A 96% confidence interval for the proportion of all students who live off campus is (0.31833,0.38067). What was the sample size of the survey

Question

A  96% confidence interval for the proportion of all students who live off campus is  (0.31833,0.38067). What was the sample size of the survey

Bytelearn · Accepted Answer

Calculate Margin of Error: To calculate the sample size (n) for a confidence interval for a proportion, we can use the formula for the margin of error (E) in a proportion confidence interval, which is given by: $$ E = z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} $$ where: - $ E $ is the margin of error - $ z $ is the z-score corresponding to the confidence level - $ \hat{p} $ is the point estimate of the population proportion - $ n $ is the sample size First, we need to find the margin of error (E) from the confidence interval provided. The confidence interval is given as (0.31833, 0.38067). The margin of error is half the width of the confidence interval. $$ E = \frac{0.38067 - 0.31833}{2} $$ $$ E = \frac{0.06234}{2} $$ $$ E = 0.03117 $$Find Z-Score: Next, we need to find the z-score corresponding to a 96% confidence level. The z-score for a 96% confidence level is not commonly known, so we would typically use a z-table or statistical software to find it. However, for a 95% confidence level, the z-score is approximately 1.96, and for a 99% confidence level, it is approximately 2.576. Since 96% is closer to 95%, we can estimate the z-score to be slightly higher than 1.96. Let's use 2.05 as an approximation for the z-score for a 96% confidence level.Calculate Point Estimate: Now we need to calculate the point estimate of the population proportion ($ \hat{p} $). The point estimate is the midpoint of the confidence interval. $$ \hat{p} = \frac{0.31833 + 0.38067}{2} $$ $$ \hat{p} = \frac{0.699}{2} $$ $$ \hat{p} = 0.3495 $$Solve for Sample Size: With the margin of error (E), the z-score (z), and the point estimate of the population proportion ($ \hat{p} $), we can now solve for the sample size (n). Rearrange the margin of error formula to solve for n: $$ n = \frac{z^2 \times \hat{p}(1 - \hat{p})}{E^2} $$ Plug in the values we have: $$ n = \frac{2.05^2 \times 0.3495(1 - 0.3495)}{0.03117^2} $$ $$ n = \frac{4.2025 \times 0.3495 \times 0.6505}{0.0009715889} $$ $$ n = \frac{4.2025 \times 0.22732275}{0.0009715889} $$ $$ n = \frac{0.955073037625}{0.0009715889} $$ $$ n = 982.8 $$ Since the sample size must be a whole number, we round up to the nearest whole number. $$ n = 983 $$

A $96\%$ confidence interval for the proportion of all students who live off campus is $(0.31833,0.38067)$ . What was the sample size of the survey

Full solution

More problems from Interpret confidence intervals for population means

Related topics

A 96%96\%96% confidence interval for the proportion of all students who live off campus is (0.31833,0.38067)(0.31833,0.38067)(0.31833,0.38067). What was the sample size of the survey

A $96\%$ confidence interval for the proportion of all students who live off campus is $(0.31833,0.38067)$ . What was the sample size of the survey