Question

Suppose a simulation of 10 replications has an average number in queue of 15.5 with a 95% half-width 2.5. How many replications are needed if we want that estimate of the average to have a 95% confidence interval within 10% of the mean?

Answer 1

To determine the number of replications needed for the estimate of the average to have a 95% confidence interval within 10% of the mean, we can use the formula for half-width: Half-width = Z * (Standard Deviation / √n) Where: Z is the critical value from the standard normal distribution for a 95% confidence level (approximately 1.96). Standard Deviation is the standard deviation of the estimate (unknown in this case). n is the number of replications. We want the half-width to be within 10% of the mean, which means the half-width is 0.1 * Mean. Given the information in the question: Half-width = 2.5 (from 95% half-width). Mean = 15.5 (average number in queue). Z = 1.96 (for a 95% confidence level). Let's solve for n: 0.1 * Mean = Z * (Standard Deviation / √n) 0.1 * 15.5 = 1.96 * (Standard Deviation / √n) 1.55 = (1.96 * Standard Deviation) / √n √n = (1.96 * Standard Deviation) / 1.55 n = [(1.96 * Standard Deviation) / 1.55]^2 Since we don't have the exact value of the standard deviation, we cannot compute the exact number of replications needed without that information. The standard deviation of the estimate should be available from the simulation results or historical data to proceed with the calculation. Once you have the standard deviation, you can plug it into the equation above to find the required number of replications for the desired 95% confidence interval within 10% of the mean.

Answer 2