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Out of 100 people sampled, 42 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places.

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Answer:

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

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{(\pi(1-\pi))/(n)}

In which

z is the z-score that has a p-value of
1 - (\alpha)/(2).

Out of 100 people sampled, 42 had kids.

This means that
n = 100, \pi = (42)/(100) = 0.42

99% confidence level

So
\alpha = 0.01, z is the value of Z that has a p-value of
1 - (0.01)/(2) = 0.995, so
Z = 2.575.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.42 - 2.575\sqrt{(0.42*0.58)/(100)} = 0.293

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.42 + 2.575\sqrt{(0.42*0.58)/(100)} = 0.547

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

User Nijogeorgep
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