Question

A social scientist claims that the average adult watches less than 26 hours of television per week. She collects data on 50 individuals’ television viewing habits and finds that the mean number watching television was 22.4 hours. If the population is normally distributed with a standard deviation of 8 hours, can we conclude at the 1% statistical significance level that she is right? Use the critical value and the test statistic to test the null hypothesis. Then calculate the p-value to determine the level of statistical significance at which we could reject the null hypothesis.

Answer 1

The provided sample mean is 22.4 and the known population standard deviation is 8, and the sample size is n = 50

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ=26

Ha: μ<26

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the critical value for a left-tailed test is z_c = -2.33

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z = -3.182 < z_c = -2.33, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0007, and since p = 0.0007 < 0.01, 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 the population mean μ is less than 26, at the 0.01 significance level.