Question

In a random sample of 22 ​people, the mean commute time to work was 34.5 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results

Answer 1

Given that,

= 34.5

s =7.2

n =  22 ​

Degrees of freedom = df = n - 1 = 22 ​- 1 = 21

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,21 = 1.721    ( using student t table)

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

= 1.721* ( 7.2/ 22) = 2.6418

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

- E < < + E

34.5 - 2.6418 < < 34.5+ 2.6418

31.8582 < < 37.1418

(31.8582 , 37.1418 )

Margin of error = E = 2.6418