Question

Assume that a sample is used to estimate a population proportion p. Find the 95% confidence for a sample of size 246 with 52% successes. Enter your answer as an open -interval using decimals

Answer 1

Using the z-distribution, the 95% confidence interval for the proportion is given as follows:

(0.4576, 0.5824).

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 other parameters for the interval are given as follows:

$n = 246, \pi = 0.52$.

The lower and upper bound of the interval, respectively, are given by:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 - 1.96\sqrt{\frac{0.52(0.48)}{246}} = 0.4576$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 + 1.96\sqrt{\frac{0.52(0.48)}{246}} = 0.5824$

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

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