Question

In: Statistics and Probability

In a random sample of 29 people, the mean commute time to work was 32.5 minutes...

In a random sample of 29 people, the mean commute time to work was 32.5 minutes and the standard deviation was 7.3 minutes. Assume the population is normally distributed and use a ​   t-distribution to construct a 95​% 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.

Solutions

Expert Solution

Solution :

Given that,

= 32.5

s = 7.3

n = 29

Degrees of freedom = df = n - 1 = 29 - 1 = 28

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,28 = 2.048

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

= 2.048 * ( 7.3 / 29)

= 2.8

Margin of error = E = 2.8

The 95% confidence interval estimate of the population mean is,

- E < < + E

32.5 - 2.8 < < 32.5 + 2.8

29.7 < < 35.3

(29.7 , 35.3)


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