Question

1. The following data set contains miles per gallon (MPG) for U.S. cars (sample 1) and for Japanese cars (sample 2). Find the sample mean and sample standard deviation of each sample. Then perform a t-test to identify whether there is a statistically significant difference in mean MPG between the two samples. Use a significance level of 0.05. Using the result comment whether the country of manufacture is a significant effect for the response (MPG) U.S Cars Japanese Cars 18 24 15 27 18 27 16 25 17 31 15 35 14 24 14 19 14 28 15 23 15 27 14 20 15 22. 18 22 20

Answer 1

us cars

Count, N: 15 Sum, Ex: 236 Mean, 15.733333333333 Variance, s2: 4.9238095238095 Steps 1 N S = (x; – T), N-1 i=1 52-5(xi - )2 11 N-1 (18 - 15.733333333333)2 + ... + (22 - 15.733333333333)? 15-1 68.933333333333 14 = 4.9238095238095 S= 14.9238095238095 = 2.2189658681038

Japanese cars

Count, N 15 Sum, Ex: 370 Mean, X 24.666666666667 Variance, s2: 21.809523809524 Steps N 1 S = (Xi – T)? N-1 i=1 S2 (X; - x)2 N-1 (24 - 24.666666666667)2 + ... + (20 - 24.666666666667)2 15 - 1 305.33333333333 14 = 21.809523809524 S= N21.809523809524 = 4.6700667885507 II

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 12 Ha: Mi + M2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 20.015. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: Hence, it is found that the critical value for this two-tailed test is to = 2.086, for a = 0.05 and df = 20.015. The rejection region for this two-tailed test is R= {t: 1t| > 2.086).

(3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: X1 - X2 t= V + 112 15.7333 - 24.6667 -6.692 2.21892 15 +4672 15 (4) Decision about the null hypothesis Since it is observed that |t| = 6.692 > te = 2.086, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0, and since p=0 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean ji is different than u2, at the 0.05 significance level.