Question

8. (4 pts) Researcher measured the walking speed of travelers in San Francisco International Airport and Cleveland Hopkins International Airport. The standard deviation speed of the 35 travelers who were departing was 53 feet per minute. The standard deviation speed of the 35 travelers who were arriving was 34 feet per minute. Assuming walking speed is normally distributed, does the evidence suggest the standard deviation walking speed is different between the two groups? Use the a = 0.05 level of significance. IE

Answer 1

that were were S, = 53 feet Solution so We have given X: The speed of the travelers who deporting (in minute ) Y! The speed of the travelers who arriving (in minute) .: The sample standard deviation of Ž for size 9,335 is minute per and is The sample standard deviation of Ÿ for size ng=35 5 = 34 feet per minute. Here have to test whether the standard deviahon walking speed is different between the the two That is, we have to test Ho population Standard deviation speed is same agaist groups ? ? for both groups HI: 0 € 2 و from the alternative hypothesis Hi, this test is two sided test for equality of standard deviation. Test statistic: Here have ust using F-test statistic as of F = Sospeso » F -clismibution with (n;!, na!) degrees freedom

Thus 2 FE Si? S² 53 342 - 2.4299 ie -7 Calculated F = 2.4299 0 Significance level: Q = 0.05 level we have using of signi fican here Thus, the catalane tabulared vaue for test statistic is fy., - F (0.025, 34, 34) D = 2 = 35 2=0.05 by using excel, F.INV. RT (0.025, 34, 34) It gives F (0.025, 34, 34) - 1-9811 11 Conclusion! Rejection Region ! The rejection region is given by R: { Reject > tabulared FC 14, Ho if calculared F Thus, foom eg? O and @ calculared' F > Flere , N-1) :: We have to reject Ho: 09-02 at d= 0.05 Significance level. Result: There is sufficient evidente to thest standard deviahon sugest walking speed between two groups me different. 2.42gg > 1.9811