Question

In a random sample of 17 ​people, the mean commute time to work was 30.1 minutes and the standard deviation was 7.3 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results.

Answer 1

Solution :

Given that,

= 30.1

s = 7.3

n = 17

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

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,17 = 2.898

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

= 2.898 * (7.3 / 17) = 5.13

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

- E < < + E

30.1 - 5.13 < < 30.1 + 5.13

24.97 < < 36.23

(24.97 , 36.23)