Question

A random sample of 40 students taken from a university showed that their mean GPA is 2.94 and the standard deviation of their GPAs is .30. Construct a 99% confidence interval for the mean GPA of all students at this university

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 2.94

Population standard deviation =    = 0.30
Sample size = n =40

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* ( 0.30/ 40)

= 0.1222

At 99% confidence interval estimate of the population mean is,

- E < < + E

2.94-0.1222 < <2.94+ 0.1222

2.8178< < 3.0622

(2.8178,3.0622)