Statistics Evaluation Task

Solution 1

Step – 1
From the given data, we have

Step – 2
Null and alternate hypothesis:
Null hypothesis: There is no significant difference in the averages miles drivers by SUV owners and other one owners
Alternate hypothesis: There is a significant difference in the averages miles drivers by SUV owners is more than the other car owners

This is a right tailed test

Step – 3
t-test for equality of means for equality of variance:

Now,

Step – 4
Test statistics:

Step – 5
Rejection region:
For alpha = 0.01 and df = 35, the initial value of t for a right tailed test is p-value = P (t > 2.438) = 0.3768
Decision: Since, t = 0.8952 < t xss=removed xss=removed>

Step – 6
So, we fail to reject the null hypothesis at alpha 0.01

Conclusion: We conclude that there is no significant difference in the average miles driven by
SUV owners and co-owners.

Solution 2

Step – 1
Given data is
For Current Year,

For New Year,

Level of significance = 5% i.e.

Step – 2
Null and alternate hypothesis:
Null hypothesis:
Alternate hypothesis:
This is a Left tailed test

Step – 3
State the test to be used and verify the requirements:
We use a 2 sample t-test, pooled variances, with

The requirements are
(1) Samples are simple random samples
(2) The samples are independent of each other

Step – 4
Compute the standardized test statistic
The Pooled Variance:

Test Statistics:

Step – 5
p-value and compare it to the level of significance:
p-value at t = – 0.17 is 0.5665.
p-value (0.5665) is larger than the significance level (0.05)

Step – 6
Make a decision, and then write the final conclusion 
Fail to reject the null hypothesis.
Therefore, there is not sufficient evidence at the 95% significance level to conclude that the new process will save PLE money.