Question

You are the automotive specialist for the monthly magazine Consumer Beware. Part of your job is to investigate claims made by automakers in their TV commercials. You are particularly suspicious of a new economy car that the automaker claims will get 78 miles/gallon. After checking the mileage figures for a random sample of 70 car owners, you find average miles per gallon is 75.5 with a standard deviation of 3.7.

this is what i put

1. At the 99% confidence level, do your results tend to confirm or refute the manufacturer’s claim?

At the 99% CI, the results do tend to refute the manufacturer’s claim that the new car will get 78 miles/gal. When I did the calculations, it came to about 74.39 – 76.61.

Answer 1

sample mean, xbar = 75.5
sample standard deviation, s = 3.7
sample size, n = 70
degrees of freedom, df = n - 1 = 69

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.649

ME = tc * s/sqrt(n)
ME = 2.649 * 3.7/sqrt(70)
ME = 1.171

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (75.5 - 2.649 * 3.7/sqrt(70) , 75.5 + 2.649 * 3.7/sqrt(70))
CI = (74.39 , 76.61)

As the confidence interval does not contain 78 so, reject the null hypothesis

yes, results tend to refute the manufacturer’s claim that the new car will get 78 miles/gal.