Respuesta :
Answer:
The 95% confidence interval for the difference in population proportions is (0.001, 0.079).
Step-by-step explanation:
The (1 - α)% confidence interval for the difference in population proportions is:
[tex]CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha /2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}[/tex]
The information provided is as follows:
[tex]\hat p_{1}=0.95\\\hat p_{2}=0.91\\n_{1}=300\\n_{2}=350\\[/tex]
The critical value of z for 95% confidence level is 1.96.
Compute the 95% confidence interval for the difference in population proportions as follows:
[tex]CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha /2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}[/tex]
[tex]=(0.95-0.91)\pm 1.96\times\sqrt{\frac{0.95(1-0.95)}{300}+\frac{0.91(1-0.91)}{350}}\\\\=0.04\pm 0.0388\\\\=(0.0012, 0.0788)\\\\\approx (0.001, 0.079)[/tex]
Thus, the 95% confidence interval for the difference in population proportions is (0.001, 0.079).
