Question

A administrator wants to know what is the average starting salary, μ, for students graduating from her college. She is able to obtain data for 100 randomly selected students. For these 100 students, the average is $70,000, and the SD is $10,000. What is a 99% confidence interval for μ?

Answer 1

Solution :

Given that,

= 70000

s = 10000

n = 100

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

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,99 =2.626

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

= 2.626 * (10000 / 100)

= 2626.40

Margin of error = 2626.40

The 99% confidence interval estimate of the population mean is,

- E < < + E

70000 - 2626.40< < 70000 + 2626.40

67373.59 < < 72626.40