Question

From a sample of 23 graduate students, the mean number of months of work experience prior to entering an MBA program was 33.24. the national standard deviation is known to be 19 months. what is a 99% confidence interval for the population mean?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 33.24

sample standard deviation = s = 19

sample size = n = 23

Degrees of freedom = df = n - 1 = 22

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,24 = 2.819

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

= 2.819 * (19/ 23)

= 11.168

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

- E < < + E

33.24 - 11.168 < < 33.24 + 11.168

22.072 < < 44.408

(22.072 , 44.408)