16. The "spring-like" effect in a golf club could be determined by measuring the coefficient of restitution (the ratio of the outbound velocity to the inbound velocity of a golf ball fired at the club head). Drivers are randomly selected from two club makers and the coefficient of restitution is measured. The data are as follows, and are summarized in the Minitab output below: Club 1: 0.8906, 0.8104, 0.8234, 0.8198, 0.8235, 0.8562, 0.8123, 0.7976, 0.8184, 0.8265, 0.7173, 0.7871, Club 2: 0.8305, 0.7905, 0.8352, 0.8380, 0.8145, 0.8465, 0.8244, 0.8014, 0.8309, 0.8405 Test and CI for Two Variances: C1 vs C2 Method: Null hypothesis Alternative hypothesis (1) (2) (1) 1 (2) 1 a 0.05 Significance level F method was used. This method is accurate for normal data only. Statistics C2 N St Dev Variance 1 12 0.041 2 10 0.018 95% CI for StDevs 0.002 (0.029, 0.069) 0.000 (0.012, 0.033) Ratio of standard deviations = 2.278 Ratio of variances 5.188 95% Confidence Intervals CI for StDev CI for Variance Method Ratio Ratio F (1.152, 4.315) (1.326, 18.616) Tests Test 5.19 P-Value 0.020 Method DF1 DF2 Statistic F 11 9 Descriptive Statistics: C1 Variable C2 N N C1 1 12 0 2 10 Mean 0.8153 0 0.82524 SE Mean 0.0118 St Dev 0.0409 Minimum 01 Median 0.7173 0.8008 0.8191 0.00567 0.01794 0.79050 0.81123 0.83070 Variable C2 C1 1 2 Q3 0.8258 0.83863 Maximum 0.8906 0.84650 Is an assumption of equal variances justified? (a) Yes, since the p-value of .02 is less than .05. (b) No, since the p-value of .02 is less than .05. (c) No, since .0409 is not equal to .01794 (d) No, since .002 is not equal to .000 (e) Yes, because .0409 is close to .01794 17. (Continuation of 16.) Test the hypothesis that the clubs from brand 2 have a greater mean coefficient of restitution than the clubs from brand 1. Use a = .05. I (a) Do not reject Ho since -0.76 is not less than -2.086 (b) Do not reject Ho since -0.71 is not less than -1.725 (c) Do not reject Ho since -0.76 is not less than -1.725 (d) Do not reject Ho since -0.76 is not less than -1.753 (e) Do not reject Ho since -0.71 is not less than -1.753 18. (Continuation of 16.) Find the p-value for the test in 17.. (a) .2<P <.5 (b) .25 < P <.4 (c) .2236 (d) .4472 (e) 0.1 < P<0.25 19. (Continuation of 16.) Find a 95% two-sided confidence interval on the mean difference in coefficient of restitution between the two brands of golf clubs. (a) (-.1339, .1141) (b) (-.0379, .0179) (c) (-.0499, .0301) (d) (-.0586, .0388) (e) (-.1020, .0822)