The paired t test is used to detect treatment differences when measurements from one group of subjects are correlated with measurements from another.

**PAIRED t TEST**

Previously, we covered statistical tests (e.g., the
independent groups *Z* test and *t* test) for assessing differences
between group means derived from independent sam-ples. In some medical
applications, we use measures that are paired; examples are comparison of
pre–post test results from the same subject, comparisons of twins, and
comparisons of littermates. In these situations, there is an expected
correlation (relationship) between any pair of responses. The paired *t* test looks at treatment differences in
medical studies that have paired observations.

The paired *t*
test is used to detect treatment differences when measurements from one group
of subjects are correlated with measurements from another. You will learn about
correlation in more detail in Chapter 12. For now, just think of correla-tion
as a positive relationship. The paired *t*
test evaluates within-subject compar-isons, meaning that a subject’s scores
collected at an earlier time are compared with his own scores collected at a
later time. The scores of twin pairs are analogous to within-subject
comparisons.

The results of subjects’ responses to pre- and
posttest measures tend to be relat-ed. To illustrate, if we measure children’s
gains in intelligence over time, their later scores are related to their
initial scores. (Smart children will continue to be smart when they are
remeasured.) When such a correlation exists, the pairing can lead to a mean
difference that has less variability than would occur had the groups been
com-pletely independent of each other. This reduction in variance implies that
a more powerful test (the paired *t*
test) can be constructed than for the independent case. Similarly, paired *t* tests can allow the construction of
more precise confidence inter-vals than would be obtained by using independent
groups *t* tests.

For the paired *t*
test, the sample sizes *n _{t}*
and

First, we will provide background information about
the logic of the paired *t* test and
then give some calculation examples using the data from Tables 9.1 and 9.2.
Matching or pairing of subjects is done by patient; i.e., the difference is
taken be-tween the first treatment for patient A and the second treatment for
patient A, and so on for patient B and all other patients. The differences are
then averaged over the set of *n*
patients.

As implied at the beginning of this section, we do
not compute differences be-tween treatment 1 for patient A and treatment 2 for
patient B. The positive correla tion between the treatments exists because the
patient himself is the common factor. We wish to avoid mixing
patient-to-patient variability with the treatment effect in the computed paired
difference. As physicians enjoy saying, “the patient acts as his own control.”

Order effects refer to the order of the
presentation of the treatments in experi-mental studies such as clinical
trials. Some clinical trials have multiple treatments; others have a treatment
condition and a control or placebo condition. Order effects may influence the
outcome of a clinical trial. In the case in which a patient serves as his own
control, we may not think that it matters whether the treatment or con-trol
condition occurs first. Although we cannot rule out order effects, they are
easy to minimize; we can minimize them by randomizing the order of presentation
of the experimental conditions. For example, in a clinical trial that has a
treatment and a control condition, patients could be randomized to either leg
of the trial so that one-half of the patients would receive the treatment first
and one-half the control first.

By looking at paired differences (i.e., differences
between treatments *A* and *B* for each patient), we gain precision
by having less variability in these paired differences than with an
independent-groups model; however, the act of pairing discards the individual
observations (there were 2*n* of them
and now we are left with only *n*
paired differences). We will see that the resulting *t* statistic will have only *n*
– 1 degrees of freedom rather than the 2*n*
– 2 degrees of freedom as in the *t*
test for differences between means of two independent samples.

Although we have achieved less variability in the
sample differences, the paired *t *test
cuts the sample size by a factor of two. When the correlation between treatments
A and B is high (and consequently the variability is reduced considerably),
pairing will pay off for us. But if the observations being paired were truly
independent, the pairing could actually weaken our analysis.

A paired *t*-test
(two-sided test) consists of the following steps:

1. Form the paired differences.

2. State the null hypothesis *H*_{0}: μ* _{t}* = μ

3. Choose a significance level α = α_{0} (often we take α_{0} = 0.05 or 0.01).

4. Determine the critical region;
that is, the region of values of *t* in
the upper and lower α/2 tails of the sampling
distribution for Student’s *t*
distribution with *n* - 1 degrees of
freedom when μ* _{t}*/μ

5. Compute the *t* statistic: *t* = for the given sample and sample size *n*
for the paired differences, where ^{–}*d ^{-}* is the sample mean differ-ence between groups and

6. Reject the null hypothesis if the test statistic
*t* (computed in step 4) falls in the
rejection region for this test; otherwise, do not reject the null hypothe-sis.

Now we will now look at an example of how to
perform a paired *t* test. A striking
example where the correlation between two groups is due to a seasonal effect
follows. Although it is a weather example, these kinds of results can occur
easily in clinical trial data as well. The data are fictitious but are
realistic temperatures for the two cities at various times during the year. We
are considering two temperature readings from stations that are located in
neighboring cities such as Washington, D.C., and New York. We may think that it
tends to be a little warmer in Washing-ton, but seasonal effects could mask a
slight difference of a few degrees.

We want to test the null hypothesis that the
average daily temperatures of the two cities are the same. We will test this
hypothesis versus the two-sided alternative that there is a difference between
the cities. We are given the data in Table 9.1, which shows the mean
temperature on the 15th of each month during a 12-month period.

Now let us consider the two-sample *t* test as though the data for the cities
were independent. Later we will see that this is a faulty assumption. The means
for Washington (_{1})
and New York (_{2})
equal 56.16°F and 52.5°F, respectively. Is the difference (3.66) between these
means statistically significant? We test *H*_{0}:
μ_{1} – μ_{2 }= 0 against the alternative *H*_{1}:
μ_{1} – μ_{2} ≠ 0, where μ_{1} is the population mean temperature for Washington
and μ_{2} is the population mean temperature for New York.
The respective sample standard deviations, *S*_{1}
and *S*_{2}, equal 23.85 and
23.56. These sample standard deviations are close enough to make plausible the
assumption that the population standard deviations are equal.

Consequently, we use the pooled variance *S _{p}*

But let us look more closely at the data. The
independence assumption does not hold. We can see that temperatures are much
higher in summer months than in win-ter months for both cities. We see that the
month-to-month variability is large and dominant over the variability between
cities for any given day. So if we pair tem-peratures on the same days for
these cities we will remove the effect of month-to-month variability and have a
better chance to detect a difference between cities. Now let us follow the
paired *t* test procedure based on data
from Table 9.2.

Here we see that the mean difference is again 3.66 but the standard
deviation *S _{d} *= 1.614,
which is a dramatic reduction in variation over the pooled estimate of

We are beginning to see the usefulness of pairing: *t* = ( – (μ_{1} – μ_{2}))/(*S _{d}*/√n) = (3.66
– 0)/(1.614/√12) = 3.66/0.466 = 7.86. This

This outcome is truly astonishing! Using an
unpaired test with this temperature data we were not even close to a
statistically significant result, but with an appropri-ate choice for pairing,
the significance of the paired differences between the cities is extremely
high. These two opposite findings indicate how wrong one can be when using
erroneous assumptions.

There is no magic to statistical methods. Bad
assumptions lead to bad answers. Another indication that it was warmer in
Washington than in New York is the fact that the average temperature in
Washington was higher for all twelve days.

In Section 14.4, we will consider a nonparametric
technique called the sign test. Under the null hypothesis that the two cities
have the same mean temperatures each

Finally, let us go through the six steps for the
paired *t* test using the temperature
data:

1. Form the paired differences
(the far right column in Table 9.2).

2. State the null hypothesis *H*_{0}: μ_{1} = μ_{2} or μ_{1} – μ_{2} = 0 versus the alternative hypothesis *H*_{1}: μ_{1 }≠_{ }μ_{2 }or μ_{1} – μ_{2} ≠ 0.

3. Choose a significance level α = α_{0} = 0.01.

4. Determine the critical region,
that is, the region of values of *t* in
the upper and lower 0.005 tails of the sampling distribution for Student’s *t* distribution with *n *– 1 = 11 degrees of freedom when* *μ_{1}* *=* *μ_{2}* *(i.e., the sampling distribution* *when the null hypothesis is true) and
when *n* = *n*_{1} = *n*_{2}.

5. Compute the *t* statistic: *t* = { – (μ_{1} – μ_{2})}/[*S _{d}*/√n] for the
given sample and sample size

6. Reject the null hypothesis if
the test statistic *t* (computed in
step 5) falls in the rejection region for this test; otherwise, do not reject
the null hypothesis. For a *t* with 11
degrees of freedom and α = 0.01, the critical value is
3.1058. Be-cause the test statistic *t*
is 7.86, we reject *H*_{0}.

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