In telephone interviews with a random sample of 1,100 online shoppers, shoppers said they expected to spend an average of $976 for online purchases this year.Using a sample standard deviation of $180, you build a 95% confidence interval estimate of the average spending for the population of online shoppers represented by the sample and determine the margin of error for your 95% interval.Now suppose you want to reduce the margin of error in your interval by 20% without sacrificing the 95% confidence level.How big a sample would be required?
A)1096
B)1718
C)1461
D)1528
E)1637

Question

Quizplus · Accepted Answer

The margin of error (MOE) in a confidence interval is inversely proportional to the square root of the sample size (n). To reduce the MOE by 20%, we need to increase the sample size so that the square root of the new sample size is 1/0.8 (or 1.25) times the square root of the original sample size. If the original sample size is 1100, the new sample size (n') can be calculated as follows: n' = 1100 * (1.25)^2 = 1100 * 1.5625 = 1718.75. Since sample size must be a whole number, we round to the nearest whole number, which gives us 1718.

In Telephone Interviews with a Random Sample of 1,100 Online