Question

For a random sample of 17 recent business school graduates beginning their first job, the mean starting salary was found to be $39,500, and the sample standard deviation was $8,500. Assuming the population is normally distributed, find the margin of error of a 90% confidence interval for the population mean.

Answer 1

Solution :

Given that,

= $39,500

s = $8,500

n = 17

Degrees of freedom = df = n - 1 = 17 - 1 = 16

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,16 = 1.746

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

= 1.746 * (8,500 / 17)

= 3599.471

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

- E < < + E

39,500 - 3,599.471 < < 39,500 + 3599.471

35900.529 < < 43099.471

(35900.529, 43099.471)