Question

Construct and interpret a 95​% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without. Select the correct choice below and fill in any answer boxes within your choice.

Answer 1

Answer:

The 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without is (0.500, 0.562).

Step-by-step explanation:

The complete question is:

A random sample of 1014 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1014 adults surveyed, 538 indicated that televisions are a luxury they could do without.Construct and interpret a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without.

Solution:

In a sample of 1014 adults, 538 indicated that televisions are a luxury they could do without.

Compute the sample proportion of adults who indicated that televisions are a luxury they could do without as follows:

$\hat p=\frac{538}{1014}=0.531$

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The critical value of z for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*use a z-table.

Compute the 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

     $=0.531\pm 1.96\times \sqrt{\frac{0.531(1-0.531}{1014}}}\\=0.531\pm 0.0307\\=(0.5003, 0.5617)\\\approx (0.500, 0.562)$

Thus, the 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without is (0.500, 0.562).