Paired T-Test

Introduction

The paired t-test in statistics is a tool that compares the data between two groups of samples and determines whether the mean difference between given measurements is zero or not. In statistics, two-sample t-tests are tools used to compare the means of two datasets. These tests are also known as Student’s t-tests, The results from the tests are used to evaluate the difference between the mean of two samples. The difference is not likely to be due to sampling error or by chance. Paired t-test is used to compare means of two datasets where we are made available with two samples, and compared to determine if the difference is zero.

Definition

A paired t-test is used when each sample to be evaluated by a pair of measurements, where paired t-test determines the mean change for these samples from zero. A paired t-test is used as a statistics procedure that uses observations to draw conclusions among datasets.

For example

• When the same subject has been observed two times before and after (e.g. a person’s before and after diagnostic test results for a particular observation).

• Two sets of observations for a kind of measurement, where both the observations belong to the same subject. (e.g. Body mass index using height, weight, and age).

Formula

For using the formula used for paired t-test, two hypotheses are followed, namely the null hypothesis and the alternative hypothesis.

In the null hypothesis, the difference of the means of the paired samples is assumed to be zero. On the other hand, in the alternative hypothesis, the assumption of the difference between the paired samples is zero. There are further extensions to the alternative hypothesis based on low or high value of the result.

The above mentioned hypothesis may be represented as −

Null Hypothesis, H₀: d₁ = d₂ or H₀: d₁ - d₂ = 0.

Alternative hypothesis, H₁: d₁≠d₂ or H₁: d₁ - d₂ ≠ 0.

Here,

• d₁ is the mean of the sample 1

• d₂ is the mean of the sample 2 in the population

The formula for calculating the Paired t-test −

$$\mathrm{t = \frac{m}{\frac{s}{√n}}}$$

Where, m is the mean, s denotes the standard deviation of the mean difference (d), and n represents the size of “d”

Table

Below is the table for a Paired T-test that allows the t-value from a t-test to be interpreted in the form of a statement. The table is given below −

Two Tailed Significance
Degrees of freedom(n-1)	$$\mathrm{\alpha=0.20}$$	0.1	0.05	0.02	0.01	0.002
1	3.078	6.314	12.706	31.821	63.657	318.3
2	1.886	2.92	4.303	6.965	9.925	22.327
3	1.638	2.353	3.182	4.541	5.841	10.214
4	1.533	2.132	2.776	3.747	4.604	7.173
5	1.476	2.015	2.571	3.305	4.032	5.893
6	1.44	1.943	2.447	3.143	3.707	5.208
7	1.415	1.895	2.365	2.998	3.499	4.785
8	1.397	1.86	2.306	2.896	3.355	4.501
9	1.383	1.833	2.262	2.821	3.25	4.297
10	1.372	1.812	2.228	2.764	3.169	4.144
11	1.363	1.796	2.201	2.718	3.106	4.025
12	1.356	1.782	2.179	2.681	3.055	3.93
13	1.35	1.771	2.16	2.65	3.012	3.852
14	1.345	1.761	2.145	2.624	2.977	3.787
15	1.341	1.753	2.131	2.602	2.947	3.733

Also known as the dependent samples t-test, which tells us that the two observations are paired or dependent because they contain the data with the same subject.

Paired vs Unpaired T-test

Paired and unpaired t-tests have some distinctions which can be listed as −

Paired t-Test	Unpaired t-Test
It is used to compare the mean differences of the same group or item in two similar cases.	An unpaired t-test is used when the mean differences between two groups which are independent.
In an unpaired t-test, the variance between two datasets is assumed to be equal.	In a paired t-test, the variance between two datasets is not assumed to be equal.
Uses two dependent variables.	Uses one dependent and one independent variables

How to find Paired T-test

Let us assume a population of n students was given a diagnostic test before performing some study on a module, and then the same diagnosis is repeated for the second time after performing the module. Now we want to learn specific improvements in the abilities of the students in the module like knowledge and skills. The results from our population of students can be used to compare the effects of this module.

Let us assume the dataset of results from the module 1 and 2 to be A and B with $\mathrm{x_i \in A, y_i \in B}$. where i = 1, 2,……., n. Following are the steps to apply a paired t-Test −

• We assume the actual mean difference to be zero according to the null hypothesis.

• Difference (d_i = y_i- x_i) between observations in the two sets is calculated.

• Determine the mean difference,$\mathrm{\underline{d}}$.

• Evaluate the standard deviation (SD), and then determine the standard error, $\mathrm{SE(\underline{d}) =SD/\sqrt{n}}$.

• t-statistic value is determined by $\mathrm{T=\underline{d}\: SE(\underline{d})}$ . Considering the null hypothesis, the degree of freedom is taken as “n − 1”.

• Referring to the tables of the t-distribution and comparing it for T to the t_n-1 distribution, we get the p-value for the paired t-test.

Solved Examples

Let us assume the following scores by students in two different tests.

Student	Score in Test 1	Score in Test 2	Difference
A	63	69	6
B	65	65	0
C	56	62	6
D	100	91	-9
E	88	78	-10
F	83	87	4
G	77	79	2
H	92	88	-4
I	90	85	-5
J	84	92	8
K	68	69	1
L	74	81	7
M	87	84	-3
N	64	75	11
O	71	84	13
P	88	82	-6

Assumptions

Let H₀ be the mean difference between the scores, which is assumed to be zero.

Let H₁ be the mean difference between the scores, which is not considered zero.

Mean difference; $\mathrm{\overline{d}=1.31}$

Standard error − $\mathrm{(\overline{d}) =SD/\sqrt{n}=1.75}$

Assuming our degree of significance is α = 0.05.

Now we calculate test statistic t= $\mathrm{(\overline{d})/SE=1.31/1.75=0.750}$

Degrees of freedom df=n-1=15.

Referring to the table above the t-value with α = 0.05 and 15 degrees of freedom is 2.131.

By comparing the t-statistic value to the t-value, we observe that 0.750 is less than 2.131 Therefore, we can agree to our assumption that the mean difference is zero. Both the exams seem to be equally difficult.

Conclusion

In statistics, t-tests for two samples are tools employed to analyse the means of two populations. These tests are also called Student’s t-tests. The results from the tests are used to evaluate the variation between the mean of two datasets.

For example, the same subject has been observed two times before and after (e.g. a person’s before and after diagnostic test results for a particular observation). For using the formula used for paired t-test, two hypotheses are followed, namely the null hypothesis and the alternative hypothesis. A paired t-test can be used in comparing the means of the same dataset or population under two cases. An unpaired t-test on the other hand compares the means of two independent datasets.

FAQs

1. Is a Paired t-test better than unpaired t-test? Why?

Paired t-tests are believed to be more accurate than the unpaired t-tests because in the paired t-tests the same dependent variables eliminate significant deviation between the samples while such is not the case with unpaired t-tests.

2. In paired t-tests, can the deviation between groups be equal?

No, in the paired t-tests the Deviation between groups in a paired t-test is not equal.

3. In unpaired t-tests, can the deviation between groups be equal?

yes, in the unpaired t-tests the Deviation between groups in a paired t-test is equal.

4. What can be done if the deviation between groups in an unpaired t-test is not equal?

For an unpaired t-test, in the case of non-matching deviations, a Welch’s test is conducted.

5. What is the difference between null hypothesis and alternative hypothesis?

In the case of the null hypothesis, the mean difference of the population is equal to zero; whereas the mean difference of the population is not equal to zero in the alternative hypothesis.