Question

Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 30 mpgwith a standard deviation of 7 mpg. Thirty-one non-hybrid sedans get a mean of 20 mpg with a standard deviation of three mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test at the 5% level to evaluate the manufacturers claim.
NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
State the null hypothesis
State the alternative hypothesis.
In words, state what your random variable Xhybrid − Xnon−hybrid represents.
State the distribution to use for the test.
What is the test statistic? (If using the z distribution round your answer to two decimal places, and if using the t distribution round your answer to three decimal places.)
What is the p-value?
what is the Alpha?
Explain how you determined which distribution to use.

Answer 1

99.7% of the Manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.

On Subtracting 1 from your sample size, we get

100 – 1 = 99.

On Subtracting confidence level from 1, and then divide by two.

(1 – .997) / 2 = 0.0015

Now, df = 99 and α = 0.0015

from the table at df = 99 we got 2.262.

Divide your sample standard deviation by the square root of your sample size.

28 / √(100) = 2.8

Now, 2.262 × 2.8 = 6.33

So, the confidence interval be,

(100 - 6.33 , 100 + 6.33) = (44.67 , 56.33)

Hence, 99.7% of the Manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.

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