Question

1. In a random sample of 23 people, the mean commute time to work was 32.2 minutes and the standard deviation was 7.2 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? Interpret the results. Round to one decimal place as needed.

Answer 1

Solution :

Given that,

= 32.2 mintues

s = 7.2 m=minutes

n = 23

Degrees of freedom = df = n - 1 = 23 - 1 = 22

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

  / 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,22 = 2.508

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

= 2.508 * (7.2 / 23)

= 3.1

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

- E < < + E

32.2 - 3.1 < < 32.2 + 3.1

29.1 < < 35.1

( 29.1, 35.1)