Answer to Question #108687 in Statistics and Probability for steve

Question

AOL Time Warner Inc.'s CNN has been the longtime ratings leader of cable television news. Nielsen Media Research indicated that the mean CNN viewing audience was 600,000 viewers per day during 2002 (The Wall Street Journal, March 10, 2003). Assume that for a sample of 40 days during the first half of 2003, the daily audience was 612,000 viewers with a sample standard deviation of 65,000 viewers.

Answer 1

Given:

$n=40$

$\bar{X}=612000$

$s=65000$

(a) what are the hypotheses if CNN management would like information on any change in the CNN viewing audience?

$H_0:\mu=600000$

$H_1: \mu\not=600000$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(b) what is the p-value?

The t-statistic is computed as follows:

$$t={\bar{X}-\mu \over s/\sqrt{n}}={612000-600000\over 62000/\sqrt{40}}\approx1.1676$$

The p-value is $p=0.250057$

(c) Select your own level of significance. What is your conclusion?

$\alpha=0.05$

Using the P-value approach: The p-value is $p=0.250057,$ and since $p=0.250057>0.05=\alpha,$

it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population $\mu$ is different than 600000, at the 0.05 significance level.

d. Provide a recommendation to CNN management in this application

The larger sample is not necessary.