Hypothesis Tests about the Variance
of a Normally Distributed Population

VARTEST.83p
VARTEST.86p
vartest.89p

Suppose we are studying the variance Var of a normally distributed population and we wish to test the following three null hypotheses:

1. Ho: Var <= v
2. Ho: Var >= v
3. Ho: Var = v


We accept or reject each null hypothesis based on the value of the sample variance (Sx)^2 from a random sample of size n. In this setting, we define the test statistic by x = (n - 1)*(Sx)^2 / v, which follows a Chi Square Distribution with n - 1 degrees of freedom: X ~ X2(n - 1). We use this distribution to define the left and right tail probability values of the test statistic, P(X <= x) and P(X >= x).

If a is our pre-chosen level of significance, then we reject the first null hypothesis Var <= v if (Sx)^2 is too large; that is, if the right tail value is too small: P(X >= x) < a. In this case, we would be saying that there is too small a chance of the sample variance being as large as it is if the null hypothesis were true. This test is equivalent to the test Ho: Var = v with a one-sided alternative Ha: Var > v, and the right tail value is also called the p-value.

We reject the second null hypothesis Var >= v if (Sx)^2 is too small; i.e., if the left tail value is too small: P(X<= x) < a. In this case, we would be saying that there is too small a chance of the sample variance being as small as it is if the null hypothesis were true. This test is equivalent to the test Ho: Var = v with a one-sided alternative Ha: Var < v,

We reject the third null hypothesis if P(X <= x) < a / 2 or P(X >= x) < a / 2. For this two-sided test, the p-value is twice the smallest tail value.



Using the VARTEST Program

The VARTEST program can be used to perform these hypothesis tests. To execute the program, we enter the variance value to be tested, followed by the sample size, the sample variance, and the level of significance. The program displays the conclusion for each test, the test statistic, and the tail probability values.



Example. IQ tests are often designed to have an average of around 100 with a standard deviation of around 15. In the article "The malleability of IQ as judged from adoption studies" in the journal Intelligence, 14, (1990), the IQ's of a sample of 29 couples who had adopted children yielded an average IQ of 118 with a sample deviation of Sx = 9. Using the data from these 58 people and a = 0.05, test the following three hypotheses about the true variance in IQ of similar couples who have adopted children.

1. Ho: Var <= 225
2. Ho: Var >= 225
3. Ho: Var = 225


Solution. After calling up the VARTEST program, enter 225 for TEST VARIANCE, 58 for SAMPLE SIZE, 9^2 (or 81) for SAMPLE VARIANCE, and .05 for LEVEL OF SIG.

We obtain a test statistic of 20.52, a left tail value of 0, and we reject the second and third hypotheses.

Based on this data, we have accepted this first hypothesis and rejected the third; thus, we conclude that Var < 225. Equivalently, we can say that the standard deviation in IQ of all similar couples is less than 15.

If Var >= 225 or if Var = 225, then there would be no chance of (Sx)^2 being as small as 81 with a sample of size 58, which is why we have rejected the second and third hypotheses.



Exercises


1. Consider the following random sample of combined SAT scores from a sophomores at a random university.

Random Collection of SAT Scores

1270
1070
1440
880
970
1400
860
980
1220
1110
850
1080
1210
720
920
800
1160
1220
1280
1200
1030
1170
1350
1070
950
1150
1020
1120
1020
1070
910
1320

Assuming that the SAT scores are normally distributed, use a 0.04 level of significance to test the following hypotheses about the true standard deviation s of SAT scores of all sophomores at this university.

1. Ho: s <= 160
2. Ho: s >= 160
3. Ho: s = 160


2. A study had shown that the standard deviation in Math SAT scores among all college bound high school seniors is 90. A researcher believes that the deviation is less for males. A random sample of the Math SAT scores of 40 males yielded a sample deviation of 85.64. State and test an appropriate hypothesis test for the researcher.



Solutions

1. We first note that these tests about s are equivalent to tests about the variance with a test variance of (160)^2. Now we enter the data into an appropriate list and compute the compute the sample deviation. Upon doing so, we obtain Sx = 177.9532 with this sample of size 32.

Next, call up the VARTEST program. For TEST VARIANCE enter 160^2, for SAMPLE SIZE enter 32, for SAMPLE VARIANCE enter Sx^2 (or 177.9532^2), for LEVEL OF SIG. enter .04.

We receive a right tail value of 0.1707 and we do not have evidence to reject any of the claims at the 0.04 level of significance.

If s <= 160, there would still be as much as a 17.07% chance of Sx being as large as 177.953 with a sample of size 32. This (one-sided) p-value of 0.1707 is not too small to reject the first hypothesis. (The p-value for the third test is 2*.1707 = 0.3414.)


2. We shall test Ho: s = 90 with a one-sided alternative Ha: s < 90 at the 0.05 level of significance.

Call up the VARTEST program. For TEST VARIANCE enter 90^2, for SAMPLE SIZE enter 40, for SAMPLE VARIANCE enter 85.64 ^ 2, for LEVEL OF SIG. enter .05.

We receive a (left-tail) p-value of 0.3612. Based on this data, there is not significant evidence to reject the hypothesis that s = 90 for male Math SAT scores. If s = 90, there would still be a 36.12% chance of obtaining a sample deviation as low as 85.64 with a sample of size 40.



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