A publisher is interested in estimating the proportion of textbooks that students resell at the end of the semester. He is interested in making this estimate using a confidence level of 95 percent and a margin of error of ±0.02. Based upon his prior experience, he believes that π is somewhere around 0.60. Given this information, the required sample size is over 2,300 students.

Question

Quizplus · Accepted Answer

To estimate the required sample size for a proportion with a specific margin of error (E) and confidence level, the formula is $n = \frac{Z^2 \times \hat{p} \times (1 - \hat{p})}{E^2}$, where $\hat{p}$ is the estimated proportion (0.60 in this case), and $Z$ is the Z-score corresponding to the confidence level (approximately 1.96 for 95%). Plugging in the values: $n = \frac{(1.96)^2 \times 0.60 \times (1 - 0.60)}{(0.02)^2}$, which calculates to approximately 2,304, thus requiring a sample size of over 2,300 students.

A Publisher Is Interested in Estimating the Proportion of Textbooks