If the variance of a national accounting examination is 900 how large a sample is needed to estimate the true mean score within 5 points with 95 percent confidence?

To determine the required sample size, you can use the formula for sample size ( n = \left( \frac{Z \cdot \sigma}{E} \right)^2 ), where ( Z ) is the Z-score for the desired confidence level (1.96 for 95% confidence), ( \sigma ) is the standard deviation (the square root of variance), and ( E ) is the margin of error. Given a variance of 900, the standard deviation ( \sigma ) is 30. Plugging in the values: ( n = \left( \frac{1.96 \cdot 30}{5} \right)^2 \approx 216.64 ). Therefore, you would need a sample size of at least 217.