Question

A random sample of 80 people was selected, and 22 of the selected people indicated that it would be a good idea to eliminate the penny from circulation. What is the 99 percent confidence interval constructed from the sample proportion

Answer 1

Answer:

The 99% confidence interval for the population proportion is (0.146, 0.404).

Explanation:

The confidence interval for true proportion is:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }$

Given:

X = number of people who thought it would be a good idea to eliminate the penny from circulation = 22.

n = sample size = 80.

α = 1 - 0.99 = 0.01

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{22}{80}= 0.275$

The critical value of z for 99% confidence level is:

$z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58$

Compute the 99% confidence interval for the population proportion as follows:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }\\=0.275\pm2.58\sqrt{\frac{0.275(1-0.275)}{80} }\\=0.275\pm0.129\\=(0.146,0.404)$

Thus, the 99% confidence interval for the population proportion is (0.146, 0.404).