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Natalija [7]
3 years ago
11

a major metroplitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. they aske

d wether the paper should increase its coverage of local news. forty percent of the sample wanted mre local news. WHat is the 99% confidence interval for the proportion of readers who would like more coverage of local news
Mathematics
1 answer:
Arisa [49]3 years ago
7 0

Answer:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 1600, \pi = 0.4

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.3685

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.4315

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

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