12.1 Test of two variances  (Page 2/12)

To determine the critical points we have to find F _α , _df1 , _df2 . See Appendix A for the F table. The F distribution has two different degrees of freedom, one associated with the numerator, _df1 , and one associated with the denominator, _df2 and to complicate matters the F distribution is not symmetrical and changes the degree of skewness as he degrees of freedom change. The degrees of freedom in the numerator is n ₁ -1, where n ₁ is the sample size for group 1, and the degrees of freedom in the denominator is n ₂ -1, where n ₂ is the sample size for group 2. F _α , _df1 , _df2 will give the critical value on the upper end of the F distribution.

To find the critical value for the lower end of the distribution, reverse the degrees of freedom and divide the F-value into one.

• Upper tail critical value : F _α , _df1 , _df2
• Lower tail critical value : 1/F _α , _df2 , _df1

When the calculated value of F is between the critical values we cannot reject the null hypothesis that the two variances came from a population with the same variance. If the calculated F-value is in either tail we cannot accept the null hypothesis just as we have been doing for all of the previous tests of hypothesis.

An alternative way of finding the critical values of the F distribution makes the use of the F-table easier. We note on the F-table that all the values of F are greater than one therefore the critical F value for the left hand tail will always be less than one because to find the critical value on the left tail we divide an F value into the number one. We also note that if the sample standard deviation in the numerator of the test statistic is larger than the sample standard deviation in the denominator, the resulting F value will be greater than one. The shorthand method for this test is thus to be sure that the larger of the two sample standard deviations is placed in the numerator to calculate the test statistic. This will mean that only the right had tail critical value will have to be found in the F-table.

Two college instructors are interested in whether or not there is any variation in the way they grade math exams. They each grade the same set of 30 exams. The first instructor's grades have a variance of 52.3. The second instructor's grades have a variance of 89.9. Test the claim that the first instructor's variance is smaller. (In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors.) The level of significance is 10%.

Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively.

n ₁ = n ₂ = 30.

H ₀ : ${\sigma }_1^2={\sigma }_2^2$ and H _a : ${\sigma }_1^2\text{ < }{\sigma }_2^2$

Calculate the test statistic: By the null hypothesis $\text{(}{\sigma }_{\text{1}}^{\text{2}}\text{ = }{\sigma }_{\text{2}}^{\text{2}}\text{)}$ , the F statistic is:

$F=\frac{{s_2}^2}{{s_1}^2}=\frac{89.9}{52.3}=1.719$

Critical value for the test: F _29,29 where n ₁ – 1 = 29 and n ₂ – 1 = 29 = 1.65.

Make a decision: Since the calculated F value is not in the tail we cannot accept H ₀ .

Conclusion: With a 10% level of significance, from the data, there is sufficient evidence to conclude that the variance in grades for the first instructor is smaller.