Answer:
The 99% confidence interval for the population proportion is (0.146, 0.404).
Explanation:
The confidence interval for true proportion is:
[tex]CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
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:
[tex]\hat p=\frac{X}{n}=\frac{22}{80}= 0.275[/tex]
The critical value of z for 99% confidence level is:
[tex]z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58[/tex]
Compute the 99% confidence interval for the population proportion as follows:
[tex]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)[/tex]
Thus, the 99% confidence interval for the population proportion is (0.146, 0.404).
