Answer:
The degrees of freedom are:
And the p value would be:
Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are significantly higher than those in Dallas
Step-by-step explanation:
Data given
represent the sample size for group Boston
represent the sample size for group Dallas
represent the sample mean for the group Boston
represent the sample mean for the group Dallas
represent the sample standard deviation for group Boston
represent the sample standard deviation for group Dallas
We can assume that we have independent samples from two normal distributions with equal variances and that is:
Let the subindex 1 for Boston and 2 for Dallas we want to check the following hypothesis:
Null hypothesis:
Alternative hypothesis:
The statistic is given by this formula:
Where t follows a t student distribution with degrees of freedom and the pooled variance
is given by this formula:
Replacing we got:
And the deviation would be just the square root of the variance:
The statitsic would be:
The degrees of freedom are:
And the p value would be:
Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are significantly higher than those in Dallas