Question

In a sample of n=3,600 students, the average GPA was x⎯=3.5 with a sample variance of s2=0.25. Calculate the 98% confidence interval for the population GPA μ

What would be the upper limit of the 90% confidence interval if the sample size had only been 16 (round to two digits)?

	3.29
	3.39
	3.71
	3.82

Answer 1

 

1)

Point estimate = sample mean = = 3.5

sample standard deviation = s = 0.5

sample size = n = 3600

Degrees of freedom = df = n - 1 = 3599

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,3599 = 2.327

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

= 2.327 * (0.5 / √ 3600)

= 0.019

The 98% confidence interval estimate of the population mean is,

- E < < + E

3.5 - 0.019 < < 3.5 + 0.019

3.48 < < 3.52

2)

sample size = n = 16

Degrees of freedom = df = n - 1 = 15

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,15 = 1.753

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

= 1.753* ( 0.5/ √ 16)

= 0.21

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

- E < < + E

3.5 - 0.21 < < 3.5 + 0.21

3.29 < < 3.71

upper limit of the 90% confidence interval is 3.71