1
BA 2606
Section 13.4 Inferences on Two Population Variances
In addition to comparing two means, researchers may be interested in
comparing two population variances. For example, is the variation in
two quality control processes different? Another reason may be in
determining which t test to use when comparing two means: the
pooled variance case, or the case of unequal variances.
For the comparison of two variances or standard deviations, an F test
is used. Note that when comparing means, we look at the difference
between the two means. When comparing variances we look to the
s 12
 12
ratios of two variances, or 2 . Not surprisingly, the ratio 2 will be
s2
2
used as the test statistic, where s12 is the larger variance. The
sampling distribution of
s 12
s 22
is the F distribution.
Properties of the F distribution
 The values of F cannot be negative, because variances are
always positive or zero
 The distribution is positively skewed
 The F distribution is a family of curves based on the degrees of
freedom of the variance in the numerator and the degrees of
freedom of the variance in the denominator.
 The assumptions here are that the two independent random
samples come from normally distributed populations
Test Statistic:
F
s 12
s 22
with numerator df  1  n1 1 and denominator
df,  2  n2 1, where s12 is the larger variance.
 12
A 1   100% Confidence Interval for 2 has

2
 s2
LCL   12
 s2

1

F
  2 , 1 , 2
and
 s2
UCL   12
 s2

 F
 2 , 2 , 1

2
Table 6 in Appendix B gives the critical values for the F distribution for
=0.05, 0.025 and 0.01. It is limited since the requirement of two sets
of degrees of freedom for the numerator and the denominator means
1
a lot of numbers. We may use the relationship F1 , , 
for the
2 1 2
F , ,
2
2
1
lower tailed critical value. Choosing the larger of the two sample
variances as the numerator, may save some work.
Example:
Find the critical values for a
a.
right tailed F test when =0.05 and  1  15, 2  21
(2.18)
b,
two tailed F test when =0.05 and  1  20, 2  12
1
(3.07,
)
2.68
When testing the equality of two variances, these hypotheses are
used:
H 0 :  12   22
H 0 :  12   22
H 0 :  12   22
H A :  12   22
H A :  12   22
H A :  12   22
Two-tailed
=
Right-tailed
Left-tailed
Test Statistic:
F
s 12
s 22
with numerator df  1  n1 1 and denominator
df,  2  n2 1
Rejection Region:
a)
two tailed test:Reject H 0 if F > F
2
, 1 , 2
b)
right tailed test:Reject H 0 if F > F , 1 , 2
c)
left tailed test:
or if F < F1
Reject H 0 if F < F1 , 1 , 2 
2
, 1 , 2

1
F
2
, 2 , 1
1
F , 2 , 1
Calculations:
No p-value necessary for the F test due to the limitations of the table.
Conclusion:
3
Exercises 481-482
13.59 b.
 s12 
1
 28  1
LCL   2 

 .6492

 s2  F.025,24,24  19  2.27
 s2 
 28 
UCL   12  F.025,24,24  
 2.27  3.34526
 19 
 s2 
A 95 % Confidence Interval for
 12
is (.6492, 3.34526)
2
2
We are 95% confident that
13.60


2
1
2
falls in this interval.
2
H 0 :  12   22
H A :  12   22
=.05
Test Statistic:
F
s 12
s 22
with numerator df  1  n1 1 and
denominator df,  2  n2 1
Rejection Region:
Reject H 0 if F > F.025,9,10  3.78 or if
F < F.975,9,10 
1
F.025,10,9

1
 .2525
3.96
Calculations:
Machine 1: s = .002394438
Machine 2: s = .0033709993
.002394438

2
.0033709993
2
F

.00000573333
 .5045
.0000113636
Note that the reciprocal is F = 1.982 and the RR for this
would be reject the null hypothesis if F > 3.96 or if
1
F 
 .26455
3.78
Conclusion: Fail to reject H0 . There is insufficient evidence
to conclude that the two machines differ in the consistency
of their fills.
4
13.63
H 0 :  12   22
H A :  12   22
=.05
Test Statistic:
F
s 12
s 22
with numerator df  1  n1 1 and
denominator df,  2  n2 1
Rejection Region:
Reject H 0 if F > F.05,99,99  1.39 or if
Calculations:
Week 1: s12  19.38, n1  100
Week 2: s22  12.70, n2  100
19.38
 1.526
12.70
Conclusion: Reject H0 . There is enough evidence to infer
that limiting both minimum and maximum speeds reduces
the variation in speeds.
F 
Section 13.5
Inference about the Difference between two
Population Proportions
Sampling Distribution of pˆ 1  pˆ 2
X
X
Let pˆ 1  1 and pˆ 2  2 where X 1 and X 2 are the number of
n1
n2
successes in their respective samples. If n1 pˆ 1 , n1 qˆ1 , n2 pˆ 2 , n2 qˆ 2  5 then
  pˆ 1  pˆ 2  is approximately normally distributed
 The mean or expected value of  pˆ 1  pˆ 2  is  p1  p 2 
pq
p q
 The variance of  pˆ 1  pˆ 2  is 1 1  2 2
n1
n2
 The standard error of  pˆ 1  pˆ 2  is
 Therefore Z 
 pˆ 1  pˆ 2    p1  p 2 
p1 q1 p 2 q 2

n1
n2
has a standard normal
p1 q1 p 2 q 2

n1
n2
distribution.
However, this is not exactly the test statistic we will use. There are
two possible cases with two separate test statistics, although both are
5
Z’s with standard normal distributions, there are slight differences in
each of the two.
Case 1:
We test that there is no difference between the
population proportions.
There are three possible sets of hypotheses.
H 0 :  p1  p 2   0
H A :  p1  p 2   0
Two tailed
tailed
H 0 :  p1  p 2   0
H A :  p1  p 2   0
right or upper tailed
H 0 :  p1  p2   0
H A :  p1  p2   0
left or lower
We would like to use the Z statistic from above, but the standard error
p1 q1 p 2 q 2
of  pˆ 1  pˆ 2  is unknown:
and so it must be estimated from

n1
n2
the sample data. When the two population parameters are as
hypothesized, equal, we can pool the data from the 2 samples to come
X  X2
up with a pooled proportion estimate pˆ  1
n1  n 2
 pˆ 1  pˆ 2 
Test Statistic for Case 1:
Z
1
1 
pˆ qˆ   
 n1 n 2 
Case 2:
We test that there exists a specified difference
between the population proportions.
The three possible sets of hypotheses are as follows.
H 0 :  p1  p 2   D0
H A :  p1  p 2   D0
Two tailed
tailed
H 0 :  p1  p 2   D0
H A :  p1  p 2   D0
right or upper tailed
Where D0  0 .
Test Statistic for Case 2:
Z
 pˆ 1  pˆ 2   D0
pˆ 1 qˆ1 pˆ 2 qˆ 2

n1
n2
H 0 :  p1  p 2   D0
H A :  p1  p 2   D0
left or lower
6
There is only one case for interval estimation for the difference in
population proportions:
A (1-)100% Confidence Interval for  p1  p 2  is:
 pˆ 1  pˆ 2   z
2
pˆ 1 qˆ1 pˆ 2 qˆ 2

n1
n2
Exercises pages 493-497
13.69 b.
A 90% Confidence Interval for  p1  p 2  is:
.48  .52   1.645
.48  1  .48   .52  1  .52 
100
100
.04  .1162
 .1562,.0762 
We are 90 % confident that  p1  p 2  falls in this interval.
13.72
Let Population 1 be those who score under 600 on a credit scorecard
and Population 2 be those who score 600 or more
H0 :  p1  p2   0
HA :  p1  p2   0
  .10
Test Statistic:
Z
 pˆ 1  pˆ 2 
1
1 
pˆ qˆ   
 n1 n 2 
Rejection Region: Reject H0 if Z > 1.28
Calculations:
11
7
p1 
 .01957
p2 
 .0087
562
804
11  7
18
p

 .013177
562  804 1366
.01957  .0087
Z 
 1.7336
1 
 1
.013177 1  .013177 


 562 804 
p-value = 1-.9582=.0418
7
Conclusion:
Reject H0. There is enough evidence to conclude that
those who score under 600 are more likely to default than those who
score 600 or more.
13.73 a.
Let Population 1 be voter preference six months ago and Population 2
be voter preference this month.
Also p1 represents the proportion of voters who support this politician
6 months ago and p2 represents the proportion of voters who support
this politician this month.
H 0 :  p1  p2   0
H A :  p1  p2   0
  .05
Test Statistic:
Z
 pˆ 1  pˆ 2 
1
1 
pˆ qˆ   
 n1 n 2 
Rejection Region: Reject H0 if Z < -1.645
Calculations:
X1
p1 
 .56, X1  616
1100
X
p2  2  .46, X2  368
800
616  368
984
p

 .51789
1100  800 1900
.56  .46 
Z 
 4.31
1 
 1
.51789 1  .51789  


 1100 800 
p-value ≈ 0
Conclusion:
Reject H0 at .05 level of significance. There is
enough evidence to conclude that the politician’s popularity has
decreased significantly.
8
13.73 b.
H0 :  p1  p2   0.05
HA :  p1  p2   0.05
  .05
Test Statistic:
Z
 pˆ 1  pˆ 2   D0
pˆ 1 qˆ1 pˆ 2 qˆ 2

n1
n2
Rejection Region: Reject H0 if Z < -1.645
Calculations:
Z 
.56  .46   .05
.56  .44  .46  .54
 2.16
1100
800
p-value = 1 - .9846 = .0154
Conclusion:
Reject H0 at .05 level of significance. There is
enough evidence to conclude that the politician’s popularity has
decreased by more than 5 %.
c.
A 95 % Confidence Interval for  p1  p 2  is .10  .045 ….we are 95
% confident that  p1  p 2  falls in this interval.
Chapter Exercises pages 499 - 502