Psych 5500/6500
Introduction to the F Statistic
(Segue to ANOVA)
Fall, 2008
1
Overview of the F test
The F test is used in many contexts. We will begin
by taking a general look at how the F test works.
In its most general form, the F test is used to
determine whether two populations have the
same variance.
2
Example
We know that the mean height of men is greater
than the mean height of females, but what about
their respective variances? (two-tail example)
H0: σ²Female = σ²Male
HA: σ²Female  σ²Male
3
Test Statistic
The test statistic is:
2
1
2
2
est.σ
Fobt 
est.σ
If H0 is true, then both estimates are independently
estimating the same thing, and thus should roughly
equal each other (they won’t exactly equal each
other due to random bias), and thus if H0 is true
then the value of Fobt should be around 1.
4
est.σ12
Fobt 
2
est.σ 2
Degrees of Freedom
There are two different degrees of freedom in the
F test, one for the numerator and one for the
denominator. Remember that:
SS1
est. 
N1  1
2
1
SS 2
est. 
N2 1
2
2
The numerator has df1 = N1-1 and the
denominator has df2 = N2-1
5
Expected (Mean) Value of F
2
1
2
2
est.σ
Fobt 
est.σ
Again, if H0 is true and both populations have the same
variance then we would expect est.σ²1 to approximately
equal est.σ²2 and thus F should be around 1. Random
bias in the value of the denominator has a strange effect
on the value of F, however, when est.σ²2 is way larger
than est.σ²1 it drives the value of F from 1 down towards
0, but when est.σ²2 is way smaller than est.σ²1 it drives
the value of F from 1 up towards infinity. The end result
of this is that...
6
Expected (Mean) Value of F
est.σ12
Fobt 
est.σ 22
df 2
WhenH0 is true: μ F 
(df 2  2)
Remember that df2 is the df for est.σ²2 (i.e. N2-1). Thus if
sample 2 has 30 scores in it, then df2 would equal 29, and
the mean value of F when H0 is true would be:
29
μF 
 1.07
27
7
Expected (Mean) Value of F
est.σ12
Fobt 
est.σ 22
df 2
WhenH0 is true: μ F 
(df 2  2)
As the N of group 2 gets larger then est.σ²2
becomes more accurate and the expected
value of F gets closer to 1. For example, if
N2= 500 then when H0 is true μF=1.004.
Bottom line: if H0 is true then est.σ²1 est.σ²2
and μF  1 rather than the more intuitively
reasonable μF=1.
8
Sampling Distribution
Now that we have a test statistic (F) we can look at
the ‘Sampling Distribution of F assuming H0 is
true’.
The mean value of F will be close to 1...actually
df2/(df2-2)...if H0 is true. The sample
distribution is not a normal distribution, or a t
distribution, it is not even symmetrical, as it has
a mean close to ‘1’, but the lowest value F can
take on is zero and the highest value is infinity.9
Shape of the F distribution
The shape of the F distribution is dependent upon the degrees of
freedom of both the numerator and denominator. Red has df1=2 and
df2=3, blue has df1= 4 and df2=30, and black has df1= 20 and df2=20.
10
Hypotheses
Two-tail test:
H0: σ²1 = σ²2
HA: σ²1  σ²2
One-tail test predicting σ²1 < σ²2
H0: σ²1  σ²2
HA: σ²1 < σ²2
One-tail test predicting σ²1 > σ²2
H0: σ²1  σ²2
HA: σ²1 > σ²2
11
Fc values
As the shape of the F distribution changes with
different degrees of freedom, you need to know
both df to find the Fc values.
Remember:
df1 (i.e. for the numerator of F)= N-1 for est. σ²1
df2 (i.e. for the denominator of F) = N-1 for est. σ²2
12
Fc values
Because of the way the F test is used in ANOVA
(which we will get to later) Fc tables rarely
have the left-tail Fc value. The F distribution
tool I provide makes it easy to find the Fc
values (enter a p of .975 and then a p of .025).
The left-tail Fc value can also computed fairly
easily from a table that only has right-tail Fc
values.
13
Calculating Fc Left Tail
Fcritical,left tail:df1 ,df 2 
1
Fcritical,right tail:df 2 ,df1
Note the switch of df in the Fc right tail.
Example:
df1  15, df2  10, α  .05, two - tailedtest
Fcritical,right tail, .025  3.52
Fcritical,left tail,.025
1

 0.33
3.06
14
Back to Our Example
We know that the mean height of men is greater
than the mean height of females, but what about
their respective variances?
H0: σ²Female = σ²Male
HA: σ²Female  σ²Male
NFemale=16, NMale=11
Set up the sampling distribution of F assuming H0
is true. μF=10/8=1.25 if H0 true.
Fc = 0.33 and 3.52
15
Sampling Distribution of F
16
Computations (by hand)
Females: NFemale=16, SSFemale=46
Males: NMale=11, SSMale=25
SS
46


 3.06
N  1 15
SS
25
2
est. Male 

 2.5
N  1 10
2
est. Female
3.06
Fobt 

 1.22
2
2.5
est. Male
2
est. Female
17
Computations SPSS
While SPSS doesn’t provide this use of the F test it
will provide the ‘Variance’ of each group.
Remember that in SPSS the ‘Variance’ of a group
is actually the est. σ² of the population from
which the sample was drawn, which is just what
we need to compute F. You will still, however,
need an F table to come up with the Fcritical
values.
18
Decision
H0: σ²Female = σ²Male
HA: σ²Female  σ²Male
If H0 is true then we would expect F to approximately
equal 1.25. If H0 is false we would expect F to not
equal 1.25 In this case Fobtained = 1.22, does this differ
enough from what H0 predicted to reject H0? Mark the
approximate location of F=1.22 on the ‘sampling
distribution of F assuming H0 is true’ to see if you can
reject H0. In this case we ‘do not reject H0’, we were
unable to determine whether or not the two population
variances differ.
19
One-Tail Test
If we were testing a theory which predicted that
women have a greater variance:
H0: σ²Female  σ²Male
HA: σ²Female > σ²Male
We need to look up the one-tail Fcritical value
(upper tail in this case). If H0 is true then we
would expect F to be less than or equal to 1.25.
If H0 is false we would expect F to be greater
than 1.25 (which is where we will put the
rejection region).
20
Sampling Distribution of F
21
One-Tail Test
If we were testing a theory which predicted that
women have lesser variance:
H0: σ²Female  σ²Male
HA: σ²Female < σ²Male
We need to look up the one-tail Fcritical value
(lower tail in this case). If H0 is true then we
would expect F to be greater than or equal to
1.25. If H0 is false we would expect F to be
less than 1.25 (which is where we will put the
rejection region).
22
Sampling Distribution of F
23
Assumptions of this Use of F
1. The two variance estimates are independent of
each other.
2. Both populations are normally distributed.
Monte Carlo studies have shown that this
assumption is quite important for the validity of
this test.
24
Back to the Assumptions of the t Test
One of the assumptions of the t test for independent means
is that the variances of the two populations are equal.
The F test we have just covered can test that assumption.
But remember, due to the nature of null hypothesis
testing, we can prove two variances are different but we
can’t prove two variances are equal, because we can’t
prove that H0 is true (unless we can show we have a
powerful experiment, which would make beta small).
The affect that non-normality has on the validity of this
F test has led to it not being used as much as Levene’s
test (mentioned next).
25
Levene’s Test
Levene’s test is another way to determine whether or not
the population variances are the same. Levene’s test has
two advantages over the F test we just covered. First, it
is less dependent upon the populations being normally
distributed. Second, it can be used to test whether
several groups all have the same variance.
H0: σ²1 = σ²2 = σ²3 …
HA: at least one σ² is different than the rest.
We will cover Levene’s test and how it works soon.
26