Question

use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n

Answer 1

Using the z-distribution, the confidence interval for the population proportion is given by:

(0.2153, 0.4633).

What is the missing information?

This problem is incomplete, but researching it on a search engine, we have that the information is given as follows:

n = 56, x = 19, 95% confidence.

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 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so the critical value is z = 1.96.

The sample size and the estimate are given as follows:

$n = 56, \pi = \frac{19}{56} = 0.3393$

Hence the bounds of the interval are given as follows:

• $\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3393 - 1.96\sqrt{\frac{0.3393(0.6607)}{56}} = 0.2153$

• $\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3393 + 1.96\sqrt{\frac{0.3393(0.6607)}{56}} = 0.4633$

The confidence interval for the population proportion is given by:

(0.2153, 0.4633).

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

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