In: Statistics and Probability
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.
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)