Question

A simple random sample of size n=18 is drawn from a population that is normally distributed. The sample mean is found to be x overbar =50 and the sample standard deviation is found to be s=14. Construct a 90​% confidence interval about the population mean.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 50

sample standard deviation = s = 14

sample size = n = 18

Degrees of freedom = df = n - 1 = 18-1 =17

t _/2,df = 1.74

Margin of error = E = t_/2,df * (s /n)

= 1.74* (14 / 18)

Margin of error = E = 5.74

The 90% confidence interval estimate of the population mean is,

- E < < + E

50 - 5.74 < < 50 + 5.74

44.26 < < 55.74

(44.26,55.74)