Question

In a random sample of25 ​people, the mean commute time to work was 32.8 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 μ. What is the margin of error of μ​? Interpret the results.

Answer 1

Solution :

Given that,

= 32.8

s =7.3

n =25

Degrees of freedom = df = n - 1 = 25- 1 =24

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t _/2,df = t_0.01,24 = 2.492 ( using student t table)

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

= 2.492* ( 7.3/ 25)

E=3.6383

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

- E < < + E

32.8-3.6383   < < 32.8+ 3.6383

29.1617 < < 36.4383

( 29.1617, 36.4383)