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

Holding everything constant, what would be the upper limit of the 90% confidence interval for the population mean? (round to two digits)

	3.48
	3.49
	3.51
	3.52

Answer 1

 

sample size = n = 3600

Degrees of freedom = df = n - 1 = 3599

tα /2,df = 1.645

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

= 1.645 * (0.5 / √ 3600)

Margin of error = E = 0.01

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

- E < μ < + E

3.5 - 0.01 < μ < 3.5 + 0.01

3.49 < μ < 3.51

Upper limit = 3.51