T-test overview

Often in social science situations, we want to see if there is a statistical difference between two groups.

That is, are any recorded differences due to "chance" (not significant), or are they NOT due to chance (significant)?

To determine if the differences are significant, we use a simple inferential statistical test called the t-test.

Here's how we solve for t:

t = X1 - X2/ Sm1 - Sm2

where X1 = mean (average) of the first group and X2 = mean (average) of the second group, and Sm = standard error of the mean.

So, for example, let's say we have two groups of students in an experiment where we're trying to test whether or not kids can learn their multiplication facts better via a TV show than in school.

Let's say we bring in 20 kids and we randomly assign them to two groups. The first group of 10 learns their multiplication facts via TV show, and the second group of 10 learns via the traditional classroom approach.

So we have something like this:

# of TV kids = 10
# of class kids = 10
Overall N = 20

We examine their scores and we see that the TV kids averaged a 2/10 on a post-test quiz measuring multiplication facts and the traditional class kids averaged a 6/10 on the same test.

Normally, you'd have to solve for the Sm, but-- since this is an 11-day online course-- I'm just going to provide it to you to simplify the process (plus, it's simple to solve and most statistical packages provide it as a matter of course, so almost never do you solve for it by hand). Let's say that the Sm = 1.05.

Ok, so here's what we do--

We solve for t like this:

t = 2 - 6/1.05

t = -4/1.05

t = 3.81 (we always take the absolute value)

This value, in and of itself, tells us nothing.

Just like chi-square, however, we have to be concerned about degrees of freedom (df).

The df for a t-test is simple-- you take the N for the group and subtract one. So, for the first group, the df is 9 (10-1), and for the second group it's also 9 (10-1).

9 + 9 = 18, so the df = 18.

So, now armed with this info, we can check out the t-value chart (either the one in the book, or one we find easily online-- like here for example-- t-test table), and we see that in order to be significant with 18 df (at the .05 level), the t-value needs to be greater than 2.10.

Since our t-value of 3.81 is higher than 2.10, we say that there is a significant difference between the groups.

We then look back at our original data and we see that the traditional kids scored, on average, much better than the TV kids, so we conclude that it's better to use the traditional method.

(Typically, in experimental settings, we never make such a claim without several iterations of the experiment).

Get it?