Question

A recently televised broadcast of a popular television show had a 15 share, meaning that among 5000 monitored households with TV sets in use, 15% of them were tuned to the show. A 0.01 significance level is used to test an advertiser's claim that among the households with TV sets in use, less than 20% were tuned in to the show. Find the P-value.

Answer 1

The answer is 0.0001.

Step-by-step explanation:

I promise!

Answer 2

The p-value of the test is approximately 0.

Step-by-step explanation:

The hypothesis can be defined as:

H₀: The proportion of TV sets tuned in to the show is not less than 20%, p ≥ 0.20.

Hₐ: The proportion of TV sets tuned in to the show is less than 20%, p < 0.20.

Given:

`n=5000\\\hat p=0.15\\\alpha =0.01`

The test statistic is:

`z=\frac{\hat p-p}{\sqrt{(p(1-p))/(n)} }=\frac{0.15-0.20}{\sqrt{(0.20(1-0.20))/(5000) }}= -8.77`

The p-value of test is:

`p-value=P(Z<-8.77)\approx0`

*Use a standard normal table.

Thus, the p-value of the test is approximately 0.