Question

A random sample of 41 car owners was taken to try to determine the mean length of car ownership for all owners.  The sample resulted in a mean of 3.98 years and a standard deviation of 1.5 years.  The data is normally distributed.  Form a 95% confidence interval for the mean length of car ownership.  The confidence interval would be 3.98  what margin of error?

Answer 1

Given that

n=41

df=41-1=40

critical t value=tinv(0.05,40)=2.021

a)95% confidence interval for mean

=3.98+/-2.021*1.5/sqrt(41)

=3.98+/0.47

=(3.51, 4.45)

(as,n>30,we can also use z multiplier of 1.96,then margin of error will be 0.46 and CI will be (3.52,4.44))

b)3.98+/-4.47

margin of error=0.47