Question

In a random sample of 21 people, the mean commute time to work was 34.1 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean μ. What is the margin of error of μ​? Interpret the results.

The confidence interval for the population mean μ is __,__

​(Round to one decimal place as​ needed.)

The margin of error of μ is __,__

​(Round to one decimal place as​ needed.)

Interpret the results.

A. It can be said that 99​% of people have a commute time between the bounds of the confidence interval.

B. With 99​% confidence, it can be said that the commute time is between the bounds of the confidence interval.

C. If a large sample of people are taken approximately 99​% of them will have commute times between the bounds of the confidence interval.

D. With 99​% confidence, it can be said that the population mean commute time is between the bounds of the confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 34.1

sample standard deviation = s = 7.2

sample size = n = 21

Degrees of freedom = df = n - 1 = 21 - 1 = 20

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,20 = 2.845

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

= 2.845 * ( 7.2 / 21)

Margin of error = E = 4.5

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

  ± E  

= 34.1  ± 4.5

= ( 29.6, 38.6 )

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

= 2.845 * ( 7.2 / 21)

Margin of error = E = 4.5

D. With 99​% confidence, it can be said that the population mean commute time is between the bounds of the confidence interval.