11.3 Ftests for Equality of Two Variances
Learning Objectives
 To understand what Fdistributions are.
 To understand how to use an Ftest to judge whether two population variances are equal.
FDistributions
Another important and useful family of distributions in statistics is the family of Fdistributions. Each member of the Fdistribution family is specified by a pair of parameters called degrees of freedom and denoted df1 and df2. Figure \(\PageIndex{1}\) shows several Fdistributions for different pairs of degrees of freedom. An F random variable is a random variable that assumes only positive values and follows an Fdistribution.
Figure \(\PageIndex{1}\): Many FDistributions
The parameter df1 is often referred to as the numerator degrees of freedom and the parameter df2 as the denominator degrees of freedom. It is important to keep in mind that they are not interchangeable. For example, the Fdistribution with degrees of freedom df1=3 and df2=8 is a different distribution from the Fdistribution with degrees of freedom df1=8 and df2=3.
Definition
The value of the F random variable F with degrees of freedom df1 and df2 that cuts off a right tail of area c is denoted Fc and is called a critical value. See Figure \(\PageIndex{2}\).
Figure \(\PageIndex{2}\) Fc Illustrated
Tables containing the values of Fc are given in Chapter 11. Each of the tables is for a fixed collection of values of c, either 0.900, 0.950, 0.975, 0.990, and 0.995 (yielding what are called “lower” critical values), or 0.005, 0.010, 0.025, 0.050, and 0.100 (yielding what are called “upper” critical values). In each table critical values are given for various pairs (df1,df2). We illustrate the use of the tables with several examples.
Example \(\PageIndex{1}\)
Suppose F is an F random variable with degrees of freedom df1=5 and df2=4. Use the tables to find
 F_{0.10}
 F_{0.95}
Solution:

The column headings of all the tables contain df1=5. Look for the table for which 0.10 is one of the entries on the extreme left (a table of upper critical values) and that has a row heading df2=4 in the left margin of the table. A portion of the relevant table is provided. The entry in the intersection of the column with heading df1=5 and the row with the headings 0.10 and df2=4, which is shaded in the table provided, is the answer, F0.10=4.05.
F Tail Area df1 1 2 ⋅ ⋅ ⋅ 5 ⋅ ⋅ ⋅ df2 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0.005 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 22.5 ⋅ ⋅ ⋅ 0.01 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 15.5 ⋅ ⋅ ⋅ 0.025 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 9.36 ⋅ ⋅ ⋅ 0.05 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 6.26 ⋅ ⋅ ⋅ 0.10 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 4.05 ⋅ ⋅ ⋅ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ Look for the table for which 0.95 is one of the entries on the extreme left (a table of lower critical values) and that has a row heading df2=4 in the left margin of the table. A portion of the relevant table is provided. The entry in the intersection of the column with heading df1=5 and the row with the headings 0.95 and df2=4, which is shaded in the table provided, is the answer, F0.95=0.19.
F Tail Area df1 1 2 ⋅ ⋅ ⋅ 5 ⋅ ⋅ ⋅ df2 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0.90 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 0.28 ⋅ ⋅ ⋅ 0.95 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 0.19 ⋅ ⋅ ⋅ 0.975 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 0.14 ⋅ ⋅ ⋅ 0.99 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 0.09 ⋅ ⋅ ⋅ 0.995 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 0.06 ⋅ ⋅ ⋅ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
Example \(\PageIndex{2}\)
Suppose F is an F random variable with degrees of freedom df1=2 and df2=20. Let α=0.05. Use the tables to find
 Fα
 Fα/2
 F1−α
 F1−α/2
Solution
The column headings of all the tables contain df1=2. Look for the table for which α=0.05 is one of the entries on the extreme left (a table of upper critical values) and that has a row heading df2=20 in the left margin of the table. A portion of the relevant table is provided. The shaded entry, in the intersection of the column with heading df1=2 and the row with the headings 0.05 and df2=20 is the answer, F0.05=3.49.
F Tail Area  df1  1  2  ⋅ ⋅ ⋅ 

df2  
⋮  ⋮  ⋮  ⋮  ⋮ 
0.005  20  ⋅ ⋅ ⋅  6.99  ⋅ ⋅ ⋅ 
0.01  20  ⋅ ⋅ ⋅  5.85  ⋅ ⋅ ⋅ 
0.025  20  ⋅ ⋅ ⋅  4.46  ⋅ ⋅ ⋅ 
0.05  20  ⋅ ⋅ ⋅  3.49  ⋅ ⋅ ⋅ 
0.10  20  ⋅ ⋅ ⋅  2.59  ⋅ ⋅ ⋅ 
⋮  ⋮  ⋮  ⋮  ⋮ 

Look for the table for which α/2=0.025 is one of the entries on the extreme left (a table of upper critical values) and that has a row heading df2=20 in the left margin of the table. A portion of the relevant table is provided. The shaded entry, in the intersection of the column with heading df1=2 and the row with the headings 0.025 and df2=20 is the answer, F0.025=4.46.
F Tail Area df1 1 2 ⋅ ⋅ ⋅ df2 ⋮ ⋮ ⋮ ⋮ ⋮ 0.005 20 ⋅ ⋅ ⋅ 6.99 ⋅ ⋅ ⋅ 0.01 20 ⋅ ⋅ ⋅ 5.85 ⋅ ⋅ ⋅ 0.025 20 ⋅ ⋅ ⋅ 4.46 ⋅ ⋅ ⋅ 0.05 20 ⋅ ⋅ ⋅ 3.49 ⋅ ⋅ ⋅ 0.10 20 ⋅ ⋅ ⋅ 2.59 ⋅ ⋅ ⋅ ⋮ ⋮ ⋮ ⋮ ⋮

Look for the table for which 1−α=0.95 is one of the entries on the extreme left (a table of lower critical values) and that has a row heading df2=20 in the left margin of the table. A portion of the relevant table is provided. The shaded entry, in the intersection of the column with heading df1=2 and the row with the headings 0.95 and df2=20 is the answer, F0.95=0.05.
F Tail Area df1 1 2 ⋅ ⋅ ⋅ df2 ⋮ ⋮ ⋮ ⋮ ⋮ 0.90 20 ⋅ ⋅ ⋅ 0.11 ⋅ ⋅ ⋅ 0.95 20 ⋅ ⋅ ⋅ 0.05 ⋅ ⋅ ⋅ 0.975 20 ⋅ ⋅ ⋅ 0.03 ⋅ ⋅ ⋅ 0.99 20 ⋅ ⋅ ⋅ 0.01 ⋅ ⋅ ⋅ 0.995 20 ⋅ ⋅ ⋅ 0.01 ⋅ ⋅ ⋅ ⋮ ⋮ ⋮ ⋮ ⋮

Look for the table for which 1−α/2=0.975 is one of the entries on the extreme left (a table of lower critical values) and that has a row heading df2=20 in the left margin of the table. A portion of the relevant table is provided. The shaded entry, in the intersection of the column with heading df1=2 and the row with the headings 0.975 and df2=20 is the answer, F0.975=0.03.
F Tail Area df1 1 2 ⋅ ⋅ ⋅ df2 ⋮ ⋮ ⋮ ⋮ ⋮ 0.90 20 ⋅ ⋅ ⋅ 0.11 ⋅ ⋅ ⋅ 0.95 20 ⋅ ⋅ ⋅ 0.05 ⋅ ⋅ ⋅ 0.975 20 ⋅ ⋅ ⋅ 0.03 ⋅ ⋅ ⋅ 0.99 20 ⋅ ⋅ ⋅ 0.01 ⋅ ⋅ ⋅ 0.995 20 ⋅ ⋅ ⋅ 0.01 ⋅ ⋅ ⋅ ⋮ ⋮ ⋮ ⋮ ⋮
A fact that sometimes allows us to find a critical value from a table that we could not read otherwise is:
If Fu(r,s) denotes the value of the Fdistribution with degrees of freedom df1=r and df2=s that cuts off a right tail of area u, then
Fc(k,ℓ)=1F1−c(ℓ,k)
Example \(\PageIndex{3}\)
Use the tables to find
F0.01 for an F random variable with df1=13 and df2=8
F0.975 for an F random variable with df1=40 and df2=10
Solution:
There is no table with df1=13, but there is one with df1=8. Thus we use the fact that
Using the relevant table we find that F0.99(8,13)=0.18, hence F0.01(13,8)=0.18−1=5.556.
There is no table with df1=40, but there is one with df1=10. Thus we use the fact that
Using the relevant table we find that F0.025(10,40)=3.31, hence F0.975(40,10)=3.31−1=0.302.
FTests for Equality of Two Variances
In Chapter 9 we saw how to test hypotheses about the difference between two population means μ1 and μ2. In some practical situations the difference between the population standard deviations σ1 and σ2 is also of interest. Standard deviation measures the variability of a random variable. For example, if the random variable measures the size of a machined part in a manufacturing process, the size of standard deviation is one indicator of product quality. A smaller standard deviation among items produced in the manufacturing process is desirable since it indicates consistency in product quality.
For theoretical reasons it is easier to compare the squares of the population standard deviations, the population variances σ21 and σ22. This is not a problem, since σ1=σ2 precisely when σ21=σ22, σ1<σ2 precisely when σ21<σ22, and σ1>σ2 precisely when σ21>σ22.
The null hypothesis always has the form H0:σ21=σ22. The three forms of the alternative hypothesis, with the terminology for each case, are:
Form of Ha  Terminology 

Ha:σ21>σ22  Righttailed 
Ha:σ21<σ22  Lefttailed 
Ha:σ21≠σ22  Twotailed 
Just as when we test hypotheses concerning two population means, we take a random sample from each population, of sizes n_{1} and n_{2}, and compute the sample standard deviations s_{1} and s_{2}. In this context the samples are always independent. The populations themselves must be normally distributed.
Test Statistic for Hypothesis Tests Concerning the Difference Between Two Population Variances
If the two populations are normally distributed and if H0:σ21=σ22 is true then under independent sampling F approximately follows an Fdistribution with degrees of freedom df1=n1−1 and df2=n2−1.
A test based on the test statistic F is called an Ftest.
A most important point is that while the rejection region for a righttailed test is exactly as in every other situation that we have encountered, because of the asymmetry in the Fdistribution the critical value for a lefttailed test and the lower critical value for a twotailed test have the special forms shown in the following table:
Terminology  Alternative Hypothesis  Rejection Region 

Righttailed  Ha:σ21>σ22  F≥Fα 
Lefttailed  Ha:σ21<σ22  F≤F1−α 
Twotailed  Ha:σ21≠σ22  F≤F1−α/2 or F≥Fα/2 
Figure \(\PageIndex{3}\) illustrates these rejection regions.
Figure \(\PageIndex{3}\) Rejection Regions: (a) RightTailed; (b) LeftTailed; (c) TwoTailed
The test is performed using the usual fivestep procedure described at the end of Section 8.1.
Example \(\PageIndex{4}\)
One of the quality measures of blood glucose meter strips is the consistency of the test results on the same sample of blood. The consistency is measured by the variance of the readings in repeated testing. Suppose two types of strips, A and B, are compared for their respective consistencies. We arbitrarily label the population of Type A strips Population 1 and the population of Type B strips Population 2. Suppose 15 Type A strips were tested with blood drops from a wellshaken vial and 20 Type B strips were tested with the blood from the same vial. The results are summarized in Table 11.16. Assume the glucose readings using Type A strips follow a normal distribution with variance σ21 and those using Type B strips follow a normal distribution with variance with σ22. Test, at the 10% level of significance, whether the data provide sufficient evidence to conclude that the consistencies of the two types of strips are different.
Strip Type  Sample Size  Sample Variance 

A  n1=16  s21  =2.09
B  n2=21  s22=1.10 
Solution:
Step 1. The test of hypotheses is
Step 2. The distribution is the Fdistribution with degrees of freedom df1=16−1=15 and df2=21−1=20.
Step 3. The test is twotailed. The left or lower critical value is F1−α/2=F0.95=0.43. The right or upper critical value is Fα/2=F0.05=2.20. Thus the rejection region is [0,−0.43]∪[2.20,∞), as illustrated in Figure \(\PageIndex{4}\).
Figure \(\PageIndex{4}\) Rejection Region and Test Statistic for Note 11.27 "Example 6"
Step 4. The value of the test statistic is
Step 5. As shown in Figure \(\PageIndex{4}\), the test statistic 1.90 does not lie in the rejection region, so the decision is not to reject H0. The data do not provide sufficient evidence, at the 10% level of significance, to conclude that there is a difference in the consistency, as measured by the variance, of the two types of test strips.
Example \(\PageIndex{5}\)
In the context of Note 11.27 "Example 6", suppose Type A test strips are the current market leader and Type B test strips are a newly improved version of Type A. Test, at the 10% level of significance, whether the data given in Table 11.16 provide sufficient evidence to conclude that Type B test strips have better consistency (lower variance) than Type A test strips.
Solution:
Step 1. The test of hypotheses is now
Step 2. The distribution is the Fdistribution with degrees of freedom df1=16−1=15 and df2=21−1=20.
Step 3. The value of the test statistic is
Step 4. The test is righttailed. The single critical value is Fα=F0.10=1.84. Thus the rejection region is [1.84,∞), as illustrated in Figure \(\PageIndex{5}\).
 Step 5. As shown in Figure \(\PageIndex{5}\), the test statistic 1.90 lies in the rejection region, so the decision is to reject H_{0}. The data provide sufficient evidence, at the 10% level of significance, to conclude that Type B test strips have better consistency (lower variance) than Type A test strips do.
Key Takeaways
 Critical values of an Fdistribution with degrees of freedom df1 and df2 are found in tables in Chapter 12.
 An Ftest can be used to evaluate the hypothesis of two identical normal population variances.