Question

In a random sample of eleven ​people, the mean driving distance to work was 23.3 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 95​% confidence interval for the population mean μ. Interpret the results. Identify the margin of error.

(round to one decimal place)

Answer 1

Solution :

Given that,

= 23.3

s =5.4

n =11

Degrees of freedom = df = n - 1 =11 - 1 = 10

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,10 =2.228 ( using student t table)

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

= 2.228* (5.4 / 11)

= 3.6

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

- E < < + E

23.3 - 3.6 < < 23.3+ 3.6

19.7 < < 26.9

(19.7 , 26.9 )