Question

Construct a 99​% confidence interval to estimate the population proportion with a sample proportion equal to 0.36 and a sample size equal to 100 .

Answer 1

Using the z-distribution, the 99​% confidence interval to estimate the population proportion is: (0.2364, 0.4836).

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem, we have a 99% confidence level, hence$\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so the critical value is z = 2.575.

The estimate and the sample size are given by:

$\pi = 0.36, n = 100$.

Then the bounds of the interval are:

• $\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 - 2.575\sqrt{\frac{0.36(0.64)}{100}} = 0.2364$

• $\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 + 2.575\sqrt{\frac{0.36(0.64)}{100}} = 0.4836$

The 99​% confidence interval to estimate the population proportion is: (0.2364, 0.4836).

More can be learned about the z-distribution at brainly.com/question/25890103

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