Hypothesis Test: Difference Between Paired Means
This lesson explains how to conduct a hypothesis test for the
difference between
paired means.
The test procedure, called the
matchedpairs ttest,
is appropriate when the following conditions are met:
 The sampling distribution is approximately
normal, which is generally true if any of the following conditions apply.
 The population distribution is normal.
 The population data are
symmetric,
unimodal, without
outliers,
and the sample size is 15 or less.
 The population data are slightly
skewed,
unimodal, without outliers,
and the sample size is 16 to 40.
 The sample size is greater than 40, without outliers.
This approach consists of four steps: (1) state the hypotheses,
(2) formulate an analysis plan, (3) analyze sample data, and
(4) interpret results.
State the Hypotheses
Every hypothesis test requires the analyst
to state a
null hypothesis
and an
alternative hypothesis. The hypotheses are stated in such
a way that they are mutually exclusive. That is, if one is
true, the other must be false; and vice versa.
The hypotheses concern a new variable d, which is based on the
difference between paired values from two data sets.
d = x_{1}  x_{2}
where x_{1} is the value of variable x in the first data set,
and x_{2} is the value of the variable from the second data set
that is paired with x_{1}.
The table below shows three sets of null and alternative hypotheses.
Each makes a statement about how the true difference in
population values μ_{d} is related to some hypothesized value D.
(In the table, the symbol ≠ means " not equal to ".)
Set 
Null hypothesis 
Alternative hypothesis 
Number of tails 
1

μ_{d}= D 
μ_{d} ≠ D 
2 
2

μ_{d} > D 
μ_{d} < D 
1 
3

μ_{d} < D 
μ_{d} > D 
1 
The first set of hypotheses (Set 1) is an example of a
twotailed test, since an extreme value on either side of the
sampling distribution would cause a researcher to reject the null
hypothesis. The other two sets of hypotheses (Sets 2 and 3) are
onetailed tests, since an extreme value on only one side of the
sampling distribution would cause a researcher to reject the
null hypothesis.
Formulate an Analysis Plan
The analysis plan describes
how to use sample data to accept or reject the null
hypothesis. It should specify the following elements.
 Significance level. Often, researchers choose
significance levels
equal to
0.01, 0.05, or 0.10; but any value between 0 and
1 can be used.
 Test method. Use the
matchedpairs ttest
to determine whether the difference between sample means
for paired data is
significantly different from the hypothesized difference
between population means.
Analyze Sample Data
Using sample data, find the standard deviation, standard error,
degrees of freedom,
test statistic, and the Pvalue associated with the test statistic.
 Standard deviation. Compute the
standard deviation (s_{d}) of the differences
computed from n matched pairs.
s_{d} = sqrt [
(Σ(d_{i}  d)^{2}
/ (n  1) ]
where d_{i} is the difference for pair i,
d is the sample mean of the differences,
and n is the number of paired values.
 Standard error. Compute the
standard error (SE)
of the sampling distribution of d.
SE =
s_{d} *
sqrt{ ( 1/n ) * [ (N  n) / ( N  1 ) ] }
where s_{d} is the standard deviation of the sample
difference, N is the number of matched pairs in the population, and
n is the number of matched pairs in the sample. When the population size is much
larger (at least 20 times larger) than the sample size, the
standard error can be approximated by:
SE =
s_{d} / sqrt( n )
 Degrees of freedom. The
degrees of freedom (DF) is: DF = n  1 .
 Test statistic. The test statistic is a tscore (t) defined by
the following equation.
t = [ (x_{1}
 x_{2})  D ]
/ SE
= (d  D) / SE
where
x_{1} is the mean of sample 1,
x_{2} is the mean of sample 2,
d is the mean difference between
paired values in the sample,
D is the hypothesized difference between population means,
and SE is the standard error.
 Pvalue. The Pvalue is the probability of observing a
sample statistic as extreme as the test statistic. Since the
test statistic is a tscore, use the
t Distribution Calculator
to assess the probability associated with the tscore, having
the degrees of freedom computed above. (See the
sample problem at the end of this lesson for guidance on how this
is done.)
Interpret Results
If the sample findings are unlikely, given
the null hypothesis, the researcher rejects the null hypothesis.
Typically, this involves comparing the Pvalue to the
significance level,
and rejecting the null hypothesis when the Pvalue is less than
the significance level.
Test Your Understanding
Problem
Fortyfour sixth graders were randomly selected from a school district.
Then, they were divided into 22 matched pairs, each
pair having equal IQ's. One member of each pair was randomly
selected to receive special training. Then,
all of the students were given an IQ test. Test results are
summarized below.
Pair 
Training 
No training 
Difference, d 
(d  d)^{2} 
1 
95 
90 
5 
16 
2 
89 
85 
4 
9 
3 
76 
73 
3 
4 
4 
92 
90 
2 
1 
5 
91 
90 
1 
0 
6 
53 
53 
0 
1 
7 
67 
68 
1 
4 
8 
88 
90 
2 
9 
9 
75 
78 
3 
16 
10 
85 
89 
4 
25 
11 
90 
95 
5 
36 

Pair 
Training 
No training 
Difference, d 
(d  d)^{2} 
12 
85 
83 
2 
1 
13 
87 
83 
4 
9 
14 
85 
83 
2 
1 
15 
85 
82 
3 
4 
16 
68 
65 
3 
4 
17 
81 
79 
2 
1 
18 
84 
83 
1 
0 
19 
71 
60 
11 
100 
20 
46 
47 
1 
4 
21 
75 
77 
2 
9 
22 
80 
83 
3 
16 

Σ(d  d)^{2} = 270
d = 1
Do these results provide evidence that the special training
helped or hurt student performance? Use an 0.05 level of
significance. Assume that the mean differences are approximately
normally distributed.
Solution
The solution to this problem takes four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) interpret results.
We work through those steps below:
State the hypotheses. The first step is to
state the null hypothesis and an alternative hypothesis.
Null hypothesis: μ_{d} = 0
Alternative hypothesis:
μ_{d} ≠ 0
Note that these hypotheses constitute a twotailed test.
The null hypothesis will be rejected if the
difference between sample means
is too big or if it is too small.
Formulate an analysis plan. For this analysis,
the significance level is 0.05. Using sample data, we will
conduct a
matchedpairs ttest
of the null hypothesis.
Analyze sample data. Using sample data, we
compute the standard deviation of the differences (s),
the standard error (SE) of the mean difference,
the degrees of freedom (DF),
and the tscore test statistic (t).
s = sqrt [
(Σ(d_{i}  d)^{2}
/ (n  1) ]
= sqrt[ 270/(221) ] = sqrt(12.857) = 3.586
SE = s / sqrt(n) = 3.586 / [ sqrt(22) ] = 3.586/4.69 = 0.765
DF = n  1 = 22 1 = 21
t = [ (x_{1}
 x_{2})  D ]
/ SE
= (d  D)/ SE
= (1  0)/0.765 = 1.307
where
d_{i} is the observed difference for pair i,
d is mean difference between
sample pairs,
D is the hypothesized mean difference between population pairs,
and n is the number of pairs.
Since we have a
twotailed test, the Pvalue is the probability that a
tscore having 21 degrees of freedom is more extreme than 1.307;
that is, less than 1.307 or greater than 1.307.
We use the
t Distribution Calculator
to find P(t < 1.307) = 0.103, and
P(t > 1.307) = 0.103. Thus, the
Pvalue = 0.103 + 0.103 = 0.206.
Interpret results. Since the Pvalue (0.206) is
greater than the significance level (0.05), we cannot reject the
null hypothesis.
Note: If you use this approach on an exam, you may also want to mention
why this approach is appropriate. Specifically, the approach is
appropriate because the sampling method was simple random sampling,
the samples consisted of paired data, and the mean differences were
normally distributed. In addition, we used the approximation formula
to compute the standard error, since the sample size was small
relative to the population size.