Question

In a random sample of 28 ​people, the mean commute time to work was 30.1 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 95​% confidence interval for the population mean mu. What is the margin of error of mu​?

Interpret the results.

The confidence interval for the population mean mu is:

What is the margin of error of mu​?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 30.1

sample standard deviation = s = 7.2

sample size = n = 28

Degrees of freedom = df = n - 1 = 2 8 - 1 = 27

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,27 = 2.052

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

= 2.052 * (7.2 / 28)

= 2.8

Margin of error = E = 2.8

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

- E < < + E

30.1 - 2.8 < < 30.1 + 2.8

27.3 < < 32.9

(27.3 , 32.9)