Question

A 96% confidence interval for GPA values is (2.67, 3.15). What is the estimated mean GPA of the population, and what is the margin of error?

Answer 1

Solution :

Confidence Interval is Mean $\pm$ ME

For 96% confidence Interval, the value of Z is 2.054

Upper Limit is 3.15

Lower Limit is 2.67

$UL = \mu + z\times \frac{\sigma }{\sqrt{n}} = 3.15$ ........................(1)

$LL = \mu - z\times \frac{\sigma }{\sqrt{n}} = 2.67$......................(2)

Adding (1) and (2), we get

$2\mu = 2.67 +3.15 = 5.82$

$\therefore \mu= \frac{5.82}{2} =2.91$

b) Margin of error is

$\Rightarrow \mu + z\times \frac{\sigma }{\sqrt{n}} = 3.15$

$\Rightarrow 2.91+ z\times \frac{\sigma }{\sqrt{n}} = 3.15$

$\Rightarrow z\times \frac{\sigma }{\sqrt{n}} = 3.15 -2.91 =0.24$

ME = 0.24