Question

EXPA media research reported that adult women watch TV an average of 5.3 hrs/day, compared to an average of 4.6 hrs/day for adult men. Assume that those results are found from a sample of 120 men and 160 women and that two population standard deviations are minimum 0.48 and 0.60 respectively.

a) test the belief that women on average spend more time watching TV than men. Use a 0.05 significance level.

H0:

H1:

Test statistic:

b) construct a 90% confidence interval for the difference of the means

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ1=μ2

​ Ha:μ1>μ2
​ This corresponds to a right-tailed test, for which a z-test for two population means, with known population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is zc=1.64.

The rejection region for this right-tailed test is R={z:z>1.64}
(3) Test Statistics :The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z=10.84>zc​=1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.00, and since p=0.000<0.05, it is concluded that the null hypothesis is rejected.

b) The 90% confidence interval for the difference of the means is 0.594<μ1​−μ2​<0.806.