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.
Determine the assumptions needed to construct a confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without. Check all that apply (only choose three).
$n> 30$
SRS
$n< 0.05 N$
population normal
$n p(1-p)> 10$
data normal without outliers

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

Step 2 ：First, we calculate the sample proportion (p̂) which is the number of successes (people who think TV is a luxury) divided by the total number of trials (total people surveyed).

Step 3 ：Using the given data, we have n = 1003 and x = 521. Therefore, the sample proportion p̂ = x/n = 521/1003 = 0.519.

Step 4 ：Next, we calculate the standard error of the proportion. The formula for the standard error of a proportion is \(\sqrt{\frac{p̂(1 - p̂)}{n}}\), where n is the sample size.

Step 5 ：Substituting the values, we get the standard error se = \(\sqrt{\frac{0.519(1 - 0.519)}{1003}}\) = 0.0158.

Step 6 ：We then use the Z-score for a 95% confidence interval, which is approximately 1.96, to calculate the margin of error. The margin of error is the Z-score times the standard error.

Step 7 ：Substituting the values, we get the margin of error me = 1.96 * 0.0158 = 0.0309.

Step 8 ：We can now calculate the confidence interval by subtracting and adding the margin of error from/to the sample proportion.

Step 9 ：Subtracting the margin of error from the sample proportion, we get the lower limit of the confidence interval ci_lower = 0.519 - 0.0309 = 0.488.

Step 10 ：Adding the margin of error to the sample proportion, we get the upper limit of the confidence interval ci_upper = 0.519 + 0.0309 = 0.550.

Step 11 ：Therefore, the 95% confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without is approximately (0.488, 0.550). This means that we are 95% confident that the true population proportion lies within this interval.

Step 12 ：For the assumptions needed to construct a confidence interval for the population proportion, we need to check the following: n>30, SRS (Simple Random Sample), and n p(1-p)>10.

Step 13 ：Given that our sample size is 1003 which is greater than 30, the assumption n>30 is satisfied.

Step 14 ：We assume that the sample of 1003 adult Americans was randomly selected, so the SRS assumption is satisfied.

Step 15 ：We also assume that the sample size is less than 5% of the population size, so the assumption n<0.05 N is satisfied.

Step 16 ：Finally, we check the assumption n p(1-p)>10. Substituting the values, we get n*p(1-p) = 1003*0.519*(1-0.519) = 249.8 which is greater than 10. So, this assumption is also satisfied.

Step 17 ：Thus, the assumptions that apply are n>30, SRS, and n p(1-p)>10.

Step 18 ：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 \(\boxed{(0.488, 0.550)}\). The assumptions needed to construct this confidence interval are n>30, SRS, and n p(1-p)>10.