related-samples t-test

There are two types of research designs that produce related samples:

matched-samples designs : each participant is paired with another participant in the other condition. They need to be related in some meaningful way that isn’t the independent/dependent variable.

repeated-measures designs : same participants are measured under all conditions of the independent variable

related-samples t-tests have these requirements:

1. The samples have to be dependent
2. The dependent scores are normally distributed interval or ratio scores
3. The populations have homogenous variance
4. Because they must be paired, $n$s must be equal.

The general process:

1. grab the difference of before and after.
2. Use that as a one-sampled t-test.

Formulating hypotheses: If we have a two-tailed test, we’re looking at $H_0:{\mu }_D=0$ and $H_a:{\mu }_D\ne 0$ One-tailed tests look like $H_0:{\mu }_D\le 0$ and $H_a:{\mu }_D>0$

Steps:

1. Compute $s_D^2$ ($s_D^2=\frac{\Sigma D^2-\frac{(\Sigma D{)}^2}{N}}{N-1}$)
2. Compute the standard error of the mean difference ($s_{\bar{D}}=\sqrt{\frac{s_D^2}{N}}$)
3. Determine $t_{obt}=\frac{\bar{D}-{\mu }_D}{s_{\bar{D}}}$

$df=N-1$

Example

Find the difference that therapy makes in a phobia study.

Data:

	before therapy	after therapy	D	D^2
Millie	11	8	+3	9
Archie	16	11	+5	25
Jerome	20	15	+5	25
Althea	17	11	+6	36
Leon	10	11	-1	1
n=5	mean: 14.80	mean: 11.20	sum = 18	sum = 96
			mean = +3.6

From here, it’s basically the a one-sample t-test, but using the differences.

$$\begin{array}{rl}{s}_D^2& =\frac{\Sigma D^2-\frac{(\Sigma D{)}^2}{N}}{N-1}\\ & \\ & =\frac{96-\frac{(18^2}{5}}{4}\\ & \\ & =\frac{96-\frac{(324}{5}}{4}\\ & \\ & =\frac{96-64.8}{4}\\ & \\ & =\frac{31.2}{4}\\ & \\ & =7.8\end{array}$$ $$\begin{array}{rl}{s}_{\bar{D}}& =\sqrt{\frac{s_D^2}{N}}\\ & \\ & =\sqrt{\frac{7.8}{5}}\\ & \\ & =\sqrt{1.56}\\ & \\ & =1.249\end{array}$$ $$\begin{array}{rl}{t}_{obt}& =\frac{\bar{D}-{\mu }_D}{s_{\bar{D}}}\\ & \\ & =\frac{3.6-0}{1.249}\\ & \\ & =2.882\end{array}$$

$t_{crit}=\pm 2.776$ with $df=4$, so we can reject the null hypothesis at $\alpha =.05$.