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Gekata [30.6K]
3 years ago
10

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

Mathematics
1 answer:
Tpy6a [65]3 years ago
5 0

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) .

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