814 CHAPTER 39. STATISTICAL TESTS

=Γ( r1+r2

2

)( r1r2

)r1/2

Γ(r1/2)Γ(r2/2)f (r1/2)−1(

f r1r2

+1) r1+r2

2

, f > 0

Note that if r1 = r2 = r, this is much less ugly. It then reduces to

Γ(r)

Γ(r/2)2f (r/2)−1

( f +1)r

You can probably see that this F distribution could be used to test the ratio of variancescoming from two normal densities and obtain a confidence interval for this ratio. If thisinterval did not contain 1, then you could conclude that with a certain probability the twovariances are different. Here is a graph of F (x) ≡ P(X ≤ x) where X is an F randomvariable with r = r1 = r2 = 10.

0 5 100

0.5

1

39.2.5 Confidence Intervals for the Ratio of Two VariancesSuppose you have two random samples of length r taken from two normal distributionshaving variances σ2

1 and σ22. The symbols which represent such normal distributions are

n(µ i,σ

2i). Let these be {X1, · · · ,Xr} from n

(µ1,σ

21)

and {Y1, · · · ,Yr} from n(µ2,σ

22). Let

rS21 ≡

r

∑k=1

(Xk− X̄)2, rS2

2 ≡r

∑k=1

(Yk− Ȳ )2

Then the rS2i /σ2

i is X 2 (r−1) . It follows from the above that

rS21/σ2

1

rS22/σ2

2=

σ22

σ21

rS21

rS22

is distributed as an F random variable with the parameter equal to r−1.

Example 39.2.9 Find a .9 confidence interval for the ratio σ22

σ21

where you have two random

samples, of length 11 taken respectively from two normal distributions. n(µ1,σ

21)

andn(µ2,σ

22). These samples are {−3,2,−1,0,1,−2, .5, .4,−.2,−.5, .3} and

{−4,−7,7,10,15,5,−8,11,12,−12,−5} . You can see that the second sample is muchmore spead out than the first. Thus, they should have different variances. Does the confi-dence interval predict this?

Some computations show that 11S21 = 19.22 and 11S2

2 = 906.64. Now, from the graphof F (x)≡ P(X ≤ x) where X has F distribution with r1 = r2 = r = 10 given above, using