Overview
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In statistics, a t-test is a method used to check if there is a significant difference between the averages (means) of two groups. It helps us find out if the differences we see in the data are real or just happened by chance. This test is useful when comparing small groups of data and is often used in experiments or surveys.
There are three main types of t-tests:
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In this mathematics article, we will learn about the paired t-test, paired t-test formula, paired t-test table, difference between paired and unpaired t-test, how to find the paired t-test and solve problems based on the paired t-test.
A paired t-test is a statistical method used to compare the averages (means) and standard deviations of two related sets of data. These two sets are usually from the same group of people or items, measured at two different times or under two different conditions. The goal is to check if there is a significant difference between them.
This test helps us decide if the change we observe is real or just happened by chance. For example, we might measure students' test scores before and after a training session. If the scores are different, the paired t-test tells us whether that difference is meaningful.
We calculate the differences between each pair, then check if those differences have a consistent pattern, using their mean and standard deviation. If the differences are large compared to their spread, we say there is a significant change.
There are two possible hypotheses in a paired t-test.
The null hypothesis (H₀) states that there is no significant difference between the means of the two groups.
The alternative hypothesis (H₁) states that there is a significant difference between the two population means, and that this difference is unlikely to be caused by sampling error or chance.
To learn about the mean, median and mode, please click here.
The assumptions for a paired t-test are given below:
Paired t-test is a test which is based on the differences between the values of a single pair, i.e., one deducted from the other. In the formula for a paired t-test, this difference is denoted by ‘d’. The formula of the paired t-test is defined as the sum of the differences of each pair divided by the square root of n times the sum of the differences squared minus the sum of the squared differences, overall n−1.
The formula for the paired t-test is given by
\(t=\frac{\sum d}{\sqrt{\frac{n(\sum d^{2})-(\sum d)^{2}}{n-1}}}\).
Here, \(\sum d\) is the sum of the differences.
Paired T-test table enables the t-value from a t-test to be converted to a statement about significance. The table is given below:
Two Tailed Significance |
||||||
Degree of freedom (n-1) |
α=0.20 |
0.10 |
0.05 |
0.02 |
0.01 |
0.002 |
1 |
3.078 |
6.314 |
12.706 |
31.821 |
63.657 |
318.300 |
2 |
1.886 |
2.920 |
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 |
1.440 |
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 |
1.860 |
2.306 |
2.896 |
3.355 |
4.501 |
9 |
1.383 |
1.833 |
2.262 |
2.821 |
3.250 |
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 |
3.930 |
13 |
1.350 |
1.771 |
2.160 |
2.650 |
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 |
The difference between a paired t-test and unpaired t-test are tabulated below:
Feature |
Paired T-Test |
Unpaired T-Test |
Definition |
A statistical test that compares the means and standard deviations of two related samples. |
A statistical test that compares the means and standard deviations of two unrelated or independent samples. |
Also known as |
Dependent t-test |
Independent t-test |
Hypothesis |
1. Null hypothesis \(H_{0}\): no significant difference between the means of two related groups. 2. Alternate hypothesis \(H_{1}\): there is a significant difference between the means of two related groups. |
1. Null hypothesis \(H_{0}\): no significant difference between the means of two unrelated groups. 2. Alternate hypothesis \(H_{1}\): there is a significant difference between the means of two unrelated groups. |
Variance |
Does not assume equal variance between groups. |
Assumes equal variance between groups, in case of unequal variance use Welch’s test. |
Assumptions |
1. The dependent variable is normally distributed. 2. Independently sampled observations. 3. Dependent variable measured in ratios or intervals. 4. Independent variables have two related or matched groups. |
1. The dependent variable is normally distributed. 2. Independently sampled observations. 3. Dependent variable measured in ratios or intervals. 4. The variance of data is the same between groups. 5. The same standard deviation for groups. 6. Independent variables have two unrelated or independent groups. |
When to use |
When an item or a group is to be tested twice. |
When comparing mean between two independent groups with equal variance. |
Example scenarios |
1. Effect of a drug on the same sample of people. 2. Different courses for the same subject on a group of students. |
1. Effect of drug on \(\frac{1}{2}\) of the patients assigned to a treatment group and other \(\frac{1}{2}\) assigned to a control group. 2. Measuring the level of glucose for two independent groups like men and women. |
Advantages |
1. Requires small sample size. 2. Two groups have the same sample with the same abilities. |
1. If an individual drops the study it won’t affect the sample size of the other group. 2. As the samples are assigned randomly, the chances of the individual getting the same test are reduced and hence minimizes the effects of the potential of order with the test. |
Disadvantages |
1. If one individual drops from the study, both the groups lose one sample as the individual shares both the groups. 2. The order in which the treatment is assigned can affect the performance of the individual. He may get more used to the process and might end up performing better in the second test. |
1. Requires a larger sample size. 2. The sample of two groups may differ in abilities, hence providing biased results. |
Let us take two sets of data that are related to each other, say X and Y with xᵢ ∈ X, yᵢ ∈ Y, where i = 1, 2, ..., n. Follow the steps given below to find the paired t-test.
Assume the null hypothesis that the actual mean difference is zero.
Determine the difference dᵢ = yᵢ − xᵢ between the set of observations.
Compute the mean difference.
Calculate the standard error of the mean difference, which is equal to S_d / √n, where n is the total number, and S_d is the standard deviation of the difference.
Determine the t-statistic value.
Refer to the t-distribution table and compare it with the tₙ₋₁ distribution. It gives the p-value.
Here are a few important notes on the paired t-test:
A paired t-test is designed to compare the means of the same group or item under two separate scenarios.
There are two possible hypotheses in a paired t-test. The null hypothesis (H₀) states that there is no significant difference between the means of the two groups. The alternative hypothesis (H₁) states that there is a significant difference between the two population means, and that this difference is unlikely to be caused by sampling error or chance.
The formula for the paired t-test is given by
t = (Σ d) / √[(n(Σ d²) − (Σ d)²) / (n − 1)],
where Σ d is the sum of the differences.
The data is taken from subjects who have been measured twice.
95% confidence interval is obtained from the difference between the two sets of joined observations.
To learn about the formula of standard deviation, please click here.
Before using a paired sample t-test, we need to check some important conditions:
Example 1: What conclusion should be made with respect to an experiment when the p-value is 0.0680.0680.068 and the significance level α=0.05\alpha = 0.05α=0.05?
Solution:
Since the p-value 0.0680.0680.068 is greater than the significance level 0.050.050.05, we fail to reject the null hypothesis.
This means that there is not enough statistical evidence to support the alternative hypothesis at the 5% significance level.
Example 2: In which of the following cases would you use a paired-sample t-test?
(a). When comparing the same participant’s performance before and after training.
(b). When comparing two separate groups of people.
Solution: A T-test can be used in making observations on the same sample before and after an event.
In option (b) the data does not involve observations before and after an event for the same set of people.
Thus, the correct answer is (a): when comparing the same participant’s performance before and after training.
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