Question

The Fox TV network is considering replacing one of its prime-time crime investigation shows with a new family-oriented comedy show. Before a final decision is made, network executives commission a sample of 400 viewers. After viewing the comedy, 250 indicated they would watch the new show and suggested it replace the crime investigation show a. Estimate the value of the population proportion.b. Develop a 99 percent confidence interval for the population proportion.c. Interpret your findings.

Answer 1

Answer:

a. 0.625

b. $0.5625\leq p\leq 0.6875$

Explanation:

The estimate value of the population proportion p' is calculated using the data from the sample, so it is equal to:

$p'=\frac{250}{400} =0.625$

Because the sample have 400 viewers and 250 indicated that they would watch the new show and suggested.

It mean that 0.625 is an estimation of the population proportion.

On the other hand, the confidence interval for the population proportion p is calculated as:

$p'-z_{\alpha /2}\sqrt{\frac{p'(1-p')}{n} } \leq p\leq p'+z_{\alpha /2}\sqrt{\frac{p'(1-p')}{n} }$

Where $p'$ is the sample proportion, $1-\alpha$ is the confidence level, and n is the size of the sample.

Now, we need to replace p' by 0.625, n by 400 and $1-\alpha$ by 0.99.

If $1-\alpha$ is equal to 0.99, $\alpha$ is equal to 0.01 and, using the standard normal distribution table, $z_{\alpha /2}$ is equal to: $z_{\alpha /2}=z_{0.005}=2.58$

Then, the 99 percent confidence interval for the population proportion is calculated as:

$0.625-(2.58)\sqrt{\frac{0.625(1-0.625)}{400} } \leq p\leq 0.625+(2.58)\sqrt{\frac{0.625(1-0.625)}{400} }\\0.625-0.0625\leq p\leq 0.625+0.0625\\0.5625\leq p\leq 0.6875$

It means that with a 99% of confidence level, the value of population proportion of people that would watch the new show and suggested is between 0.5625 and 0.6875