A random sample of 1001 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 1001 adults surveyed, 511 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (bace 2). can be assumed to be less than or equal to $5 \%$ of the population size. (Round to three decimal places as needed.) (c) 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. (Type integers or decimals rounded to three decimal places as needed. Use ascending order.) A. We are $95 \%$ confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between and B. There is a $\%$ probability the proportion of adults in the country who believe that televisions are a luxury they could do without is between and (d) Is it possible that a supermajority (more than $60 \%$ ) of adults in the country believe that television is a luxury they could do without? Is it likely? It is possible, but not likely that a supermajority of adults in the country believe that television is a luxury they could do without because the $95 \%$ confidence interval does not contain 0.6 . (Type an integer or a decimal. Do not round.) (e) Use the results of part (c) to construct a $95 \%$ confidence interval for the population proportion of adults in the country who believe that televisions are a necessity. Lower bound: $\square$, Upper bound: (Round to three decimal places as needed.)
Solution
Step 1 ：Given that the sample size (n) is 1001 and the number of adults who believe that televisions are a luxury they could do without is 511, we can calculate the sample proportion (p̂) as \(\frac{511}{1001} = 0.510\).
Step 2 ：We can calculate the standard error (se) using the formula \(se = \sqrt{\frac{p̂(1-p̂)}{n}} = \sqrt{\frac{0.510(1-0.510)}{1001}} = 0.016\).
Step 3 ：For a 95% confidence interval, the Z-score (Z) is 1.96.
Step 4 ：We can then calculate the lower and upper bounds of the confidence interval using the formula \(p̂ ± Z*se\). The lower bound is \(0.510 - 1.96*0.016 = 0.480\) and the upper bound is \(0.510 + 1.96*0.016 = 0.542\).
Step 5 ：\(\boxed{\text{Final Answer: We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.480 and 0.542.}}\)
