Question

Car buyers often add accessories to their new cars. A sample of 179 Mini Cooper purchasers yielded a sample mean of $5,000 worth of accessories added to the purchase above the $20,200 base sticker price. Suppose the cost of accessories purchased for all Minis has a standard deviation of $1,500. a) Calculate a 95% confidence interval for the average cost of accessories on Minis.

b) Determine the margin of error in estimating the average cost of accessories on Minis.

c) What sample size would be required to reduce the margin of error by 50%?

Answer 1

Given that,

Sample size = n = 179

Sample mean = xbar= 5000

Population standard deviation = = 1500

a) a 95% confidence interval for the average cost of accessories on Minis ::

we know that

confidence interval =(xbar-E , xbar+E) ...........1

Here E is a margin of error

for 95% confidence the critical value = Zc = 1.96

then,

E = {1.96*1500}{\sqrt{179}}=219.7459

from equation 1

So confidence interval is ( 5000 - 1219.7459 , 5000 + 219.7459) = ( 4780.25 , 5219.75)

b) margin of error in estimating the average cost of accessories on Minis. :

we already calculated in above part

for 95% confidence the critical value = Zc = 1.96

then,

E = {1.96*1500}{\sqrt{179}}=219.7459

c) sample size would be required to reduce the margin of error by 50% :

Confidence level = 0.95

Zc = 1.96

Margin of error reduces 50% so E = 109.873

Population standard deviation = = 1500

We have to find sample size (n)