A large Midwestern university is interested in estimating the mean time that students spend at the student recreation center per week. A previous study indicated that the standard deviation in time is about 40 minutes per week. If the officials wish to estimate the mean time within ± 10 minutes with a 90 percent confidence, what should the sample size be?
A) 44
B) 62
C) 302
D) Can't be determined without knowing how many students there are at the university.

Question

Quizplus · Accepted Answer

To estimate the mean time within ±10 minutes with 90% confidence and a standard deviation of 40 minutes, the sample size can be calculated using the formula for the sample size of a mean: $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $Z$ is the Z-score corresponding to the confidence level (for 90% confidence, $Z = 1.645$), $\sigma$ is the standard deviation (40 minutes), and $E$ is the margin of error (10 minutes). Plugging in the values: $n = \left(\frac{1.645 \cdot 40}{10}\right)^2 = \left(6.58\right)^2 \approx 43.3$, which rounds up to 44. The total number of students at the university is not needed to calculate the sample size for estimating a mean with a specified confidence and margin of error.

A Large Midwestern University Is Interested in Estimating the Mean