Question

Arandom sample of 100 students had a mean grade point average (GPA) of 3.2 with a standard error of 0.02. Given below are the 90%, 95%, 98% and 99% confidence intervals (not necessarily in that order) for the mean GPA for all students. Which interval would represent the 95% confidence interval? (3.15, 3.25) (3.17.3.23) (3.16, 3.24) (3.14, 3.26)

Answer 1

Solution :

t_{$\alpha$ /2,df} = 1.984

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 1.984 * 0.02

Margin of error = E = 0.04

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

3.2 - 0.04 < $\mu$ < 3.2 + 0.04

3.16 < $\mu$ < 3.24

(3.16 , 3.24)