To determine the required sample size for estimating the true mean with a specified level of confidence, we can use the formula for sample size in the context of confidence intervals:
\[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \]
where:
- \( n \) = required sample size
- \( Z \) = Z-value corresponding to the desired confidence level
- \( \sigma \) = population standard deviation
- \( E \) = margin of error (the maximum amount by which we want our estimate to differ from the true mean)
Step 1: Identify the parameters
- Population variance \( \sigma^2 = 900 \), hence the population standard deviation \( \sigma = \sqrt{900} = 30 \).
- Margin of error \( E = 6 \) (points within which we want to estimate the true mean).
- For a 99% confidence level, the corresponding Z-value can be found using a Z-table or standard normal distribution. The Z-score for a 99% confidence level is approximately 2.576.
Step 2: Substitute the values into the formula
Now we can substitute these values into the sample size formula:
\[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] \[ n = \left( \frac{2.576 \cdot 30}{6} \right)^2 \]
Step 3: Calculate
Calculating the numerator:
\[ Z \cdot \sigma = 2.576 \cdot 30 = 77.28 \]
Now calculating \( \frac{Z \cdot \sigma}{E} \):
\[ \frac{77.28}{6} = 12.88 \]
Now squaring this result to find \( n \):
\[ n = (12.88)^2 \approx 165.71 \]
Since the sample size must be a whole number, we round up to ensure we have enough observations:
\[ n \approx 166 \]
Conclusion
Thus, the required sample size to estimate the true mean score within 6 points with 99% confidence is 166.
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