## Statistics Evaluation Task

- 15th Sep, 2021
- 01:02 AM

**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.