The F Distribution
Named in honor of R.A. Fisher who studied it in
1924.
Used for comparing the variances of two
populations.
It is defined in terms of ratio of the variances of
two normally distributed populations. So, it
sometimes also called variance ratio.
F–distribution:
where
2
2
2
2
2
1
2
1
s
s
Degrees of freedom: v
1
= n
1
– 1 , v
2
= n
2
– 1
For different values of v
1
and v
2
we will get
different distributions, so v
1
and v
2
are parameters
of F distribution.
If σ
1
2
= σ
2
2
, then , the statistic F = s
1
2
/s
2
2
follows F
distribution with n
1
– 1 and n
2
– 1 degrees of
freedom.
1
1
2
2
222
2
1
2
112
1
n
xx
s
n
xx
s
Probability density function
Probability density function of F-distribution:
where Y
0
is a constant depending on the values of
v
1 and v
2 such that the area under the curve is
unity.
21
21
2
1
12
1
F
YFf
o
Properties of F-distribution
It is positively skewed and its skewness
decreases with increase in v
1
and v
2
.
Value of F must always be positive or zero,
since variances are squares. So its value lies
between 0 and ∞.
Mean and variance of F-distribution:
Mean = v
2
/(-v
2
-2), for v
2
> 2
Variance = 2v
2
2
(v
1
+v
2
-2) , for v
2
> 4
v
1
(v
2
-2)
2
(v
2
-4)
Shape of F-distribution depends upon the
number of degrees of freedom.
Testing of hypothesis for equality of two variances
It is based on the variances in two independently
selected random samples drawn from two normal
populations.
Null hypothesis H
o
: σ
1
2
= σ
2
2
F = s
1
2
/σ
1
2
, which reduces to F = s
1
2
s
2
2
/σ
2
2
s
2
2
Degrees of freedom v
1
and v
2
.
Find table value using v
1
and v
2
.
If calculated F value exceeds table F value, null
hypothesis is rejected.
Problem 1
Two sources of raw materials are under
consideration by a company. Both sources seem
to have similar characteristics but the company is
not sure about their respective uniformity.
A sample of 10 lots from source A yields a
variance of 225 and a sample of 11 lots from
source B yields a variance of 200. Is it likely that
the variance of source A is significantly greater
than the variance of source B at α = 0.01 ?
Solution
Null hypothesis H
o
: σ
1
2
= σ
2
2
i.e. the variances of
source A and that of source B are the same. The F
statistic to be used here is
F = s
1
2
/ s
2
2
where s
1
2
= 225 and s
2
2
= 200
F = 225/200 = 1.1
Value for v
1
=9 and v
2
=10 at 1% level of
significance is 4.49. Since computed value of F is
smaller than the table value of F, the null
hypothesis is accepted.
Hence the variances of two populations are same.
Problem 2
In order to test the belief that incomes earned by
college graduates show much greater variability
than the earnings of those who did not attend
college. A sample of 21 college graduates is taken
and the sample std. deviation for the sample is s
1
=
Rs 17,000. For the second sample of 25 non-
graduates and obtains a std dev. s
2
= Rs 7,500.
s
1
= Rs 17,000 n
1
= 21
s
1
= Rs 7,500 n
2
= 25
Solution
Null hypothesis H
o
: σ
1
2
= σ
2
2
or (σ
1
2
/ σ
2
2
= 1)
Alternative hypothesis H
1
: σ
1
2
> σ
2
2
Level of significance α = 0.01
F = s
1
2
/ s
2
2
F = (17,000)
2
/ (7,500)
2
F = 5.14
Value for v
1
=20 and v
2
=24 at 1% level of
significance is 2.74. Since computed value of F is
greater than the table value of F, the null
hypothesis is rejected.