Matched pairs design is one of the experimental designs of a randomized block design which is used when the experiment has only two treatments. Additionally, as the name suggests, the subjects are grouped into pairs based on blocking variable such as gender and age. Being a randomized design, the two treatments are randomly assigned to the subjects.
Advantages | Disadvantages |
Each treatment has people with similar characteristics hence fewer participant variables | If one participant opts out,2 PPS data is lost |
No counterbalancing since there is no order effects. | Matching is a quagmire and time-consuming |
Materials/tests can be reused across the conditions | It is impossible to match people exactly unless participants are twins. |
Below is a good example of a matched pair experimental design
The table shows a matched pairs example for a hypothetical clinical experiment in which 100 subjects were involved. Each of the subjects received either a placebo or a Malaria Vaccine. The subjects were grouped into two each with 50 subjects matched pairs. Each pair was matched on gender and age. For instance, Pair 1 was two males both age 20 while Pair two was two women both age 21 and Pair 3 was two women both age 23 etc.
Treatment | ||
Pair | Placebo | Vaccine |
1 | 1 | 1 |
2 | 1 | 1 |
3 | 1 | 1 |
… | … | … |
49 | 1 | 1 |
50 | 1 | 1 |
Matched-pair t test is used to test if there is a difference in mean between two matched/related pairs. The matched –pair t test is also referred to as the paired samples t-test of the dependent t-test. The test is a parametric test whose assumptions are
The matched-pair t test statistic is calculated as below;
The calculated value is compared to the tabulated value with n-1 degrees of freedom.
#ID | x1 | x2 | – | X1-X2 | (X1-X2)² |
1 | 5.24 | 7.53 | – | -2.29 | 5.2441 |
2 | 6.28 | 9.06 | – | -2.78 | 7.7284 |
3 | 7.09 | 12.61 | – | -5.52 | 30.4704 |
4 | 8.21 | 14.76 | – | -6.55 | 42.9025 |
5 | 11.27 | 17.66 | – | -6.39 | 40.8321 |
– | – | – | Sum | -23.53 | 127.1775 |
n | 5 | – | Mean | -4.706 | – |
df=n-1: | 4 | – | sd | 2.027641 | – |
At 0.05 level, the t value with 4 degrees of freedom is 2.78.Since the calculated value is greater than the tabulated then there is no significance claimed.
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