Question

A major metropolitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers.

Answer 1

Answer: $(0.36845,\ 0.43155)$

Step-by-step explanation:

Let p be the population proportion of  readers who would like more coverage of local news.

As per given , we have

sample size :=1600

sample proportion : $\hat{p}=40\%=0.40$

The confidence interval for population proportion is given by :-

$\hat{p}\pm z^* \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

From z-table ,

For 99% confidence interval the z-value is 2.576.

Substitute all the values in the above formula , we get

The 99% confidence interval for the proportion of readers who would like more coverage of local news as

$0.40\pm(2.576) \sqrt{\dfrac{0.40(1-0.40)}{1600}}$

$0.40\pm(2.576) \sqrt{\dfrac{0.24}{1600}}$

$0.40\pm(2.576) \sqrt{0.00015}$

$0.40\pm(2.576) (0.0122474487139)$

$0.40\pm0.03155$

$(0.40-0.03155,\ 0.40+0.03155)=(0.36845,\ 0.43155)$

Hence, the 99% confidence interval for the proportion of readers who would like more coverage of local news is $(0.36845,\ 0.43155)$ .