Question

The GPA of accounting students in a university is known to be normally distributed. A random sample of 32 accounting students results in a mean of 2.64 and a standard deviation of 0.15. Construct the 95% confidence interval for the mean GPA of all accounting students at this university.

Answer 1

= Solution :

Given that,

Point estimate = sample mean = = 2.64

Population standard deviation =    = 0.15

Sample size n =32

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 * ( 0.15 /  32 )

=0.05
At 95% confidence interval
is,

- E < < + E

2.64 - 0.5 <   < 2.64 + 0.5

2.14 <   < 3.14