Question

Consider the following data from two independent samples. Construct a 99% confidence interval to estimate the difference in population proportions. x1 = 90 n1 100 x2 80 P2=100 The 99% confidence interval is ) (Round to four decimal places as needed.)

Answer 1

$\hat{p}_{2}=\frac{80}{100}=0.8$

Z critical value for 99% confidence level is 2.5758 (by using Z table) or =NORMSINV(1-(0.01/2))

Confidence interval formula

$\left ( \hat{p}_{1}-\hat{p}_{2} \right )\pm Z*\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}$

$=>(0.90-0.80)\pm 2.5758*\sqrt{\frac{0.90(1-0.90)}{100}+\frac{0.80(1-0.80)}{100}}$

=>(-0.0288,0.2288)

Therefore, 99% confidence interval is (-0.0288, 0.2288)