Question

A recent broadcast of a television show had a 15 ?share, meaning that among 5000 monitored households with TV sets in? use,15?% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in? use, less than 20% were tuned into the program. Identify the null? hypothesis, alternative? hypothesis, test? statistic, conclusion about the null?hypothesis, and final conclusion that addresses the original claim. Use the normal distribution as an approximation of the binomial distribution.

Answer 1

As we are trying to test, whether less than 20% were tuned into the program, therefore the null and the alternative hypothesis here is computed as:

The test statistic here is computed as:

Therefore -8.8388 is the required test statistic value here.

As the z value here is very very low, therefore the p-value here is computed from the standard normal tables as:

p = P(Z < -8.8388) = 0

Therefore 0 is the required p-value here.

As the p-value here is very very low, the test is significant and we can reject the null hypothesis here and conclude that we ave sufficient evidence that  less than 20% were tuned into the program