Answer to Question #346753 in Statistics and Probability for kath

Question

Compute the population proportion interval estimate given n, p, and the confidence level

a. Confidence Level= 99%, p=0.4, n=40

b. Confidence Level= 90%, p=0.15, n=55

Answer 1

a. The critical value for $\alpha = 0.01$ is $z_c = z_{1-\alpha/2} = 2.5758.$

The corresponding confidence interval is computed as shown below:

$CI(Proportion)=(\hat{p}-z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},$ $\hat{p}+z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$ $=(0.4-2.5758\times \sqrt{\dfrac{0.4(1-0.4)}{40}},$ $0.4+2.5758\times \sqrt{\dfrac{0.4(1-0.4)}{40}})$ $=(0.20048, 0.59952)$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.20048 < p < 0.59952,$ which indicates that we are 99% confident that the true population proportion $p$ is contained by the interval $(0.20048, 0.59952).$

b. The critical value for $\alpha = 0.10$ is $z_c = z_{1-\alpha/2} =1.6449.$

The corresponding confidence interval is computed as shown below:

$CI(Proportion)=(\hat{p}-z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},$ $\hat{p}+z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$ $=(0.15-1.6449\times \sqrt{\dfrac{0.15(1-0.15)}{55}},$ $0.15+1.6449\times \sqrt{\dfrac{0.15(1-0.15)}{55}})$ $=(0.06975, 0.23025)$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.06975 < p < 0.23025,$ which indicates that we are 90% confident that the true population proportion $p$ is contained by the interval $(0.06975, 0.23025).$