Question

In a random sample of twelve ​people, the mean driving distance to work was 20.8 miles and the standard deviation was 5.4 miles. Assume the population is normally distributed and use the​t-distribution to find the margin of error and construct a 99​% confidence interval for the population mean mu. Interpret the results. Identify the margin of error.

Answer 1

Given that,

= 20.8

s = 5.4

n = 12

Degrees of freedom = df = n - 1 = 12- 1 =11

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,11= 3.106 ( using student t table)

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

= 3.106* ( 5.4/ 12) = 4.84

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

- E < < + E

20.8 -4.84 < <20.8 +4.84

15.96 < < 25.64