Question

1. Suppose your boss makes a statement that American children study more hours per week than Brazilian children. You think American children study less. You do some research and find that a sample of 41 American children study an average of 6.2 hours per week and a sample of 46 Brazilian children study an average of 8.5 hours per week. The standard deviation for the American children is 3.1 hours and the standard deviation for the Brazilian children is 2.4 hours. Be 91% confident in challenging your boss. Please set up the null and the alternative hypothesis, do the calc for this data and the z critical and decide to reject or not.

What if your boss says that Brazilian children have a larger variance in their study time than American.. Do the fcalc for this problem. Hint: ANOVA.

*** I am unsure if I have the correct answer for both of them since they are intertwined with each other.

Answer 1

(1) Null and Alternative Hypothesis:

H0: American children study is not more hours per week than Brazillan Children
H1: American children stud is more hours per week than Brazillan children

i.e.

The following null and alternative hypotheses need to be tested: Ho: Ha: 2 This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. Testing for Equality of Variances A F-test is used to test for the equality of variances. The following F-ratio is obtained: 3.12 1.668 2.42 0.588 and F The critical values are Fr 1.686, and since F = 1.668, then the null hypothesis _ of equal variances is not rejected. (2) Rejection Region Based on the information provided, the significance level is a 0.09, and the degrees of freedom are df 85. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: 1.352, for a 0.09 and df Hence, it is found that the critical value for this right-tailed test is te 85 The rejection region for this right-tailed test is R 1.352 {t: t

(3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: х, — Х, (n,-1)+(ng-1)s n,+n-2 6.2 8.5 -1.414 (41-1)3.12+(46-1)2.4 41+46-2 (4) Decision about the null hypothesis -1.414 te 1.352, it is then concluded that the null hypothesis is not Since it is observed that t = rejected. Using the P-value approach: The p-value is p 0.9195, and since p 0.9195 0.09, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is greater than , at the 0.09 significance level.

Thus we conclude that American children study is not more hours per week than Brazillan Children