Question

30 students were randomly selected from a large group of students taking a certain test. The mean score for the students in the sample was 86.
Assume that σ = 11.1. Construct a 99% confidence interval for the mean score, μ, of all students

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 86

Population standard deviation =    = 11.1

Sample size = n =30

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

Margin of error = E = Z/2* ( /n)

= 2.576* ( 11.1/ 30)

= 5.22

At 99% confidence interval is,

- E < < + E

86-5.22 < < 86+5.22

80.78< < 91.22

(80.78, 91.22)