Paried t-test procedure demonstrated with an example

The idea and demo example

A common experiment design is to have a test and control conditions. While the regular two-sample t-test assumes independence, paired t-test assumes that the two samples are dependent. For example, in an experiment where the response time is measured with or without taking a drug. Each subject could have been measured twice, once in the absence of the drug (control value) and the other after taking the drug (treatment value).

The question is whether the response time differs between the two conditions? The model is

response (continuous) ~ drug (categorical: 2 levels)

The data is "response.csv".

Open the data set from SAS. Or import with the following command.

```
data response;
infile "H:\sas\data\response.csv" dlm=',' firstobs=2;
input control treatment;
run;
```

Checking assumptions

Paired sample t-test assumes that

1. The two groups of data are dependent;
2. The differences between control and treatment follow normal distribution;

When the assumptions are not met, other methods are possible based on the two samples:

• Two independent samples and follow Normal distribution, suggest two-sample t-test;
• Two independent samples and does not follow Normal distribution, suggest WMW test;
• Two dependent samples and does not follow Normal distribution, suggest Signed Rank test;

Comparing two dependent samples with paired t-test

The regular t-test cannot be used here since the groups are no longer independent. In stead, a paried t-test is more appropiate.

Compare two sample with paired t-test

```
proc ttest data=response sides=2 alpha=0.05 h0=0;
title "Paired sample t-test example";
paired Control * Treatment;
run;
```

The output is shown below.

```           Paired sample t-test example

The TTEST Procedure

Difference:  control - treatment

N        Mean     Std Dev     Std Err     Minimum     Maximum

6     -7.3333      4.1312      1.6865    -15.0000     -4.0000

Mean       95% CL Mean        Std Dev      95% CL Std Dev

-7.3333    -11.6687  -2.9979      4.1312      2.5787  10.1322

DF    t Value    Pr > |t|

5      -4.35      0.0074

```

As a significance level of 0.05, the hypothesis is whether the response time is significantly different in the treatment than control group.

• The average different between control and treatment is -7.33
• The test statistics is -4.35, and p-value=0.0074

We can state that response times are longer under the treatment than the control group

Note that SAS only perform a two side test, meaning the hypothesis is to compare a significant difference between two groups. If one wants to test whether one group is greater(smaller) than the other, p-value must be divided by 2. For example, the p-value/2=0.0074/2=0.00342 < 0.05, hence the concluse for the one side test is still to reject the hypothesis.