F-Test for Equality of Two Variances

F-Test for Equality of Two Variances
Purpose: To test if two population variances are equal.
Assumptions:
- If two populations are normal → F-test
- If two populations are not normal → Levene’s Test
Hypotheses:
Ho:σ12=σ22 “The two populations variances are equal.”
Ha:σ12=σ22 “The two populations variances are not equal.”
Test Statistic:
Important: The larger sample variance sL2 should be put in the numerator, and the smaller sample variance ss2 is put in the denominator. This forces the F-test to always be a right-sided test, which makes things a lot easier.
If the variances are equal sL2=sS2, then F=1.
The F-test has two degrees of freedom:
- numerator df=nL−1
- denominator df=ns−1
P-value method:
- If you are able to solve for the p-value using software, ensure that the p-value is doubled for two-tailed tests.
- In order to reject Ho, the p-value needs to be smaller than the significance level α.
Critical value method:
- Without software, use to F-table to find the critical value.
- For two-tailed tests, divide α by 2 to find the critical value F2α,df1,df2.
- In order to reject Ho, the test statistic needs to be bigger than the critical value.
Example
Ho:σ12=σ22 “The two populations variances are equal.”
Ha:σ12=σ22 “The two populations variances are not equal.”
α=0.10
- numerator df=9
- denominator df=20
Find the critical value.
This is a two-tailed test, so 2α=20.10=0.05
We will therefore use the 5% F-table.

CV: F0.05,9,20=2.393

Example: F-Test for Equality of Two Variances
We want to test if the variances in Group 1 and Group 2 are equal or not. Assume both population distributions are normal. The sample standard deviation in Group 1 is s1=4.764 and the sample standard deviation in Group 2 is s2=4.116.
Raw data:

(i) State the hypotheses.
Ho:σ12=σ22 “The two populations variances are equal.”
Ha:σ12=σ22 “The two populations variances are not equal.”
(ii) α=0.10. Find the critical value.
This is a two-tailed test, so 2α=20.10=0.05
We will therefore use the 5% F-table.
You need the degrees of freedoms first.
n1=9
n2=9
numerator df=nL−1=9−1=8
denominator df=ns−1=9−1=8

Critical value F0.05,8,8=3.438
In order to reject Ho, the test statistic needs to be bigger than the critical value (3.438).
(iii) Find the test statistic.
(iv) Refer to the software output below. Is the evidence that the variances differ?
P-value method:
P-value (0.689) > alpha (0.10) → Fail to reject Ho. There is no evidence that the variances differ.
Critical value method:
Test statistic (1.34) < critical value (3.438) → Fail to reject Ho. There is no evidence that the variances differ.
Minitab Output:


Note: Since this is a two-tailed test, Minitab had doubled the p-value.
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