independent-samples t-test

Independent-samples t-tests have three requirements:

1. The samples have to be independent
2. The dependent scores are normally distributed interval or ratio scores
3. The populations have homogenous variance

The alternative hypothesis ($H_0$) says that there’s a difference between the two populations ($H_a:{\mu }_1\ne {\mu }_2$). Another way to phrase that is ($H_a:{\mu }_1-{\mu }_2\ne 0$). This would then leave $H_0:{\mu }_1-{\mu }_2=0$.

To do the test:

1. estimate the variance of the raw score population
2. compute the estimated standard error of the sampling distribution
3. compute $t_{obt}$.

Estimated variance for a condition:

$$\begin{array}{r}{s}_x^2=\frac{\Sigma X^2-\frac{(\Sigma X{)}^2}{n}}{n-1}\end{array}$$

The pooled variance ($s_{pool}^2$):

$$\begin{array}{r}{s}_{pool}^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{(n_1-1)+(n_2-1)}\end{array}$$

Standard error of the difference ($s_{{\bar{X}}_1-{\bar{X}}_2}$)

$$\begin{array}{r}{s}_{{\bar{X}}_1-{\bar{X}}_2}=\sqrt{(s_{pool}^2)(\frac{1}{n_1}+\frac{1}{n_2})}\end{array}$$

df is weirder:

$$\begin{array}{r}df=(n_1-1)+(n_2-1)\end{array}$$

Example

We’re computing two samples of whether folks can recall info when hypotised better than unhyponotised.

	sample 1	sample 2
scores ($n$)	17	15
variance ($s_{pool}^2$)	9	7.5
mean score ($\bar{X}$)	23	20

The pooled variance is then

$$\begin{array}{rl}{s}_{pool}^2& =\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{(n_1-1)+(n_2-1)}\\ & \\ & =\frac{(17-1)9+(15-1)7.5}{(17-1)+(15-1)}\\ & \\ & =\frac{(17)9+(14)7.5}{16+14}\\ & \\ & =\frac{144+105}{30}\\ & \\ & =\frac{249}{30}\\ & \\ & =8.3\end{array}$$

Standard error of the difference:

$$\begin{array}{rl}{s}_{{\bar{X}}_1-{\bar{X}}_2}& =\sqrt{(s_{pool}^2)(\frac{1}{n_1}+\frac{1}{n_2})}\\ & \\ & =\sqrt{(8.3)(\frac{1}{17}+\frac{1}{15})}\\ & \\ & =\sqrt{8.3(.059+.067)}\\ & \\ & =\sqrt{8.3(.126)}\\ & \\ & =\sqrt{1.046}\\ & \\ & =1.023\end{array}$$

Calculate $t_{obt}$ (note ${\mu }_1-{\mu }_2$ is just $H_0$)

$$\begin{array}{rl}{t}_{obt}& =\frac{({\bar{X}}_1-{\bar{X}}_2)-({\mu }_1-{\mu }_2)}{s_{{\bar{X}}_1-{\bar{X}}_2}}\\ & \\ & =\frac{(23-20)-0}{1.023}\\ & \\ & =\frac{3}{1.023}\\ & \\ & =2.93\end{array}$$ $$\begin{array}{rl}df& =(n_1-1)+(n_2-1)\\ & \\ & =(17-1)+(15-1)\\ & \\ & =(16)+(14)\\ & \\ & =30\end{array}$$

$t_{crit}=\pm 2.042$ (found in the appendix for two-tailed tests for df(30)),

so because $+t_{crit}<t_{obt}$, we can reject the null hypothesis that there’s no difference at $\alpha =.05$.