A research scientist wants to know how many times per hour a certain strand of bacteria reproduces. He wants to construct the \( 85 \% \) confidence interval with a maximum error of 0.15 reproductions per hour. Assuming that the mean is 6.9 reproductions and the variance is known to be 4.84 , what is the minimum sample size required for the estimate? Round your answer up to the next integer.

To calculate the minimum sample size required for an estimate of the mean with a specified maximum error, we can use the formula: \[ n = \left( \frac{Z \times \sigma}{E} \right)^2 \] Where: - \( n \) is the required sample size, - \( Z \) is the z-score corresponding to the desired confidence level, - \( \sigma \) is the standard deviation, and - \( E \) is the maximum allowable error. For an 85% confidence level, the z-score (which can be found in a z-table or using statistical software) is approximately 1.440. Given: - Mean (\( \mu \)) = 6.9 reproducions per hour (not needed for sample size calculation), - Variance (\( \sigma^2 \)) = 4.84, hence \( \sigma = \sqrt{4.84} \approx 2.2 \), - Maximum error, \( E = 0.15 \). Now substituting the values into the formula: \[ n = \left( \frac{1.440 \times 2.2}{0.15} \right)^2 \] Calculating the numerator: \[ 1.440 \times 2.2 = 3.168 \] Now calculating \( n \): \[ n = \left( \frac{3.168}{0.15} \right)^2 = \left( 21.12 \right)^2 \approx 446.58 \] Since we round up to the next integer, the minimum sample size required is: \[ \boxed{447} \]