TEST OF HYPOTHESIS ABOUT mu d

TEST OF HYPOTHESIS ABOUT m_d

DEPENDENT SAMPLES

A manufacturer of shock absorbers would like to advertise that their shock absorbers last longer than those produced by its biggest competitor. To see if there is support for such a claim, six of the manufacturer’s shocks and six of the competitor’s shocks were randomly selected, and one of each brand was installed on the rear wheels of each of six cars. After the cars had been driven 20,000 miles, the strength of each shock absorber was measured. These data are below.

Is there sufficient evidence to conclude that the manufacturer’s shocks have a greater mean strength after 20,000 miles of driving than the competitor’s? Use a .01 level of significance.

SOLUTION

The parameter of interest is m_d, the difference in the mean strength of the manufacturer’s shocks and the competitor’s shocks after 20,000 miles of driving.

H₀: m_D = 0 H_a: m_D > 0

Decision Rule: Accept H_a if the calculated p-value < .01.

Calculations from StatCrunch: t = 7.68, p-value = 0.0003 < .01 ---> Accept H_a

Interpretation: At the .01 significance level I conclude that the manufacturer’s shock absorbers have a greater mean strength after being on cars for 20,000 miles than the competitor’s shock absorbers.

COMMENTS ABOUT THE SOLUTION

The usual structure for presenting the solution to a test of hypothesis was followed. All test of hypotheses solutions should follow this structure.

Since the two types of shock absorbers were placed on each of 6 cars, these samples are dependent.

This procedure is valid when a random sample of differences between pairs of strength measurements are obtained from a population of differences that are normally distributed.

The complete StatCrunch analyses are below.

Hypothesis test results:
μ₁ - μ₂ : mean of the paired difference between Manufacturer and Competitor
H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ > 0

Difference

Sample Diff.

Std. Err.

DF

T-Stat

P-value

Manufacturer - Competitor

0.41666666

0.054262735

5

7.678689

0.0003

A graph of the appropriate t-distribution and the resulting p-value is furnished below.

Difference	Sample Diff.	Std. Err.	DF	T-Stat	P-value
Manufacturer - Competitor	0.41666666	0.054262735	5	7.678689	0.0003