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.
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