Sample size of team oracle n2= 10 Sample size of land rover bar x1= 13.2083 Sample mean of team oracle x2= 11.67 Sample mean of land rover bar s12= 8.6808 Sample variance of team oracle s22= 1.9601 Sample variance of land rover bar α= 0.05 Level of significance To test the hypothesis Ho:μ1=μ2 VS Ha:μ1≠μ2 Test statistic t=x1-x2(n1-1)×s12+(n2-1)×s22n1+n2-2×1n1+1n2t=13.2083-11.6711×8.6808+9×1.960112+10-2×112+110t= 13.2083-11.6795.4888+17.640922-2×(0.083333+0.1)t=13.2083-11.67113.129720×0.183333t=13.2083-11.675.656485×0.183333t=13.2083-11.671.03702036t=1.53831.018342= 1.510593t= 1.5106 Test statistic t= 1.5106 To find degrees of freedom Degree of freedom =df= n1+n2-2df= 12+10-2df= 20 a) Decision using critical value method The critical values at α=0.05 are -tα/2,df= -t0.025,20=-2.086 From t table and tα/2,df= t0.025,20=2.086 From t table The critical values are -2.086 and 2.086 Rejection region of the test: Reject Ho if t> 2.086 or Reject Ho if t< -2.086 Here t= 1.5106 < 2.086 Fail to reject Ho Conclusion: There is insufficient evidence to conclude that there is significant difference in mean times. b) Decision using P value method: To find p value p value = 2×P(t> 1.5106)p value = 0.146529 From Excel p value = 0.1465 The p value for the test is 0.1465 Decision: p value > α=0.05 Fail to reject Ho Conclusion: There is insufficient evidence to conclude that there is significant difference in mean times.