T-test for Two Dependent Samples (Paired or Matched Design)

The third application of a t-test that we will consider is for two dependent (paired or matched) samples. This can be applied in either of two types of comparisons.

Example 1: Does an intervention increase knowledge about HIV transmission?

Early in the HIV epidemic, there was poor knowledge of HIV transmission risks among health care staff. A short training was developed to improve knowledge and attitudes around HIV disease. Was the training effective in improving knowledge?

Table - Mean ( ± SD) knowledge scores, pre- and post-intervention, n=15

Pre-Intervention Post-Intervention Change
18.3 ± 3.8 21.9 ± 4.4 3.53 ± 3.93

 The raw data for this comparison is shown in the next table.

Subject

Kscore1 Kscore2 difference

1

17 22 5

2

17 21 4

3

15 21 6

4

19 26 7

5

18 20 2

6

14 14 0

7

27 31 4

8

20 18 -2

9

12 22 10

10

21 20 -1

11

20 27 7

12

24 23 -1

13

17 15 -2

14

17 24 7

15

17 24 7
mean difference = 3.533333
sddiff= 3.925497
p-value=

0.003634

 

The strategy is to calculate the pre-/post- difference in knowledge score for each person and determine whether the mean difference=0.

First, establish the null and alternative hypotheses.

Then compute the test statistic for paired or matched samples

T-test of Two Dependent (paired) Samples Using R

For the example above we can compute the p-value using R. First, we compute the means.

> mean(Kscore1)
[1] 18.33333

> mean(Kscore2)
[1] 21.86667

Then we perform the t-test for two dependent samples.

> t.test(Kscore2,Kscore1,paired=TRUE)

Paired t-test

data: Kscore2 and Kscore1 t = 3.4861, df = 14, p-value = 0.003634
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
1.359466 5.707201
sample estimates:
mean of th
e differences
3.533333

The null hypothesis is that the mean change in knowledge scores from before to after is 0. However, the analyis shows that the mean difference is 3.53 with a 95% confidence interval that ranges from 1.36 to 5.7. Since the confidence interval does not include the null value of 0, the p-value must be < 0.05, and in fact it is 0.003634.