You are given the sample mean and the sample standard deviat

You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If convenient use technology to construct the confidence intervals. A random sample of 40 gas grills has a mean price of $645.90 and a standard deviation of $59.90. The 90% confidence interval is ( ). (Round to one decimal place as needed.) The 95% confidence interval is ( ). (Round to one decimal place as needed.) Which interval is wider? Choose the correct answer below. The 90% confidence interval The 95% confidence interval

Solution

xbar +- z * s / sqrt(n) where z has alpha/2 = .025. z-score = 1.645 632.7 +- 1.645 * 59.4 / sqrt(44) = (617.97, 647.43) where z has alpha/2 = .025. z-score = 1.96 632.7 +- 1.96 * 59.4 / sqrt(44) = (615.15, 650.25) the 95% CI is wider.