Question

If the mean GPA among students is 3.25 with a standard deviation of 0.75, and we select a random sample of 300 people, at what value for the sample mean would be greater than exactly 95% of possible sample means?

Answer 1

Solution :

Given that,

$\bar x$ = 3.25

s =0.75

n = 100

Degrees of freedom = df = n - 1 = 100 - 1 = 99

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,99 =1.984

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

=1.984 * (0.75 / $\sqrt$ 100)

= 0.158

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

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

3.25 - 0.158 < $\mu$ < 3.25 + 0.158

3.092 < $\mu$ < 3.409

(3.092, 3.409 )