A random sample of 1003 adult Americans was asked, "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1003 adults surveyed, 521 indicated that televisions are a luxury they could do without. Construct and interpret a $95 \%$ confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without.
Please state your conclusion below.

Step 1 ：Given that a random sample of 1003 adult Americans was surveyed, and 521 of them indicated that televisions are a luxury they could do without. We are asked to construct and interpret a 95% confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without.

Step 2 ：The formula for a confidence interval for a population proportion is given by: \(\hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(\hat{p}\) is the sample proportion, \(z\) is the z-score corresponding to the desired level of confidence, and \(n\) is the sample size.

Step 3 ：In this case, \(\hat{p} = \frac{521}{1003}\), \(z\) is approximately 1.96 for a 95% confidence interval, and \(n = 1003\).

Step 4 ：Substituting these values into the formula, we get the confidence interval as \((0.489, 0.550)\).

Step 5 ：\(\boxed{\text{Final Answer: The 95% confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without is approximately (0.489, 0.550). This means we are 95% confident that the true proportion of adult Americans who believe that televisions are a luxury they could do without is between 48.9% and 55.0%.}}\)