Question

A university of dean of students wishes to estimate the average number of hours students spend doing homework per week. The standard deviation from a previous study is 6.2 hours. How large a sample must be selected if he wants to be 99% confident of finding whether the true mean differs from the sample mean by 1.5 hours?

Answer 1

To determine the required sample size for estimating the average number of hours students spend doing homework per week, we can use the formula for the sample size \( n \) when estimating a population mean. The formula is:

\[
n = \left( \frac{Z \times \sigma}{E} \right)^2
\]

Where:
- \( Z \) is the z-score corresponding to the desired confidence level
- \( \sigma \) is the standard deviation of the population
- \( E \) is the margin of error (the maximum difference between the sample mean and the population mean that we are willing to accept)

Given the values:
- Confidence level = 99% → The z-score (from z-tables) is approximately 2.576
- \( \sigma = 6.2 \) hours
- \( E = 1.5 \) hours

Plugging in these values:

\[
n = \left( \frac{2.576 \times 6.2}{1.5} \right)^2
\]

Calculating the numerator:

\[
2.576 \times 6.2 = 15.98
\]

Now calculating the term inside the parentheses:

\[
\frac{15.98}{1.5} \approx 10.6533
\]

Now squaring this term:

\[
10.6533^2 \approx 113.52
\]

Finally, since the sample size \( n \) must be a whole number, we round up to get the required sample size:

\[
n \approx 114
\]

Thus, the university must select a sample of at least **114 students** to be 99% confident that the true mean differs from the sample mean by no more than 1.5 hours.