Question

In a random sample of 28 ​people, the mean commute time to work was 34.4 minutes and the standard deviation was 7.3 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 98​% confidence interval for the population mean mu. What is the margin of error of mu​?

The confidence interval for the population mean mu is (__, __)

Answer 1

Solution :

Given that,

= 34.4

s = 7.3

n = 28

Degrees of freedom = df = n - 1 = 28 - 1 = 27

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,27 = 2.473

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

= 2.473 * (7.3 / 28)

= 3.41

Margin of error = 3.41

The 98% confidence interval estimate of the population mean is,

- E < < + E

34.4 - 3.41 < < 34.4 + 3.41

30.99 < < 37.81

(30.99, 37.81)