Answer to Question #65423 in Statistics and Probability for Dorjee

Question

A sample of 25 items is selected from a very large shipment. It is found to have a mean weight of 310 gm and standard deviation equal to 9 gm. State and compute the 95% confidence limits for the population mean weight.

Answer 1

1 - alpha = 0.95 -> alpha = 0.05
1 - alpha/2 = 1 - 0.05/2 = 0.975
0.975 quantile z of normal distribution is equal to 1.96
n = 25, mean = 310, sd = 9
so the 95% confidence limits are [310 - 1.96*9 / sqrt(25), 310 + 1.96*9 / sqrt(25)] =
[306.472, 313.528]