According to a report in USA Today, more and more parents are helping their young adult children buy homes. You would like to construct a 90% confidence interval of the proportion of all young adults in Louisville, Kentucky, who received help from their parents in buying a home. How large a sample should you take so that the margin of error is no more than 0.06?
A) n = 114
B) n = 188
C) n = 267
D) Cannot be determined.

Question

Quizplus · Accepted Answer

To determine the sample size needed for a confidence interval with a specific margin of error, you can use the formula for the sample size of a proportion, which is $n = \left(\frac{Z^2 \cdot p \cdot (1-p)}{E^2}\right)$, where $Z$ is the Z-score corresponding to the desired confidence level, $p$ is the estimated proportion (if unknown, 0.5 is used for maximum sample size), and $E$ is the desired margin of error. For a 90% confidence interval, the Z-score is approximately 1.645. Assuming $p = 0.5$ for maximum variability, and wanting $E = 0.06$, the calculation is $n = \left(\frac{1.645^2 \cdot 0.5 \cdot (1-0.5)}{0.06^2}\right)$, which simplifies to approximately 188. Therefore, a sample size of 188 is needed.

According to a Report in USA Today, More and More