Question

A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is the 90% confidence interval for the mean score, μ, of all students taking the test. Show Work

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 29.5

Population standard deviation =    = 5.2
Sample size = n =59

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2    * ( /n)

= 1.645 * ( 5.2 /  59 )

= 1.11
At 90% confidence interval
is,

- E < < + E

29.5 - 1.11 <   < 29.5 + 1.11

28.39 <   < 30.61