Question

The standard recommendation for automobile oil changes is once every 3000 miles. A local mechanic is interested in determining whether people who drive more expensive cars are more likely to follow the recommendation. Independent random samples of 49 customers who drive luxury cars and 38 customers who drive compact lower-price cars were selected. The average distance driven between oil changes was 3178 miles for the luxury car owners and 3200 miles for the compact lower-price cars. The sample standard deviations were 41.80 and 50.60 miles for the luxury and compact groups, respectively. Assume that the population distributions of the distances between oil changes have the same standard deviation for the two populations. Using the 1% significance level, can you conclude that the mean distance between oil changes is less for all luxury cars than that for all compact lower-price cars

Answer 1

We accept H₀ we have not enough evidence to support that the mean distance between oil changes is less for all luxury cars than that for all compact lower-price cars

Explanation:

Luxury cars sample:

sample size n₁ = 49

sample mean x₁ = 3178

sample standard deviation s₁ = 41,80

Compact lower-price cars sample

sample size n₂ = 38

sample mean x₂ = 3200

sample standard deviation s₂ = 50,60

Test Hypothesis:

Null Hypothesis H₀ x₁ - x₂ = 0 or x₁ = x₂

Alternative Hypothesis Hₐ x₁ - x₂ < 0 or x₁ < x₂

CI = 99 % then significance level is α = 1 % α = 0,01

Alternative Hypothesis indicates that we have to develop a one tail-test to the left

z(c) for α 0,01 is z(c) = -2,32

To calculate z(s)

z(s) = [ ( x₁ - x₂ 9 ] / √ (s₁²)/n₁ + (s₂)²/n₂

z(s) = ( 3178 - 3200 ) / √ 35,66 + 67,38

z(s) = ( - 22 / 10,15 )

z(s) = - 2,17

Comparing z(s) and z(c)

z(s) > z(c) - 2,17 > - 2,32

Then z(s) is in the acceptance region we accep H₀