12.2 Paired Samples $T$-test

Referring to the introduction of this chapter, suppose that a physical therapist has developed a new treatment that she believes will improve flexibility for people suffering from rotator cuff injuries. How might she test the claim? A natural thing to do would be something akin to the following: Select a sample of people suffering from the same rotator cuff injuries. In a consistent fashion numerically measure flexibility of each subject. Then have each subject perform the treatment for the same amount of time. Once completed, using the same technique to measure flexibility, retest each subject. This generates two samples of numbers, but the samples are paired.

Suppose again that $X$ and $Y$ are populations of numbers with means ${\mu }_X$ and ${\mu }_Y,$ respectively. The populations are paired if for each member $x$ in $X,$ there is a corresponding member $y$ in $Y,$ and vice versa.

To execute a test on the two means, one needs samples from $X$ and $Y.$ If $X$ and $Y$ are paired, the samples will be naturally paired. If the difference ${\mu }_X-{\mu }_Y$ makes sense, then a Paired-Samples T-Test may be appropriate in testing the competing hypotheses.⁴⁷⁴⁷Populations can be paired, but taking a difference might be nonsensical. For example, if $X$ is the collection of all heights (in inches) of living Americans, and $Y$ is the population of all weights (in pounds)of living Americans, then these populations are paired, but computing ${\mu }_X-{\mu }_Y$ does not make sense.

It is important to be able to identify paired data, as treating paired data as unpaired weakens the analysis (knowledge of the samples would be discarded).

Suppose that $X$ and $Y$ are paired populations of numbers where ${\mu }_X-{\mu }_Y$ makes sense. For each pair $(x,y)$ of corresponding members $x$ and $y$ in $X$ and $Y,$ respectively, the difference $d=x-y$ makes sense as well. The population of differences $D$ consisting of all differences of pairs $d=x-y$ is well-defined and has mean ${\mu }_D={\mu }_X-{\mu }_Y.$

Thus, a hypothesis test on the difference of means ${\mu }_X-{\mu }_Y$ is equivalent to a test on the mean ${\mu }_D.$ If it is reasonable to assume the population $D$ is normally distributed⁴⁸⁴⁸If $X$ and $Y$ are normally distributed, then $D=X-Y$ is also normally distributed. or if the sample size taken from $D$ sufficiently large, then is is reasonable to use a $T$-Test on the sample of differences.

Let’s set up notation for the test. The competing hypotheses will be one of the following three cases:

Suppose that from paired populations $X$ and $Y,$ paired samples of size $n$ are generated:

For each pair $(x_i,y_i),$ compute the difference $d_i=x_i-y_i.$ The collection $d_1,$ $d_2,$ …, $d_m$ is then the samle of size $n$ from the population of differences $D.$ The data can be summarized in a table such as this:

Let $\overline{d}$ and $s$ denote the sample mean and sample standard deviation of the differences $d_i.$ The test statistic is then

with $df=n-1.$

Remark: As in Section LABEL:Sec:effect-size-single-mean, if the test yields a significant result, then it is appropriate to estimate the effect size using Cohen’s $d.$