Question

A data analyst wants to construct a confidence interval to estimate the true mean dollar amount spent per visit by a “typical” Toys-R-Us customer. Based on extensive industry data, the analyst knows that the population variance is 28.9444. The analyst draws a simple random sample of 144 recent customer cash register receipts and records the dollar amount spent on each receipt. There is a 5% chance (i.e., the probability is .05) that the resulting sample mean will have a margin of error of . (report your answer to 2 decimal places, using conventional rounding rules)

Answer 1

Population varaince = 28.9444

std.dev = sqrt(28.9444) = 5.38

z value at 95% = 1.96

margin of error = z *(s/sqrt(n))
= 1.96 * (5.38/sqrt(144))
= 0.89