Question

The GPA of accounting students in a university is known to be normally distributed. A random sample of 31 accounting students results in a mean of 3.14 and a standard deviation of 0.15. Construct the 99% confidence interval for the mean GPA of all accounting students at this university.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 3.14

Population standard deviation =    = 0.15
Sample size = n =31

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 * ( 0.15/ 31)

= 0.069

At 99% confidence interval is,

- E < < + E

3.14-0.069 < < 3.14+0.069

3.071< < 3.209