Question

If the variance of a national accounting examination σ² is 900, how large a sample is needed to estimate the true mean score within 6 points with 99% confidence?

Answer 1

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

1. **Population variance \( \sigma^2 = 900 \)**, hence the population standard deviation \( \sigma = \sqrt{900} = 30 \).
2. **Margin of error \( E = 6 \)** (points within which we want to estimate the true mean).
3. 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**.