An automobile manufacturer claims that their van has a 59.1 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the MPG for this van. After testing 160 vans, they found a mean MPG of 59.5. Assume the variance is known to be 2.89. Is there sufficient evidence at the 0.1 level that the vans have an incorrect manufacturer's MPG rating?

Step 3 of 5: Specify if the test is one-tailed or two-tailed.
- One-Tailed Test
- Two-Tailed Test

Step 4 of 5: Enter the decision rule.
- Reject [tex]H_0[/tex] if [tex]|\text{z}| >[/tex] (enter critical value).

Step 5 of 5: Enter the conclusion.
- Reject Null Hypothesis
- Fail to Reject Null Hypothesis

There is not enough evidence at the 0.01 significance level to conclude that the population variance of sample A is greater than the population variance of sample B.

To test the hypothesis H0: σ1^2 = σ2^2 vs. H1: σ1^2 > σ2^2, we can use the F-test for comparing the variances of two independent samples. The test statistic follows an F-distribution.

The F-test statistic is calculated as:
F = s1^2 / s2^2

Where:
s1^2 is the sample variance of sample A,
s2^2 is the sample variance of sample B.

In this case:
s1^2 = 4.5
s2^2 = 2.3

The degrees of freedom for sample A is n1 - 1 = 21 - 1 = 20, and for sample B is n2 - 1 = 8 - 1 = 7.

We can find the critical value from the F-distribution table or using a statistical calculator. For α = 0.01 and degrees of freedom (20, 7), the critical value is approximately 4.964.

Calculating the test statistic:
F = 4.5 / 2.3 ≈ 1.956

Since the test statistic F = 1.956 is less than the critical value 4.964, we do not reject the null hypothesis.

Conclusion: There is not enough evidence at the 0.01 significance level to conclude that the population variance of sample A is greater than the population variance of sample B.

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