Question

In a random sample of 23 people, the mean commute time to work was 30.3 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 98​% confidence interval for the population mean .What is the margin of error of f μ​? Interpret the results.

(a)The confidence interval for the population mean is _, _   ​(Round to one decimal place as​ needed.)

(b)The margin of error is _ ​(Round to two decimal places as​ needed.)

(c) Interpret the results. Choose the correct answer below:

1. It can be said that 95​% is between the bounds of the confidence interval.

2.With 95​% confidence, it can be said that the population mean is between the bounds of the confidence interval.

3.With 95​% confidence, it can be said that is between the bounds of the confidence interval.

4.If a large sample is taken approximately 95​% of them are between the bounds of the confidence interval.

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = = 30.3

sample standard deviation = s = 7.1

sample size = n = 23

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

At 98% confidence level

= 1 - 98%

=1 - 0.98 =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 * (7.1 / 23)

Margin of error = E = 3.71

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

  ± E

= 30.3 + 3.71

= ( 26.59, 34.01)  

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

= 2.508 * (7.1 / 23)

Margin of error = E = 3.71

c) With 95​% confidence, it can be said that the population mean is between the bounds of the confidence interval.