Chapter 9: Inferences Involving
Two Populations
H 0 : 1   2
H a : 1   2
1
2
Independent and Dependent Samples
• The object is to compare means of two
samples and draw conclusions about the
differences in population means.
• Two basic kinds of samples: independent
and dependent (paired).
• Which kind you have depends on the
sources of two samples and how the data
was collected.
Dependent Samples
• If one observation is collected for each sample
from the same source, the samples are dependent.
• This is often called “pair data” because you get a
pair of observations from one individual or
experimental unit.
• Examples include pretest & posttest scores, weight
before and after a diet, left eye and right eye
acuity, etc.
• There is a one-to-one correspondence between an
observation in one sample and an observation in
the other sample.
Independent Samples
• Two samples are independent if there is no
connection between an observation of one
sample with a particular observation of the
other.
• Also, there can be no connection in the
sampling procedure (an individual selected
for one in no way affects the selection of
any individual in the other, including by
exclusion)
Dependent Sample Examples
• The same test is given to all students at the
beginning and end of a course to measure
learning (one pair of scores per person).
• IQ tests are given to husband & wife pairs.
• A medical treatment is given to patients
matched for condition, age, sex, race,
weight, and other characteristics with
patients in a control group.
Independent Sample Examples
• The same test is given to all students in two
classes (no pairing of scores occurs).
• IQ tests are given to men and women
without consideration of relationship
between any of them.
• Subjects are randomly assigned to a
treatment and a control group to test a new
drug. No attempt is made to “match” them.
Difference of means for Paired Data
• When dependent samples are involved, the
data is paired data.
• Paired data results from:
– before and after studies,
– a common source, or
– from matched pairs.
• We will denote the random variables from
the two samples by X1 and X2.
The meaning of X1 and X2
• Two samples are being taken. For example,
the pretest is one sample, and the posttest is
another sample.
• Then X1 represents a pretest score and X2
represents a posttest score.
• However, there is an X1 and an X2 for each
person taking the test.
Paired Data
X1
x11
x12
x13
x14
x15
x16
x17
…
X2
x21
x22
x23
x24
x25
x26
x27
…
Here the random variable names
are given by capital letters, and
individual observations by small
letters. The data appear side-byside, each pair having the same
second subscript. You cannot
change the order of one column
without destroying the relationship between the columns. That
is what makes it “paired data.”
There are the same number of
observations, n, in each sample.
What do we want to know?
• In paired data studies, the parameter of
interest is the mean difference between the
groups.
• This is conceptually different from the
difference between the means of the groups.
• In other words, the population of interest is
actually the differences between X1 and X2 .
• We define a new value, d=x1-x2 as one
observation taken from this population.
Why is this important?
• The mean difference between the groups
and the difference between the means of the
groups are the same number.
• But their sampling distributions are
different!
• From d  x1  x2 , we calculate d , the mean
difference.
• Now, d will have a normal distribution if
X1 and X2 are normal or n>30 (approx).
Distribution of mean differences
• If we know d is normally distributed, then
we can use the same tests and confidence
intervals that we learned for x .
• We won’t bother with the “variance known”
situation this time. We will calculate the
variance from the sample and use the t
distribution.
• In other words, treat the d’s as the sample.
Find their mean and standard deviation.
Distribution of mean differences
• There is a population parameter, d , that we
are trying to estimate.
• The point estimate is d , taken from a
sample of n differences (d’s).
• The d’s have a standard deviation, sd which
is calculated in the same way as s.
sd
• The standard deviation of d is sd 
n
• This is no difference from what
we had before, except for symbols!
Confidence Interval for Paired
Differences
• A (1-α)100% CI for d is given by:
d  t n-1, 2 sd , where sd  sd .
n
Hypothesis Testing:
When testing a null hypothesis about the mean difference, the
test statistic is
d  0 d
t* 
sd n
where t* has a t distribution with df = n  1.
Example: The corrosive effects of various chemicals on
normal and specially treated pipes were tested by using a
dependent sampling plan. The data collected is summarized
by
n  17, d  5.7, sd  4.8
where d is the amount of corrosion on the treated pipe
subtracted from the amount of corrosion on the normal pipe.
Example (continued): Does this sample provide sufficient
evidence to conclude the specially treated pipes are more
resistant to corrosion? Use  = 0.05
a. Solve using the classical approach.
b. Solve using the p-value approach.
Solution:
1. State the hypotheses (you must say something about the
direction of the difference):
Test for the mean difference in corrosion, normal pipe treated pipe.
The null and alternative hypothesis:
H0: d = 0 (did not lower corrosion)
Ha: d > 0 (did lower corrosion)
2. Determine the appropriate type of test:
Assumptions: Assume corrosion measures are
approximately normal, σ unknown.
Use t-test for paired differences.
3. Define the rejection region:
a. Right tailed test, Reject H0 if t*>t(16,0.05) = 1.75.
b. Reject H0 if p<.05.
4. Calculate the value of the test statistic:
n  17, d  5.7, sd  4.8
t* 
d  0 d
sd
5.7  0.0
5.7


 4.896
n 4.8 17 1.164
p  P(t*  4.896)  .0001 by the table
p  P(t*  4.896)  0.00006731 using Excel
5. State the conclusion:
a. Decision: Reject H0 because t*=4.896>1.75
b. Decision: Reject H0 because p<.0001<α=.05.
Conclusion: The treated pipes do not corrode as much as
the normal pipes when subjected to chemicals.
Example: Salt-free diets are often prescribed for people with
high blood pressure. The following data was obtained from
an experiment designed to estimate the reduction in diastolic
blood pressure as a result of following a salt-free diet for two
weeks. Assume diastolic readings to be normally distributed.
Before
93
106
87
92
102
95
88
110
After
92
102
89
92
101
96
88
105
1
4
-2
0
1
-1
0
5
Difference
Test the hypothesis that the salt free-free diet resulted in a
reduction in blood pressure, using α=0.02.
Question: How do you decide which way to subtract?
Solution:
1. A reduction means “before”>“after” so we will use “before-after” as our
difference, which is expected to be positive.
H0: d = 0 (no reduction)
Ha: d > 0 (reduction occurred)
2. Determine the type of test: This is a t-test of paired differences.
The mean reduction (difference) in diastolic blood pressure is being tested.
Both before and after scores are assumed normally distributed, so d
is normal, and σ is unknown, so we use the t-test with 7 df.
3. Rejection Region:
Reject H0 if t*>t(7,0.02)=2.52 (from Excel) or if p<0.02.
4. Calculate the test statistic and/or p-value:
n  8, d  1.0, sd  2.39,   0.02
t* 
5.
d 0
1

 1.18 p  0.1383
sd
2.39 / 8
Conclusion:
Since t* is not in the rejection region, and p is not less than α, we do not
reject H0. We conclude there is not enough evidence to say that the diet
reduces blood pressure.
Two Independent Samples
•
•
•
•
Compare the means of two populations
Parameter of interest: (1 - 2)
Base inferences on ( x1  x2 )
The parentheses indicate that we are
thinking of the difference as one parameter
• Consider the general confidence interval
formula, P±TS. We know what P is now.
• We need to know the distribution of ( X 1  X 2 )
to find T and S.
Distribution of
( X 1  X 2 ).
• The sampling distribution of ( X 1  X 2 )
has a mean,  x1  x2  ( 1   2 )
• The point estimate of  x1  x2 is ( x1  x2 )
• The standard deviation of ( X 1  X 2 ) is
 x x 
1
2
  12

 n1
   22 


n
  2 
• Since the variances are hardly ever known,
we will have to estimate them.
Sample Standard Deviation
• The sample standard deviation of ( X 1  X 2 ) is
sx1  x2 
 s12

 n1
  s22 


n
  2
• The following assumptions are needed to use
the above formula:
– The samples are randomly selected from normally
distributed populations
– The samples are independent
– There is no reason to believe σ1=σ2
– The populations (not samples) are “large”
Distribution
• The t distribution will be used.
• Degrees of freedom:
– If n1=n2, no problem, df=n1-1.
– Otherwise, df may be calculated by a
complicated formula. Statistical computer
software will do this automatically.
– Alternatively, the smaller of n1-1 and n2-1 can
be used as an approximation. (conservative—
actual confidence level will be higher, actual pvalue will be lower)
Confidence Interval
• Now we have all the information we need.
• P= ( x1  x2 )
T=t(df,α/2)
S=
 s12

 n1
  s22 


n
  2
• A (1-α)100% confidence interval for (1-2)
is given by ( x1  x2 )  t(df , / 2)
 s12

 n1
  s22 
 
  n2 
Hypothesis Tests:
To test a null hypothesis about the difference between two
population means, use the test statistic
t* 
( x1  x2 )  (  01   02 )
 s12   s22 



n
n
 1  2
where df is the smaller of df1 or df2 when computing t*
without the aid of a computer.
Note: The hypothesized difference between the two
population means (01  02) can be any specified value. The
most common value is zero.
Example: A recent study compared a new drug to ease postoperative pain with the leading brand. Independent random
samples were obtained and the number of hours of pain relief
for each patient were recorded. The summary statistics are
given in the table below.
Pain Reliever
New Drug
Leading Brand
n
10
17
Mean
4.350
3.929
St.Dev.
0.542
0.169
Is there any evidence to suggest the new drug provides longer
relief from post-operative pain? Use  = 0.05
a. Solve using the p-value approach.
b. Solve using the classical approach.
Solution:
1. The Hypotheses:
H0: 1  2 = 0 (new drug relieves pain no longer)
Ha: 1  2 > 0 (new drug works longer to relieve pain)
2. The appropriate test:
Assumptions: Both populations are assumed to be
approximately normal. The samples were random and
independently selected.
Use t-test for difference of means, indep. samples, df = 9.
3. Rejection Region: Reject if t*>t(9, 0.05) = 1.83 or p<.05.
4. Calculations:
( x1  x 2 )  ( 1   2 ) ( 4.350  3.929)  (0.00)

 s12   s22 
 0.542 2   0.169 2 
    

  

n
n
10
17
 1  2

 

0.421
0.421


 2.39
0.0294  0.0017 0.1763
t* 
4. (cont’d)
The p-value: P  P(t*  2.39, with df  9)  0.020
5. The Conclusion:
Reject H0.
There is evidence to suggest that the new drug provides
longer relief from post-operative pain.
Sample Standard Deviation—Equal Variance
• If σ1=σ2,
X

t
1
 X 2    1   2 
sP
sP 2
2
 1
1 
 

n
n
 1
2 
n1  1  s12   n2  1  s2 2


(n1  1)  ( n2  1)
df  (n1  1)  ( n2  1)  n1  n2  2
• The formula also assumes that the populations are
normal, samples are random and independent.
Example: We modify the previous drug study
example. Suppose the statistics are as given below,
and the variances of the two populations are equal.
Pain Reliever
n
Mean
St.Dev.
New Drug
Leading Brand
10
17
4.350
3.929
0.542
0.469
Is there any evidence to suggest the new drug
provides longer relief from post-operative pain? Use
 = 0.05
a. Solve using the p-value approach.
b. Solve using the classical approach.
Solution:
1. The Hypotheses:
H0: 1  2 = 0 (new drug relieves pain no longer)
Ha: 1  2 > 0 (new drug works longer to relieve pain)
2. The appropriate test:
Assumptions: Populations normal, samples random and
independently selected, variances equal.
Use t-test for difference of means, indep. samples, df = 25.
3. Rejection Region: Reject if t*>t(25, 0.05) = 1.71 or p<.05.
4. Calculations:
sP
2
2
n

1

s

n

1

s




1
2
2
2
 1
(n1  1)  (n2  1)
t* 
( x1  x 2 )  ( 1   2 )

9  0.5422  16  0.469 2

 0.2465
9  16
(4.350  3.929)  (0.00)
1 1
 1 1
0.2465   
sP    
 10 17 
 n1 n2 
0.421
0.421


 2.18
0.0294(0.1588) 0.1979
2
The p-value: p  P(t*  2.18, with df  25)  0.019
5. The Conclusion:
Reject H0.
There is evidence to suggest that the new drug
provides longer relief from post-operative pain.
Difference of Two Proportions
If independent samples of sizes n1 and n2 are drawn randomly
from large populations with p1 = P1(success) and
p2 = P2(success), respectively, then the sampling distribution of
p1  p2 has these properties:
1. a mean  p1  p2  p1  p2
2. a standard error
 p  p 
1
2
p1q1 p2 q2

n1
n2
3. an approximately normal distribution if n1 and n2 are
sufficiently large.
Note: To ensure normality:
1. The sample sizes are both larger than 20.
2. The products n1p1, n1q1, n2p2, n2q2 are all larger than 5.
Since p1 and p2 are unknown, these products are estimated
by n1 p1 , n1q1 , n2 p2 , n2 q2
3. The samples consist of less than 10% of respective
populations.
Confidence Intervals:
1. A confidence interval for p1  p2 is based on the unbiased
sample statistic p1  p2 .
2. The confidence limits are found using the following
formula:
p1q1 p2 q2
( p1  p2 )  z ( / 2) 

n1
n2
Hypothesis Tests:
If the null hypothesis is there is no difference between
proportions, this can be written as p1 = p2, or p1  p2 = 0.
Let’s consider how we can construct a standard error term.
Now the standard deviation of p1'  p2' is actually
 p p  
1
2
p1q1 p2 q2

n1
n2
However, if the null hypothesis is true, p1 = p2, so we can say
 p p  
1
2
pq pq


n1 n2
1 1 
pq   
 n1 n2 
But we don’t know p and q!
How can we estimate these from the sample?
Under the null hypothesis, the proportions of the two samples
are the same. So simply take all of the data and pool it
together to estimate the common proportion.
x x
pp  1 2 , and qp  1  pp
n1  n2
The test statistic becomes
p1  p2
z* 
 1   1 
( pp )(qp )     
 n1   n2 
Example: The proportions of defective parts from two
different suppliers were compared. The following data were
collected.
Supplier
1
2
Sample Size
300
275
Number Defective
15
9
Is there any evidence to suggest the proportion of defectives
is different for the two suppliers? Use  = 0.01.
1. The null and alternative hypotheses:
H0: p1  p2 = 0 (proportion of defectives the same)
Ha: p1  p2  0 (proportion of defectives different)
2. The type of test:
Difference of proportions, with
Samples are larger than 20.
Products n1 p1 , n1q1 , n2 p2 , n2 q2 are larger than 5.
Sampling distribution should be approximately normal.
Use z* for difference of proportions.
3. Rejection region: Reject H0 if z* > z(.005) = 2.575 or
z* < -z(.005) = -2.575
4. Calculations:
x1  x2
15  9
24
pp 


 0.042
n1  n2 300  275 575
qp  1  pp  1  0.042  0.958
4. Calculations cont’d:
p1  p2
0.05  0.0327
z* 

 1   1 
 1   1 
(0.042)(0.958) 
( pp )(qp )     


300
275
 


 n1   n2 
0.0173
0.0173


 1.03
0.0002804 0.0167
5. Conclusion:
Do not reject H0 and conclude that there is no evidence to
suggest the proportion of defectives is different for the two
suppliers.