Question

In the year 2004, a survey was undertaken to find whether citizens of a certain political party supported their party's candidate for governor. In a sample of 300 citizens of this party, 95% of them expressed support for their party's candidate. A similar survey was conducted four years later and showed that 91% of a sample of 350 citizens of this party expressed support for their party's candidate. Construct a 95% confidence interval for the difference in population proportions of citizens of this party who supported their gubernatorial candidate in 2004 and citizens of this party who supported their gubernatorial candidate four years later. Assume that random samples are obtained and the samples are independent. (Round your answers to three decimal places.) z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

Answer 1

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:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha /2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

The information provided is as follows:

$\hat p_{1}=0.95\\\hat p_{2}=0.91\\n_{1}=300\\n_{2}=350$

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:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha /2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

     $=(0.95-0.91)\pm 1.96\times\sqrt{\frac{0.95(1-0.95)}{300}+\frac{0.91(1-0.91)}{350}}\\\\=0.04\pm 0.0388\\\\=(0.0012, 0.0788)\\\\\approx (0.001, 0.079)$

Thus, the 95% confidence interval for the difference in population proportions is (0.001, 0.079).