An insurance actuary claims that the mean footwell intrusions for small and midsize cars are equal. Crash tests at 40 mph were performed on 18 randomly selected small cars and 8 randomly selected midsize cars. The amount that the footwell intruded on the driver's legs was measured. The mean footwell intrusion for the small cars was 11.5 centimeters with a standard deviation of 4.29 centimeters. The mean footwell intrusion for the midsize cars was 11.1 centimeters with a standard deviation of 3.71 centimeters. At a= 0.20, answer parts (a) through (e). Assume the population variances are equal. If convenient, use technology to solve the problem.

We have this situation: Ho: P1 - P2 = 0, Ha: Pi – P2 # 0 a = 0.10 ni = 500 n2 = 523 What critical value forms the rejection region? O t = 1.645 O z = -1.645 O z = -1.645, z = 1.645 Ot= -1.645, t = 1.645

Use the F-Distribution Table to find the Right Tail Critical Value α=0.025,dfN=3,dfD=24.Fo=

Aviation and high-altitude physiology is a specialty in the study of medicine. Let x = partial pressure of oxygen in the alveoli (air cells in the lungs) when breathing naturally available air. Let y = partial pressure when breathing pure oxygen. The (x, y) data pairs correspond to elevations from 10,000 feet to 30,000 feet in 5000 foot intervals for a random sample of volunteers. Although the medical data were collected using airplanes, they apply equally well to Mt. Everest climbers (summit 29,028 feet). (units: mm Hg/10) (units: mm Hg/10) 3.3 x 6.7 Ly 42.2 4.9 4.2 2.1 31.7 26.2 16.2 13.9 (a) Verify that Ex = 21.2, Ey = 130.2, Ex? = 101.84, Ey? = 3927.82, Exy = 630.76, and r= 0.982. Ir = Ex 21.2 Ey 130.2 Ex2 101.84 Ey2 3927.82 Exy 630.76 r 0.982 (b) Use a 10% level of significance to test the claim that p > 0. (Use 2 decimal places.) t 9.01 critical t 1.64 Conclusion O Reject the null hypothesis, there is sufficient evidence thatp >0. O Reject the null hypothesis, there is…