Paired t test

In general mathematics, a paired t-test (also known as a correlated or dependent test) is put to use for the purpose of comparing 2 population means where we are made available with two samples. Of the two samples, observations in one sample can be paired with observations in the other sample. In everyday use, examples of where paired t test might transpire are: Before-and-after observations on similar subjects (e.g. exploratory research on rats for vaccine creation. A comparison of 2 different techniques of treatments or 2 different measurements where the treatments/measurements are entreated to the same subjects (e.g. blood sugar testing using contactless glucometer).

\[t_{calc} = \frac{\overline{d}}{\frac{s_{d}}{\sqrt{n}}}\]

Procedure for Performing Paired T-Test

Say, For example, a sample of ‘n’ employees of an organization for blood pressure test was given a diagnostic test before studying a specific system and then again after finishing off the system. We seek to identify if, usually,

The particular BP drug leads to betterment in condition/illness’ (i.e. test measurement). We can utilize the outcomes from our sample of employees to conclude about the impact of the system in general.

Let x = test measurement score before the test system, y = test measurement score after the test system

In order to measure the null hypothesis that the true mean difference is zero (0), the procedure is as given below;-

1. Compute the difference (di = yi − xi) between the two observations on each pair, ensuring that you differentiated between positive and negative distinctions.

2. Compute the mean difference, d̅

3. Compute the normal variation of the differences, sd, and consider this deviation to compute the error of the mean difference, SE (d̅) = \[\sqrt{sd}\] n.

4. Compute the t-statistic, which is mathematically expressed by T = d̅ SE (d̅). By law of null hypothesis, this statistic observes a t-distribution with n − 1 degrees of freedom.

5. Use tables of the t-distribution for setting an explicit comparison for your value for T to the tn−1 distribution.

This will ultimately render the p-value for the paired t-test.

Hypotheses of a Paired T-Test

There are feasibly 2 hypotheses in a paired t-test.

1. The null hypothesis (H0) implies that there is no significant difference between the averages/means of the two sets of sample

2. The alternative hypothesis (H1) implies that there is a significant difference between the averages/means two populations, and that this variation is improbable to have been contributed due to sampling error.

\[t = \frac{\sum d}{\sqrt{\frac{n(\sum d^{2}) - (\sum d)^{2}}{n-1}}}\]

Where d: difference per paired value

n: number of samples 

Assumptions of a Paired T-Test

1. The dependent variable is generally distributed

2. The dependent variable is measured on an incremental level, such as ratios or intervals.

3. The observations are sampled unrelated

4. The independent variables must comprise two dependent sets or equal pairs.

Solved Example 

Null Hypothesis: H0: μd = 0

Substitute Hypothesis: H1: μd  ≠ 0

Point Estimate:(the difference of sample mean) is the point estimate of .μd

Test statistic:

\[t = \frac{\overline{d} - \mu_{d}}{\frac{s_{d}}{\sqrt{n}}}\]

Remember that the standard error of \[\overline{d}\] is \[\frac{s_{d}}{\sqrt{n}}\] in which sd is the standard deviation of the differences. 

Example:

Suppose we seek to find out if the LDL (lipid profile) in the blood of the individuals in Dixon and Massey study was modified from 1952 to 1962. Now, applying the paired t-test:- 

• H0: The mean difference in lipid is 0 from 1952 to 1962

• H1: The mean difference in lipid is NOT 0 from 1952 to 1962.

• Our degree of significance is α = 0.05.

Thus,

\[\overline{d} = -69.8, s_{d} = 46.6\]        \[t = \frac{-69.8}{\frac{46.6}{\sqrt{20}}} = -6.7\]

Result: 6.70

For α = 0.05 and 19 df (degrees of freedom), the analytical value of‘t’ is 2.093. Seeing that | -6.7| is greater than 2.093, we discard H0 and conclude that we have substantial proof that the mean difference in cholesterol from 1952 to 1962 is NOT 0. In particular, there has been a standard decline of 69.8 from 1952 to 1962.

Conclusion:

On an average, levels of lipid declined 69.8 units from 1952 to 1962. For a significance level of 0.05 and 19 df, the critical value for the t-test came out to be 2.093. Since the absolute value of our test statistic (6.70) is > the critical value (2.093) we dismiss the null hypothesis and come to the conclusion that there is on average a non-zero change in lipid levels from 1952 to 1962.

Did You Know

1. Paired t-tests are contemplated to be more robust than unpaired t-tests because using the same participants or objects removes deviation between the samples that could be induced by anything excepting what’s being tested.

2. Deviation between groups in a paired t-test is NOT supposed to be equal.

3. Deviation between groups in an unpaired t-test is supposed to be equal.

4. In situations of unmatched deviations in an unpaired t-test, a Welch’s test needs to be conducted.