Question

records of 40 used passanger cars and 40 used pickup tricks were randomly selected to investigate whether there was any difference in the mean time in years that they were kept by original owner before being sold. For cars, the mean was 5.3 years with standard deviation 2.2 years. For pickup trucks, the mean was 7.1 years with standard deviation 3.0 years. Test the hypothesis that there is a difference in the means against the null hypothesis that there is no difference. Use the 1% level of significance.

Answer 1

Since our calculated value is lower than our critical value,
`z_(calc)=-3.06<-2.64=z_(critical)`, we have enough evidence to reject the null hypothesis at 1% of significance. So there is enough evidence to conclude that mean for cars it's different from the mean for pickup's.

Explanation:

1) Data given and notation

`\bar X_(1)=5.3` represent the mean for group cars

`\bar X_(2)=7.1` represent the mean for group pickup

`s_(1)=2.2` represent the sample standard deviation for the sample cars

`s_(2)=3.0` represent the sample standard deviation for the sample pickups

`n_(1)=40` sample size for the group cars

`n_(2)=40` sample size for the group pickup

t would represent the statistic (variable of interest)

`p_v` represent the p value

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:

H0:
`\mu_(1) = \mu_(2)`

H1:
`\mu_(1) \\eq \mu_(2)`

If we analyze the size for the samples both are higher than 30, but we don't know the population deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

`t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}` (1)

T-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

We have all in order to replace in formula (1) like this:

`z=\frac{5.3-7.1}{\sqrt{(2.2^2)/(40)+(3.0^2)/(40)}}=-3.06`

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking on the t distribution with df=n1+n2-1=40+40-2=78 degrees of freedom, a value that accumulates 0.005 of the area on each tail. We can use excel or a table to find it, for example the code in Excel is:

"=T.INV(1-0.005,78)", and we got
`z_(critical)=\pm 2.64`

5) Statistical decision

Since our calculated value is lower than our critical value,
`z_(calc)=-3.06<-2.64=z_(critical)`, we have enough evidence to reject the null hypothesis at 1% of significance. So there is enough evidence to conclude that mean for cars it's different from the mean for pickup's.