Question

Assume that a sample is used to estimate a population mean μ μ . Find the 99.5% confidence interval for a sample of size 36 with a mean of 41.5 and a standard deviation of 15.2. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. 99.5% C.I. = The answer should be obtained without any preliminary rounding.

Answer 1

df = n - 1 = 36 - 1 = 35

t critical value at 99.5% confidence = 2.996

99.5% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

41.5 - 2.996 * 15.2 / sqrt(36) < < 41.5 + 2.996 * 15.2 / sqrt(36)

33.910 < < 49.090

99.5% CI is ( 33.910 , 49.090 )