# Paired sample t-test -formula and practice example

### Paired sample t-test formula

The formula for the paired samples t-test is:

$${t_{paired}}= {\bar{X}_D \over \sqrt {s^2 \over {n}}}$$

We read this: t paired equals mean of the differences minus mean over standard deviation divided by the square root of sample size n.

### Paired t-test-practice example

An experimenter wanted to find the effect a new drug, Y13, on hand steadiness. He randomly selected seven subjects and tested their accuracy of hitting a target. He measured how many centimeters the shot deviated from the center of the target, He then gave each subject the drug, waited for one hour and repeated the shooting test. He recorded again the distance that the shot deviated from the center of the target. He now had two scores for each subject, one before the drug, and one after the drug. Here are the scores in a table,

Subjects | Pre Drug | Post Drug | D (Post--Pre) | D-Mean | (D-Mean) squared |
---|---|---|---|---|---|

Subject 1 Subject 2 Subject 3 Subject 4 Subject 5 Subject 6 Subject 7 |
4 2 6 4 2 3 6 |
5 3 8 5 5 6 8 |
1 1 2 1 3 3 2 |
-0.86 -0.86 0.14 -0.86 1.14 1.14 0.14 |
\0.73 0.73 0.02 0.73 1.31 1.31 0.02 |

Mean= 1.86 | SUM= 4.86 |

$${t_{paired}}= {\bar{X}_D \over \sqrt {s^2 \over {n}}}$$ $${t_{paired}}= {1.86 \over {0.33}}$$

The last step is to go to t- table and enter with degrees of freedom df=12 at the 0.05 level of significance. The t value in the t table is 1.8. We compare this with the t value of our calculations, which is 5.6. Because this is greater than the t of the table (5.6>1.8 (we conclude that we have significance. The two means are significantly different. The drug had an effect on hand stability. In our report of our experiment we include the p value as follows: p<0.05. We read this: p less than 0.05.

Note: The paired t-test is also called t-test for dependent means paired samples t-test, matched pairs t-test and matched samples, and repeated-measures t-test.

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