Question

Random samples from two normal populations produced the following statistics: s1² = 350 n1 = 30 s2² = 700 n2 = 30

a. Can we infer at the 10% significance level that the two population variances differ?

b. Repeat part (a) changing the sample sizes to n1 = 15 and n2 = 15.

c. Describe what happens to the test statistic and the conclusion when the sample sizes decrease.

Answer 1

Solution:

a) The null and alternative hypotheses will be as follows:

To test the equality of two population variances, the most appropriate test is F-test. The test statistic is given as follows:

Where, s₁² and s₂² are sample variances, n_{​​​​​​1} and n_{​​​​​​2} are sample sizes.

The degrees of freedom for the test will be (n_{​​​​​​1} - 1, n_{​​​​​​2} - 1).

We have, Where, s₁² = 350, s₂² = 700, n_{​​​​​​1} = 30, n_{​​​​​​2} = 30

The value of the test statistic is 0.5

Degrees of freedom = (n_{​​​​​​1} - 1, n_{​​​​​​2} - 1) = (30 - 1, 30 - 1) = (29, 29)

At 10% significance level we make decision rule as follows:

then, reject the null hypothesis at 10% significance level.

then we fail to reject the null hypothesis at 10% significance level.

Using F-table we get,

We have, F = 0.5 which is less than 0.5374.

Since, F < F_{(1-0.10/2, (29,29))} therefore we shall reject the null hypothesis (H_{​​​​​​0}) at 10% significance level.

Conclusion: At 10% significance level there is sufficient evidence to infer that two population variances differ.

b) The null and alternative hypotheses will be as follows:

To test the equality of two population variances, the most appropriate test is F-test. The test statistic is given as follows:

Where, s₁² and s₂² are sample variances, n_{​​​​​​1} and n_{​​​​​​2} are sample sizes.

The degrees of freedom for the test will be (n_{​​​​​​1} - 1, n_{​​​​​​2} - 1).

We have, Where, s₁² = 350, s₂² = 700, n_{​​​​​​1} = 15, n_{​​​​​​2} = 15

The value of the test statistic is 0.5

Degrees of freedom = (n_{​​​​​​1} - 1, n_{​​​​​​2} - 1) = (15 - 1, 15 - 1) = (14, 14)

At 10% significance level we make decision rule as follows:

then, reject the null hypothesis at 10% significance level.

then we fail to reject the null hypothesis at 10% significance level.

Using F-table we get,

We have, F = 0.5 which lies between 0.4026 and 2.4837

Since, F_{(1-0.10/2, (14, 14))} < F < F_{(0.10/2, (14, 14))} therefore we shall be fail to reject the null hypothesis (H_{​​​​​​0}) at 10% significance level.

Conclusion: At 10% significance level there is not sufficient evidence to infer that two population variances differ.

c) When we change the sample sizes from (n_​​1 = 30, n_{​​​​​​2} = 30) to (n_{​​​​​​1} = 15, n_{​​​​​​2} = 15) the value of the test statistic remains same. But when the sample sizes are (n_{​​​​​​1} = 30, n_{​​​​​​2} = 30), we concluded that population variances differ and when the sample sizes are (n_{​​​​​​1} = 15, n_{​​​​​​2} = 15) then, at 10% significance level we don't have sufficient evidence to concluded that population variances differ.