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:

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

Step-by-step 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

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