Question

For a random sample of 18 recent business school graduates beginning their first job, the mean starting salary was found to be $39,500, and the sample standard deviation was $9,500. Assuming the population is normally distributed, find the margin of error of a 95% confidence interval for the population mean.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 39500

sample standard deviation = s = 9500

sample size = n = 18

Degrees of freedom = df = n - 1 = 18 - 1 = 17

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,17 = 2.110

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

= 2.110 * (9500 / 18)

= 4724.652

Margin of error is 4724.652

The 95% confidence interval estimate of the population mean is,

- E < < + E

39500 - 4724.652 < < 39500 + 4724.652

34775.348 < < 44224.652