A paired samples t-test examines if 2 variables

are likely to have equal population means.

- Paired Samples T-Test Assumptions
- SPSS Paired Samples T-Test Dialogs
- Paired Samples T-Test Output
- Effect Size - Cohen's D
- Testing the Normality Assumption

## Example

A teacher developed 3 exams for the same course. He needs to know if they're equally difficult so he asks his students to complete all 3 exams in random order. Only 19 students volunteer. Their data -partly shown below- are in compare-exams.sav. They hold the number of correct answers for each student on all 3 exams.

## Null Hypothesis

Generally, the null hypothesis for a paired samples t-test is that
2 variables have equal population means.
Now, we don't have data on the entire student population. We only have a *sample* of N = 19 students and sample outcomes tend to differ from population outcomes. So even if the population means are really equal, our sample means may differ a bit. However, *very* different sample means are unlikely and thus suggest that the population means aren't equal after all. So
are the sample means different enough to draw this conclusion?
We'll answer just that by running a paired samples t-test on each *pair* of exams. However, this test requires some assumptions so let's look into those first.

## Paired Samples T-Test Assumptions

Technically, a paired samples t-test is equivalent to a one sample t-test on difference scores. It therefore requires the same 2 assumptions. These are

**independent observations**;**normality**: the difference scores must be normally distributed in the population. Normality is only needed for small sample sizes, say N < 25 or so.

Our exam data probably hold **independent** observations: each case holds a separate student who didn't interact with the other students while completing the exams.

Since we've only N = 19 students, we do require the **normality** assumption. The only way to look into this is actually computing the difference scores between each pair of examns as new variables in our data. We'll do so later on.

At this point, you should carefully inspect your data. At the very least, run some histograms over the outcome variables and see if these look plausible. If necessary, set and count missing values for each variable as well. If all is good, proceed with the actual tests as shown below.

## SPSS Paired Samples T-Test Dialogs

You find the paired samples t-test under

as shown below.In the dialog below, select each *pair* of variables and move it to “Paired Variables”. For 3 pairs of variables, you need to do this 3 times.

Clicking syntax below. We added a shorter alternative to the pasted syntax for which you can bypass the entire dialog. Let's run either version.

creates the## Paired Samples T-Test Syntax

***Syntax pasted from analyze - compare means - paired-samples t-test.**

T-TEST PAIRS=ex1 ex1 ex2 WITH ex2 ex3 ex3 (PAIRED)

/CRITERIA=CI(.9500)

/MISSING=ANALYSIS.

***Shorter version below results in exact same output.**

T-TEST PAIRS=ex1 to ex3

/CRITERIA=CI(.9500)

/MISSING=ANALYSIS.

## Paired Samples T-Test Output

SPSS creates 3 output tables when running the test. The last one -**Paired Samples Test**- shows the actual test results.

SPSS reports the mean and standard deviation of the difference scores for each pair of variables. The mean is the difference between the sample means. It should be close to zero if the populations means are equal.

The mean difference between exams 1 and 2 is **not** statistically significant at α = 0.05. This is because ‘Sig. (2-tailed)’ or p > 0.05.

The 95% confidence interval includes zero: a zero mean difference is well within the range of likely population outcomes.

In a similar vein, the second test (not shown) indicates that the means for exams 1 and 3 *do* differ statistically significantly, t(18) = 2.46, p = 0.025. The same goes for the final test between exams 2 and 3.

## Effect Size - Cohen's D

Our t-tests show that exam 3 has a lower mean score than the other 2 exams. The next question is: are the differences large or small? One way to answer this is computing an effect size measure. For t-tests, Cohen’s D is often used. Sadly, you can't get it from SPSS but it's easily computed in Excel as shown below.

The effect sizes thus obtained are

- d = -0.23 (pair 1) - roughly a
**small**effect; - d = 0.56 (pair 2) - slightly over a
**medium**effect; - d = 0.57 (pair 3) - slightly over a
**medium**effect.

## Interpretational Issues

Thus far, we compared 3 pairs of exams using 3 t-tests. A shortcoming here is that all 3 tests use the same tiny student sample. This increases the risk that at least 1 test is statistically significant just by chance. There's 2 basic solutions for this:

- apply a Bonferroni correction in order to adjust the significance levels;
- run a repeated measures ANOVA on all 3 exams simultaneously.

If you choose the ANOVA approach, you may want to follow it up with post hoc tests. And these are -guess what?- Bonferroni corrected t-tests again...

## Testing the Normality Assumption

Thus far, we blindly assumed that the normality assumption for our paired samples t-tests holds. Since we've a small sample of N = 19 students, we *do* need this assumption. The only way to evaluate it, is computing the actual difference scores as new variables in our data. We'll do so with the syntax below.

***Compute difference scores as new variables.**

compute dif_1_2 = ex1 - ex2.

compute dif_1_3 = ex1 - ex3.

compute dif_2_3 = ex2 - ex3.

execute.

## Result

We can now test the normality assumption by running

on our newly created difference scores. Since we discussed both tests in separate tutorials, we'll limit ourselves to the syntax below.

***Run normality tests from Analyze - Descriptive Statistics - Explore.**

EXAMINE VARIABLES=dif_1_2 dif_1_3 dif_2_3

/statistics none

/plot npplot.

***Note: difference score between 1 and 2 violates normality assumption.**

## Result

Conclusion: the difference scores between exams 1 and 2 are unlikely to be normally distributed in the population. This violates the normality assumption required by our t-test. This implies that
we should perhaps *not* run a t-test at all
on exams 1 and 2. A good alternative for comparing these variables is a Wilcoxon signed-ranks test as this doesn't require any normality assumption.

Last, if you compute difference scores, you can circumvent the paired samples t-tests altogether: instead, you can run **one-sample t-tests** on the difference scores with zeroes as test values. The syntax below does just that. If you run it, you'll get the exact same results as from the previous paired samples tests.

***Test if all difference scores have zero means.**

T-TEST

/TESTVAL=0

/MISSING=ANALYSIS

/VARIABLES=dif_1_2 dif_1_3 dif_2_3

/CRITERIA=CI(.95).

Right, so that'll do for today. Hope you found this tutorial helpful. And as always:

**thanks for reading!**

## THIS TUTORIAL HAS 21 COMMENTS:

## By kidest on January 23rd, 2020

Good one it helps me

## By Patrick Guziewicz on March 27th, 2020

Nice overview! Only comment would be that in standard nomenclature (at least for APA) would be not to use p=.00, it would be p<.001 since a normal distribution never touches zero. Best and thank you!

## By Ruben Geert van den Berg on March 28th, 2020

Hi Patrick, thanks for your feedback!

We actually started rewriting this tutorial from scratch yesterday because we've some more issues with it:

-it doesn't test the normality assumption required for this test -only needed for small sample sizes (say N < 25 or so);

-it doesn't mention Cohen’s D, the effect size for this test.

But anyway, we think APA recommendations usually suck and they should also be rewritten from scratch after serious debate with some good statisticians.

Regarding p-values, our view is that more accurate is always better than less accurate. You're technically right that neither the normal distribution nor the t-distribution ever comes up with exact zero probabilities because they both run from -∞ to +∞.

However, we didn't say

exactlyzero: taking rounding into regard, 0.00 may be any value between (exactly) 0 and 0.0049. That is, we don't use 0, 0.0, 0.00 and 0.000 interchangeably. Perhaps scientific notation would be appropriate here.In any case, we feel that effect size and confidence intervals deserve more attention than p-values and most researchers -as well as the APA- do a very poor job there.

Have a great weekend!

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