Question

Assume that a sample is used to estimate a population mean. Find the 98% confidence interval for a sample of size 71 with a mean of 22.8 and a standard deviation of 13.3. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

98% C.I. =

Answer 1

CI = [19.1,26.5]

Explanation:

Ok, so I'm going to assume this is a Gaussian distribution where the standard dev. is known, and n is large (>30) we're going to use the Z-distribution formula.

The setup looks like this:

`CI = [mean - (z_(\alpha /2)*(standard dev.))/(√(n) ), mean + (z_(\alpha/2)*(standard dev.))/(√(n))]`

mean = 22.8

standard dev = 13.3

alpha = 1-0.98 = 0.02

z_(alpha/2) (from z table) = 2.3263

n = 71

CI = [19.1,26.5]