Question

The standard recommendation for automobile oil changes is once every 5000 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 45 customers who drive luxury cars and 40 customers who drive compact lower-price cars were selected. The average distance driven between oil changes was 5187 miles for the luxury car owners and 5389 miles for the compact lower-price car owners. The sample standard deviations were 424 and 507 miles for the luxury and compact groups, respectively. Assume that the two population distributions of the distances between oil changes have the same standard deviation. You would like to test if the mean distance between oil changes is less for all luxury cars than that for all compact lower-price cars.

Let μ₁ denote the mean distance between oil changes for luxury cars, and μ₂ denote the mean distance between oil changes for compact lower-price cars. Suppose the test statistic for this case is -2. Calculate the p-value. Round your final answer to the nearest ten thousandth (e.g., 0.1234).

Answer 1

Answer :

we have given two indepndent variable :

first variable is : all luxuary cars and

second variable is : compact lower price cars

here

n1 = sample size for x1 = 45

x1bar = sample mean for x1 = 5187 ,

s1 = sample standard deviation = 424

and

n2 = sample size  for x2 = 40

x2bar = sample mean for x2 = 5389

and s2 = sample standard deviation = 507

# Assume that the two population distributions of the distances between oil changes have the same standard deviation ( ie population variacnces are equal )

# Claim :  You would like to test if the mean distance between oil changes is less for all luxury cars than that for all compact lower-price cars.

that is our test is one tailed , left tailed test ,

to test : Ho :  μ1 = μ2 vs H1: μ1 <  μ2

where μ1 : population mean for x1 : all luxuary cars and

μ2 : all compact lower price cars

test statistics : t = -2 ( given )

degree of freedom = n1 + n2 -2 ( here we have equal variance )

= 45 + 40 -2

= 83

## p value : we can use t statistical table and find out p value for one sided t test :

p value is equal to = 0.0243.

( we can use p value approach we reject Ho if p value is less than alpha value )