Question

A group of 76 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. Construct and interpret a 95% confidence interval for the mean score, μ, of all students taking the test. Construct and interpret a 95% confidence interval for the mean score, μ, of all students taking the test.

Answer 1

Solution :

Given that,

sample size = n = 76

Degrees of freedom = df = n - 1 = 75

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,75 = 1.992

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

= 1.992 * (5.2 / 76)

= 1.188

The 95% confidence interval estimate of the population mean is,

- E < < + E

29.5 - 1.188 < < 29.5 + 1.188

28.312 < < 30.688

( 28.312 , 30.688 )

Therefore, there is 95% confidence that the  mean score, μ, of all students taking the test. is between 28.312 and 30.688.