10.8 Paired t-Test

Gordon E. Sarty

10. Comparing Two Population Means

10.8 Paired t-Test

Here two measurements $x_{1}$ and $x_{2}$ are taken from every subject. We could say that we measure a vector $\vec{x} = [x_{1}\; x_{2}]^{T}$ as the independent variable for every subject instead of just a number as the independent variable^[1]. This is a within subject design. Within subject designs tend to be more statistically powerful than independent or between subjects designs that have two completely different bunches of people for each variable. The extra power comes because we take the difference $D = x_{1} - x_{2}$ for every subject. So any overall differences, or variances, in $x_{1}$ or $x_{2}$ due to individuals has been removed from the data.

The paired $t$ -test is a univariate test. The difference between univariate and multivariate statistics is the the independent variables are numbers for univariate statistics and vectors for multivariate statistics. For the paired $t$ -test, the vector is converted to a number by taking a difference. To convert vector data to difference data, make a table :

$x_{1}$	$x_{2}$	$D=x_{1}-x_{2}$
1	2	-1
2	3	-1
3	5	-2
1	-2	3

Note here that the differences in individuals are gone after we take differences $D$ .

The data from the $D$ column are what you will work with. Compute $\bar{D}$ and $s_{D}$ the mean and sample standard deviation of these data. With $D$ the procedure becomes a single sample $t$ -test of $D$ against zero. Specifically we can test :

Two-tailed

Left-tailed

Right-tailed

$H_0:\mu_D=0$

$H_0:\mu_D\geq0$

$H_0:\mu_D\leq0$

$H_1:\mu_D\neq0$

$H_1:\mu_D<0$

$H_1:\mu_D>0$

The test statistic is

$t_{\rm test} = \frac{\overline{D}}{\left( s_{D}/\sqrt{n} \right)}$

with $\nu = n-1$ (for finding $t_{\rm crit}$ ).

Example 10.7 : A Physical Education director claims that a vitamin will increase a weight lifter’s strength. Eight athletes are selected and tested on how much they can bench press. They are each tested once before taking the vitamin and again after taking the vitamin for two weeks. We want to test the director’s claim at $\alpha=0.05$

The data are :

Athlete	Before( $x_1$ )	After( $x_2$ )	$D=x_1-x_2$
1	210	219	-9
2	230	236	-6
3	182	179	3
4	205	204	1
5	262	270	-8
6	253	250	3
7	219	222	-3
8	216	216	0

Here we have listed the differences which is actually part of step 0 of the solution. The $x_{1}$ and $x_{2}$ columns are what you enter into SPSS as your independent variables. With SPSS you never see the differences.

Solution :

0. Data reduction.

Compute $\overline{D}=-2.375$ , $s_D=4.84$ by entering the difference data into your calculator.

1. Hypothesis.

$H_0: \mu_D\geq0$

$H_1:\mu_D<0$

Note that a negative difference, based on $x_{1} - x_{2}$ (always consistently give population 1 priority if you want to stay out of trouble without thinking), indicates an increase in strength. It is important to interpret positive or negative differences correctly by thinking about what they mean.