Hypothesis Tests about the Ratio of Variances
of Independent Normal Populations

 RATFTEST.83p RATFTEST.86p ratftest.89p

Suppose we are studying two independent normally distributed populations X and Y having unknown variances VarX and VarY respectively. We wish to test the following three hypotheses about the ratio of variances:

 1. Ho: VarX / VarY <= r 2. Ho: VarX / VarY >= r 3. Ho: VarX / VarY = r

We base the decision for each test on the ratio of sample variances (Sx)^2 and (Sy)^2 obtained from independent random samples from X and Y having sizes n and m respectively.

The test statistic we now use is x = [(Sx)^2 / (Sy)^2] / r which follows an F distribution with n - 1 and m - 1 degrees of freedom, F ~ F(n - 1, m - 1). Again, we compute the left and right tail probability values P(F <= x) and P(F >= x) in order to compare with the level of significance a.

If P(F >= x) < a, then we reject the first hypothesis. In this case, we would say that there is too small a probability (p-value) of Sx^2 / Sy^2 being as large as it is if the hypothesis were true; thus we would reject VarX / VarY <= r.

If P(F <= x) < a, then we reject the second hypothesis.

If P(F <= x) < a / 2 or if P(F >= x) < a / 2, then we reject the third hypothesis. The p-value for this two-sided test is twice the smallest tail-value.

Using the RATFTEST Program

The RATFTEST program can be used to perform these hypothesis tests. To execute the program, we enter the value of the ratio to be tested, the two sample sizes, the two sample variances, and the level of significance. The program displays the conclusion of each test, the test statistic, and the tail probability values.

Example. Suppose the ratio of the variances of weights of two different sizes of boxes of Cheer detergent is to be studied. A sample of 35 "98 oz." packages yields weights with a sample variance of 1.45 ounces. A sample of 30 "42 oz." packages yields weights with a sample variance of 1.05 ounces. Test the following three hypotheses about the ratio of true variances at the 0.05 level of significance.

 1. Ho: VarX / VarY <= 2.5 2. Ho: VarX / VarY >= 2.5 3. Ho: VarX / VarY = 2.5

Solution. Call up the RATFTEST program, enter 2.5 for TEST RATIO, 35 for X SAMPLE SIZE, 1.45 for X SAMPLE VAR., 30 for Y SAMPLE SIZE, 1.05 for Y SAMPLE VAR., and .05 for LEVEL OF SIG.

We receive a left tail value of 0.0487 from a test statistic of 0.55238, and we reject the second hypothesis. We can conclude that VarX / VarY <= 2.5.

We note that (Sx)^2 / (Sy)^2 = 1.38. If VarX / VarY >= 2.5, then there would be at most a 4.87% chance of (Sx)^2 / (Sy)^2 being as small as1.38 with samples of these sizes, which is why we reject the second claim.

However, for the two-sided test required of the third test, the p-value is 2*.0487 = 0.0974 which is more than the designated level of significance. Thus, we do not have significant evidence, at the 0.05 level, to reject the third hypothesis. Ultimaltely, we conclude VarX / VarY <= 2.5 (rather than VarX / VarY < 2.5).

Exercises

1. With the data from the preceding example, test the hypothesis that VarX = VarY at the 0.02 level of significance.

2. Consider the following random sample of Verbal SAT scores:

A Random Collection of Girls' Verbal SAT Scores
 530 570 550 410 680 470 600 660 510 520 570 490 390 500 360 760 510 320 410 400 440

A Random Collection of Boys' Verbal SAT Scores
 490 560 530 540 540 360 470 380 450 600 540 510 440 440 440 590 460 490 570 470 430

If sx and sy represent the true standard deviations of all girls' and all boys' Verbal SAT scores respectively, test the following three hypotheses at the 0.04 level of significance.

 1. Ho: sx / sy <= 1 2. Ho: sx / sy >= 1 3. Ho: sx / sy = 1

3. With the data from Exercise 2 above, test the following three hypotheses at the 0.10 level of siginificance.

 1. Ho: sy / sx <= 0.9 2. Ho: sy / sx >= 0.9 3. Ho: sy / sx = 0.9

Solutions

1. The test VarX = VarY is equivalent to testing VarX / VarY = 1. Thus, we enter the data into the program as in the original example above, but now we use 1 for TEST RATIO and .02 for LEVEL OF SIG.

We receive a right tail value of 0.1893, which gives a p-value of 0.3786 for this two-sided test, and we do not have evidence to reject.

Based on this data we can accept the claim that VarX = VarY. If VarX = VarY, then with samples of these sizes there is still a 37.86% chance of (Sx)^2 / (Sy)^2 being as far away (in either direction) from 1 as is 1.38, which is not a small enough chance to reject the hypothesis.

2. First, enter the data into appropriate lists and compute the summary statistics to obtain Sx = 110.2788673 and Sy = 65.68684727 for these samples of size 21.

These tests about the ratio of deviations being compared to 1 are equivalent to tests about the ratio of variances still being compared to 1.

So call up the RATFTEST program and enter 1 for TEST RATIO. For X SAMPLE SIZE enter 21, and for X SAMPLE VAR. enter Sx^2 (or 110.2788673^2). For Y SAMPLE SIZE enter 21, and for Y SAMPLE VAR. enter Sy^2 (or 65.68684727^2). For LEVEL OF SIG. enter .04.

We receive a right tail value of 0.0125 from a test statistic of 2.818563 and we reject the first and third hypotheses.

Based on this data, we can accept the claim that VarX / VarY > 1. That is, the variance of the girls' scores is larger than that of boys' scores. Because if VarX / VarY <= 1 were true, then there would be at most a 1.25% chance of (Sx)^2 / (Sy)^2 being as large as 2.818563 with samples of size 21 each.

3. Although the program is written in terms of the ratio VarX / VarY, we can reverse the roles of the two populations to perform tests about VarY / VarX or sy / sx.

We call up the RATFTEST program. For TEST RATIO, enter .81 (or .9^2 ). For X SAMPLE SIZE enter 21, and for X SAMPLE VAR. enter Sy^2 (or 65.68684727^2). For Y SAMPLE SIZE enter 21, and for Y SAMPLE VAR. enter Sx^2 (or 110.278867^2). For LEVEL OF SIG. enter .1.

We receive a left tail value of 0.0361 and we reject the second and third hypotheses.

Thus, we can say sy / sx < 0.9. If sy / sx >= 0.9, then with samples of these sizes there would be at most a 3.61% chance of Sy / Sx being as small as 65.68684727 / 110.2788673 = 0.595643.