Question

A recent survey of math students asked about their overall grades. 22 students were surveyed, and it was found that the average GPA of the 22 sampled students was a 3.2, with a standard deviation of 0.9 points. Calculate a 90% confidence interval for the true mean GPA of math students. Round your answer to the nearest hundredth, and choose the most correct option below: (2.88,3.52), (2.87,3.53), (.28,.94), (2.40,4.00) or none of the above...

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 3.2

sample standard deviation = s = 0.9

sample size = n = 22

Degrees of freedom = df = n - 1 = 22 - 1 = 21

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,21 = 1.721

Margin of error = E = t_/2,df * (s /n)

= 1.721 * ( 0.9 / 22)

= 0.33

The 90% confidence interval estimate of the population mean is,

- E < < + E

3.2 - 0.33 < < 3.2 + 0.33

2.87 < < 3.53

(2.87 , 3.53)