1
BA 2606
Chapter 13 Inference about the Comparison of Two Populations
The basic concepts of hypothesis testing have been explained in the last two
chapters. Here we extend our discussion to 2 populations, and we will compare two
means, two variances and two proportions. As before, we will use the Z and t test
statistic, but we will introduce a new distribution, the F test statistic, for comparing the
variances of 2 populations. There are basically 6 new test statistics that we must
familiarize ourselves with, understand the theory, the formulas and underlying
assumptions. We start with tests of hypothesis on two population means.
There are many instances when researchers wish to compare the means of two
groups. For example, the average lifetimes of two different brands of bus tires might be
compared to see whether there is any difference in tread wear. Two brands of cough
syrup might be tested to see whether one brand is more effective than the other.
Section 13-1 Inference About the Difference between two Means: Independent
Samples
Sampling Distribution of Xi  X2
 Xi  X2 is normally distributed if the populations are normal (or if the sample
sizes are sufficiently large)
 The expected value or mean of Xi  X2 is 1  2
 The variance of Xi  X2 is
12 22

n1 n2
 The standard error of Xi  X2 is
12 22

n1 n2
estimator  mean
and that confidence intervals are of form
s tan dard.error
estimate Z  s tandard.error , then it follows that in this case:
If you recall that Z 
Z=
X1  X2  1   2 
12 22

n1 n2
and
X
i

 X2  z 
2
12 22

n1 n2
We will not be using either of these since the chance that both population variances are
known is quite rare.
What do we use when population variances are unknown? The t distribution!
2
There are two cases for testing about the difference in two means when the 2 random
samples are independent of each other. For both we must assume that the two
populations are normally distributed.
Case 1:
The two populations are normally distributed. The two samples are
random and independent of each other. The variances are equal.
X

 X2  D0
with   n1  n2  2


1
1

sp2  
 n1 n2 

n1  1s12  n2  1s22
2
is the pooled estimate of the
sp 
n1  n2  2
t
Test Statistic:
1
common population variance  2
A 1  100% Confidence Interval for 1  2 :
X  X   t
i
2

Case2:
Test Statistic:
2
1
1
 with   n1  n2  2
sp2 

 n1 n2 
The two populations are normally distributed. The two samples are
random and independent of each other. The variances are unequal.
t
X
1

 X2  D0
s12 s22

n1 n2
with  
 s12 s22 



 n1 n2 
2
  s 2  2  s2  2 
 2n  
 1 n 
1
2 


 n1  1
n2  1 




And round df to the nearest integer.
Some textbooks use an approximation for this df = minimum of (n 1 – 1) or (n2 – 1)
3
A 1  100% Confidence Interval for 1  2 :
X
i

 X2  t 
 s12 s22 



n
n
2 
 1
2
2
 s12 s22 


2
2
 n  n  with    2
2 
 1
 s1 
 s22  
 n  
 n 
1
2 


 n1  1
n2  1 




There are 3 possible sets of hypotheses:
H 0 : 1   2   D0
H A : 1   2   D0
Two tailed
H 0 : 1   2   D0
H A : 1   2   D0
right or upper tailed
H 0 : 1   2   D0
H A : 1   2   D0
left or lower tailed
Use the same procedure for hypothesis testing.
The question is: how do we know whether the variances are equal or unequal? We need
to conduct another test on the ratio of the 2 variances, which we will cover in Section
13-4.
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 ratios of two variances, or
 12
. Not surprisingly, the
2
2
ratio
s 12
2
s2
will be 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
4
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. By choosing the numerator variance to be the larger of
the two sample statistics we need only find the upper tail critical value.
We will revisit this in Section 13-4 and we will learn how to find the lower tail critical
value there
Example:
a. Find the critical value for a right tailed F Test with 0.05 level of significance when the
degrees of freedom for the numerator are 15 and the degrees of freedom for the
denominator are 20 (2.20)
b. Find the upper critical value for the two tailed F test with 0.05 level of significance
when the sample size from which the variance from the numerator was obtained was 21
and the sample size from which the variance from the denominator was obtained was
12 (3.23)
Exercises pages 456 – 460
13.12 H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
α = 0.10, Note: assume both populations are normal, both samples are random
Two-tail F test: s1 = 2.424412873, s2 = 2.406010991
F = 1.015355, since this value is not > F.05, 9, 9 = 3.18, we fail to reject the hypothesis that
the population variances are equal ; use equal-variances test statistic
Test Statistic: t 
( x 1  x 2 )  ( 1   2 )
 1
1 

s 2p 

n
n
2 
 1
Rejection region: Reject H0 if t   t ,   t .10,18  1.330
t
( x 1  x 2 )  ( 1   2 )
 1
1
s 2p 

 n1 n 2



=
(5.10  7.30 )  0
 (10  1)5.88  (10  1)5.79  1 1 

  
10  10  2

 10 10 
p-value = .0283 by computer program
By table: .025 < p value < .05
 2.04,
5
Conclusion:
Reject H0 at α = .10. There is enough evidence to infer that there are fewer errors when
the yellow ball is used.
13.29 H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
α = 0.05, Note: assume both populations are normal, both samples are random
From Appendix A:
With Textbook:
x 1  63.71 s1 = 5.90, n1 = 173
No Textbook:
x 2  66.80 , s2 = 6.85, , n2 = 202
Two-tail F test: F = 1.348, since this value is not > F.025, 201, 172 ≈ 1.32, we reject the
hypothesis that the population variances are equal ; use unequal-variances test statistic
Test Statistic: t 

( x 1  x 2 )  ( 1   2 )
 s 12 s 22 



 n1 n 2 


(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 372.9862, round to 373
Rejection region: t  t ,  t.05,373  1.653
t
( x 1  x 2 )  ( 1   2 )
 s 12 s 22 



 n1 n 2 


=
(63 .71  66 .80 )  0
 5.90 2 6.85 2 



 173
202 

= –4.69, p-value < .005.
Calculations:
Reject H0 at α = .05. There is enough evidence to infer that students without textbooks
outperform those with textbooks.
6
Section 13-2 Observational and Experimental Data
We can obtain our data in two different ways:
1) Experiment (or controlled study) – where the researcher randomly divides
subjects into appropriate groups. Some treatment is applied to one or more
groups and the effect or response is observed. For example, patients may be
randomly given unmarked capsules of either aspirin or acetaminophen and the
effects of the medication may be measured for each group. Experiments often
have a treatment group and a control group where, ideally, neither the subjects
nor researchers know which group is which. In the Salk vaccine experiment of
1954, half the children received the polio vaccine while the other received a
placebo – their doctors did not even know who received what. This is an
example of a double-blind approach. Blinding occurs when the subjects and
researchers do not know who is receiving what treatments. Controlled
experiments can indicate cause and effect relationships.
2) Observational Study – Here, there is no choice as to who goes into the
treatment or control group. Specific characteristics are observed and measured
but the researcher does not attempt to modify the subjects studied. For
example, a researcher cannot ethically tell 100 people to smoke 3 packs of
cigarettes a day and another 100 to smoke only a pack a day. They can only
observe people who habitually smoke these amounts. Results of such studies can
suggest relationships, but it is difficult to conclude cause and effect.
In an experiment we impose some change or treatment and measure the result or
response. In an observational study we simply observe and measure something that
has taken place or is taking place, while not trying to cause any changes by our
presence.
Which is more appropriate, an experiment or observational study?
a) a study designed to determine whether daily calcium supplements benefit
women by increasing bone mass? Experimental
b) a study designed to examine life expectancies of tall versus short people.
Observational (examine medical records of heights and ages at time of death).
Experimental, choosing half to be short, half to be tall, makes no sense.
Exercises page 466
13.90 It is an experimental study if volunteers are randomly assigned to eating oat bran
versus another grain cereal
13.91 a. Observational, bc students choose the software package
7
b. If students are randomly assigned to a software package it becomes an
experimental study
Section 13-3 Inference About the Difference between Two Means: Matched Pairs
Experiment
In this section, our two samples are dependent samples, where the subjects are paired
or matched in some way. For example, heart rate, before and after taking a certain
medication. We will define our differences Di  X i  Yi and we are interested in the
population mean of the differences. We make the assumption that the differences are
normally distributed and we use the t distribution.
There are 3 possible sets of hypotheses:
H 0 :  D  D0
H 0 :  D  D0
H 0 :  D  D0
H A :  D  D0
Two tailed
H A :  D  D0
H A :  D  D0
right or upper tailed left or lower tailed
Use the same procedure for hypothesis testing.
Test Statistic: t 
X D  D0
sD
nD
with   n D 1
A 1   100% Confidence Interval for  D is
X D  t
sD
2
nD
Example
A dietician wishes to see if a person’s cholesterol level will change if the diet is
supplemented by a certain mineral. Six subjects were pretested and then took the
mineral supplement for a six week period. The results are shown in the following table.
(Cholesterol level is measured in milligrams per deciliter) Can it be concluded that the
cholesterol level has been changed at .10 level of significance? Assume that cholesterol
levels are normally distributed. Also construct a 90 % confidence interval for the mean
of the differences.
Subject
Before
After
1
210
190
2
235
170
3
208
210
4
190
188
5
172
173
6
244
228
8
Define D = Before – After
H0 : D  0
HA : D  0
  0.10
Test Statistic: t 
X D  D0
, 5
sD
nD
Rejection Region:
Reject H0 if t < -2.015 or t > 2.015
Calculations:
Differences = 20, 65, -2, 2, -1, 16
 Di  100,  Di2  4890, d  16.6, sD  25.39
t
16.6  0
, 2(0.05) < p-value < 2(.10)
 1.607
25.392
6
Conclusion: Fail to reject H0 at .05 level of significance. There is insufficient
evidence to support the claim that the mineral changes a person’s cholesterol
level.
Note: A 90 % Confidence Interval for μD is:
 25.39 

16.6  2.015
6 

16.6  20.89
 4.19,37.59
We are 90% confident that μD falls in this interval. Note that 0 falls within this
interval…..which implies the null hypothesis that μD = 0 cannot be rejected.
Exercises Pages 478-479
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 ratios of two variances, or
 12
. Not surprisingly, the
2
2
9
ratio
s 12
2
s2
will be used as the test statistic, where s12 is the larger variance. The sampling
distribution of
s 12
is the F distribution.
s 22
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.
A 1   100% Confidence Interval for
 12
has
2
2
 s2
LCL   12
 s2

1

F
  2 , 1 , 2
s 
and UCL    F , ,
2 1
s  2
2
1
2
2
Table 6 in Appendix B gives the critical values for the F distribution for =0.05, 0.025,
0.01 and 0.005. It is limited since the requirement of two sets of degrees of freedom for
the numerator and the denominator means a lot of numbers. We use the relationship
1
for the lower tailed critical value. Choosing the larger of the two
F1 , , 
2 1 2
F , ,
2
2
1
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)
two tailed F test when =0.05 and  1  20, 2  12
(3.07,
b,
1
)
2.68
10
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
or if F < F1
b)
right tailed test:Reject H 0 if F > F , 1 , 2
c)
left tailed test: Reject H 0 if F < F1 , 1 , 2 
2
, 1 , 2

1
F
1
F , 2 , 1
Calculations:
No p-value necessary for the F test due to the limitations of the table.
Conclusion:
Exercises on page 486-487
13.121 b.
 s2 
1
 28  1
LCL   12 

 .6492

 s2  F.025,24,24  19  2.27
 s12 
 28 
UCL   2  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


2
1
2
2
falls in this interval.
2
, 2 , 1
11
13.122
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
1
 .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.
would be reject the null hypothesis if F > 3.96 or if F 
Example:
The CEO of an airport hypothesizes that the variance for the number of passengers for
American airports is greater than the number of passengers for foreign airports. At α =
0.10 is there enough evidence to support this hypothesis? The data in millions of
passengers per year are shown for selected airports. Assume the variable is normally
distributed.
American airports
Foreign airports
36.8
73.5
60.7
51.2
72.4
61.2
42.7
38.6
60.5
40.1
12
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 successes in their
n1
n2
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
p1 q1 p 2 q 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 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 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
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 tailed
We would like to use the Z statistic from above, but the standard error of  pˆ 1  pˆ 2  is
p1 q1 p 2 q 2
and so it must be estimated from the sample data. When the

n1
n2
two population parameters are as hypothesized, equal, we can pool the data from the 2
X  X2
samples to come 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 
unknown:
13
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 0 :  p1  p 2   D0
H A :  p1  p 2   D0
Two tailed
H A :  p1  p 2   D0
right or upper tailed
H 0 :  p1  p 2   D0
H A :  p1  p 2   D0
left or lower tailed
Where D0  0 .
Test Statistic for Case 2:
Z
 pˆ 1  pˆ 2   D0
pˆ 1 qˆ1 pˆ 2 qˆ 2

n1
n2
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 498 – 499
13.133 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.
13.137 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.
14
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.
13.137 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
1100
800
p-value = 1 - .9846 = .0154
 2.16
15
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.
13.148 H 0 : (p1  p 2 ) = 0
H1 : ( p1  p 2 )  0 Note: wording is incorrect in text
Rejection region: z  z   z .05  1.645
p̂1 
z
47
64  47
64
 .3422 p̂ 2 
 .2568 p̂ 
 .3000
187
183
187  183
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



=
(. 3422  .2568 )
1 
 1
.3000 (1  .3000 )


 187 183 
 1.79 , p-value = P(Z > 1.79) = 1- .9633 =
.0367
There is enough evidence to allow us to conclude that drivers behind a male driver are
more likely to honk.