formula for hypothesis testing paired means

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Hypothesis Test: Difference Between Paired Means

This lesson explains how to conduct a hypothesis test for the difference between paired means . The test procedure, called the matched-pairs t-test , is appropriate when the following conditions are met:

• The sampling method for each sample is simple random sampling .
• The test is conducted on paired data. (As a result, the data sets are not independent .)
• The population distribution is normal.
• The population data are symmetric , unimodal , without outliers , and the sample size is 15 or less.
• The population data are slightly skewed , unimodal, without outliers, and the sample size is 16 to 40.
• The sample size is greater than 40, without outliers.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

The hypotheses concern a new variable d, which is based on the difference between paired values from two data sets.

d = x 1 - x 2

where x 1 is the value of variable x in the first data set, and x 2 is the value of the variable from the second data set that is paired with x 1 .

The table below shows three sets of null and alternative hypotheses. Each makes a statement about how the true difference in population values μ d is related to some hypothesized value D. (In the table, the symbol ≠ means " not equal to ".)

The first set of hypotheses (Set 1) is an example of a two-tailed test , since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests , since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use the matched-pairs t-test to determine whether the difference between sample means for paired data is significantly different from the hypothesized difference between population means.

Analyze Sample Data

Using sample data, find the standard deviation, standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

s d = sqrt [ (Σ(d i - d ) 2 / (n - 1) ]

SE = s d * sqrt{ ( 1/n ) * [ (N - n) / ( N - 1 ) ] }

SE = s d / sqrt( n )

• Degrees of freedom. The degrees of freedom (DF) is: DF = n - 1 .

t = [ ( x 1 - x 2 ) - D ] / SE = ( d - D) / SE

• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the t statistic, having the degrees of freedom computed above. (See the sample problem at the end of this lesson for guidance on how this is done.)

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

Forty-four sixth graders were randomly selected from a school district. Then, they were divided into 22 matched pairs, each pair having equal IQ's. One member of each pair was randomly selected to receive special training. Then, all of the students were given an IQ test. Test results are summarized below.

Σ(d - d ) 2 = 270

Do these results provide evidence that the special training helped or hurt student performance? Use an 0.05 level of significance. Assume that the mean differences are approximately normally distributed.

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

Null hypothesis: μ d = 0

Alternative hypothesis: μ d ≠ 0

• Formulate an analysis plan . For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

s = sqrt [ (Σ(d i - d ) 2 / (n - 1) ]

s = sqrt[ 270/(22-1) ] = sqrt(12.857) = 3.586

SE = s / sqrt(n) = 3.586 / [ sqrt(22) ]

SE = 3.586/4.69 = 0.765

DF = n - 1 = 22 -1 = 21

t = [ ( x 1 - x 2 ) - D ] / SE

t = ( d - D)/ SE = (1 - 0)/0.765 = 1.307

where d i is the observed difference for pair i , d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test , the P-value is the probability that the t statistic having 21 degrees of freedom is less than -1.307 or greater than 1.307. We use the t Distribution Calculator to find P(t < -1.307) is about 0.103.

• Then, we enter 1.307 as the sample mean in the t-distribution Calculator. We find the that the P(t < 1.307) is about 0.897. Therefore, P(t > 1.307) is 1 minus 0.897 or 0.103. Thus, the P-value = 0.103 + 0.103 = 0.206.
• Interpret results . Since the P-value (0.206) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples consisted of paired data, and the mean differences were normally distributed. In addition, we used the approximation formula to compute the standard error, since the sample size was small relative to the population size.

Paired T-Test

In Statistics, a t-test can be represented as a statistical hypothesis test where the test statistic supports a student’s t distribution if the null hypothesis is established. In Paired T-Test, they compare the means of two groups of observations. The observations must be randomly assigned to each of the two groups so that the difference in response seen is due to the treatment and not because of any other factors. If two samples are given, then the observation of one sample can be paired with the observation of the other sample, by using the paired t-test.

Further, this test can be used in making observations on the same sample before and after an event. Now, let us discuss what is paired t-test, its formula, table, and the procedure to perform the paired t-test in detail.

Paired T-Test Definition

The paired t-test gives a hypothesis examination of the difference between population means for a set of random samples whose variations are almost normally distributed. Subjects are often tested in a before-after situation or with subjects as alike as possible. The paired t-test is a test that the differences between the two observations are zero.

Let us assume two paired sets, such as Xi and Yi for i = 1, 2, …, n such that their paired difference is independent which is identically and normally distributed. Then the paired t-test concludes whether they notably vary from each other.

Paired T-Test Formula

Paired T-test is a test that is based on the differences between the values of a single pair, that is one deducted from the other. In the formula for a paired t-test, this difference is notated as 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

\(t = \frac{\sum d}{\sqrt{\frac{n(\sum d^{2})-(\sum d)^{2}}{n-1}}}\) Where, Σd is the sum of the differences.

Paired T-Test Table

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:

\(\begin{array}{l} \text { Paired T Test Table }\\ \begin{array}{|c|c|c|c|c|c|c|} \hline {\text { Two Tailed Significance }} \\ \hline \begin{array}{c} \text { Degrees of } \\ \text { fredom }(n-1) \end{array} & \alpha=0.20 & 0.10 & 0.05 & 0.02 & 0.01 & 0.002 \\ \hline & & & & & & \\ \hline 1 & 3.078 & 6.314 & 12.706 & 31.821 & 63.657 & 318.300 \\ \hline 2 & 1.886 & 2.920 & 4.303 & 6.965 & 9.925 & 22.327 \\ \hline 3 & 1.638 & 2.353 & 3.182 & 4.541 & 5.841 & 10.214 \\ \hline 4 & 1.533 & 2.132 & 2.776 & 3.747 & 4.604 & 7.173 \\ \hline 5 & 1.476 & 2.015 & 2.571 & 3.305 & 4.032 & 5.893 \\ \hline 6 & 1.440 & 1.943 & 2.447 & 3.143 & 3.707 & 5.208 \\ \hline 7 & 1.415 & 1.895 & 2.365 & 2.998 & 3.499 & 4.785 \\ \hline 8 & 1.397 & 1.860 & 2.306 & 2.896 & 3.355 & 4.501 \\ \hline 9 & 1.383 & 1.833 & 2.262 & 2.821 & 3.250 & 4.297 \\ \hline 10 & 1.372 & 1.812 & 2.228 & 2.764 & 3.169 & 4.144 \\ \hline 11 & 1.363 & 1.796 & 2.201 & 2.718 & 3.106 & 4.025 \\ \hline 12 & 1.356 & 1.782 & 2.179 & 2.681 & 3.055 & 3.930 \\ \hline 13 & 1.350 & 1.771 & 2.160 & 2.650 & 3.012 & 3.852 \\ \hline 14 & 1.345 & 1.761 & 2.145 & 2.624 & 2.977 & 3.787 \\ \hline 15 & 1.341 & 1.753 & 2.131 & 2.602 & 2.947 & 3.733 \\ \hline \end{array} \end{array}\)

Paired Vs Unpaired T-Test

The similarity between paired and unpaired t-test is that both assume data from the normal distribution.

Characteristics of Unpaired T-Test:

• The two groups taken should be independent.
• The sample size of the two groups need not be equal.
• It compares the mean of the data of the two groups.
• 95% confidence interval for the mean difference is calculated.

Characteristics of Paired T-Test:

• 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.

How to Find the Paired T-Test?

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 Sd /√n, where n is the total number, and Sd is the standard deviation of the difference.
• Determine the t-statistic value.
• Refer to the T-distribution table and compare it with t he t n-1  dis tribution. It gives the p-value.  

Important Notes on Paired T-Test

Here are a few important notes on the Paired T-Test:

Related Topics 

• T-Distribution
• Sample Standard Deviation

Solved Examples on Paired T-Test

What conclusion should be made with respect to an experiment when the significance level is 0.068?

Since the p-value of 0.068 is greater than α = 0.05, it would fail to reject the null hypothesis.

Answer: As the value of p < 0.05, the null hypothesis is rejected.

In which of the following cases could you use a paired-samples t-test?

(a) When comparing the same participant's performance before and after training (b) When comparing two separate groups of people

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.

Answer: a. When comparing the same participant's performance before and after training

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Practice Questions

Faqs  on paired t-test, how do you know if paired or unpaired t-test.

A paired t-test is designed to compare the means of the same group or item under two separate scenarios. An unpaired t-test compares the means of two independent or unrelated groups. In an unpaired t-test, the variance between groups is assumed to be equal. In a paired t-test, the variance is not assumed to be equal.

Should I Use Paired or Unpaired T-Test?

If the data are paired or matched, then you should choose a paired t-test instead. If the pairing is effective in controlling for experimental variability, the paired t-test will be more powerful than the unpaired test.

What Is the Difference Between a Paired T-Test and a 2 Sample T-Test?

A two-sample t-test is used when the data of two samples are statistically independent, while the paired t-test is used when data is in the form of matched pairs. ... To use the two-sample t-test, we need to assume that the data from both samples are normally distributed and have the same variances.

Why Would You Use a Paired T-Test?

A paired t-test is used when we are interested in the difference between two variables for the same subject. Often the two variables are separated by time. ... Since we are ultimately concerned with the difference between two measures in one sample, the paired t-test reduces to the one-sample t-test.

How Do You Interpret a Paired T-Test?

Complete the following steps to interpret a paired t-test. ...

• Step 1: Determine a confidence interval for the population mean difference. First, consider the mean difference, and then examine the confidence interval. ...
• Step 2: Determine whether the difference is statistically significant. ...
• Step 3: Check your data for problems.

How Do You Know if Data Is Paired?

Two data sets are "paired" when the following one-to-one relationship exists between values in the two data sets.

• Each data set has the same number of data points.
• Each data point in one data set is related to one, and only one, data point in the other data set.

T-test for Paired Samples

Instructions: This calculator conducts a t-test for two paired samples. This test applies when you have two samples that are dependent (paired or matched). Please select the null and alternative hypotheses, type the sample data (or paste it from Excel) and the significance level, and the results of the t-test for two dependent samples will be displayed for you.

If you need a larger sample size click in the button below, or paste directly from Excel

The T-Test For Paired Samples

More about the t-test for two dependent samples so you can understand in a better way the results delivered by the solver.

How do you calculate a paired t-test?

A t-test for two paired samples is a hypothesis test that attempts to make a claim about the population means (\(\mu_1\) and \(\mu_2\)). More specifically, a t-test uses sample information to assess how plausible it is for difference \(\mu_1\) - \(\mu_2\) to be equal to zero.

The test has two non-overlapping hypotheses, the null and the alternative hypothesis. The null hypothesis is a statement about the population parameter which indicates no effect, and the alternative hypothesis is the complementary hypothesis to the null hypothesis. The idea of the test is to assess whether or not there is statistical significance. The main properties of the t-test for two paired samples are:

• The test required two dependent samples, which are actually paired or matched or we are dealing with repeated measures (measures taken from the same subjects)
• As with all hypotheses tests, depending on our knowledge about the "no effect" situation, the t-test can be two-tailed, left-tailed or right-tailed
• The main principle of hypothesis testing is that the null hypothesis is rejected if the test statistic obtained is sufficiently unlikely under the assumption that the null hypothesis is true
• The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true
• In a hypothesis tests there are two types of errors. Type I error occurs when we reject a true null hypothesis, and the Type II error occurs when we fail to reject a false null hypothesis

How do you manually calculate a paired t-test? What formula do you use?

The formula for a t-statistic for two dependent samples is:

where \(\bar D = \bar X_1 - \bar X_2\) is the mean difference and \(s_D\) is the sample standard deviation of the differences \(\bar D = X_1^i - X_2^i\), for \(i=1, 2, ... , n\).

How to use paired t-test formula

• Step 1: First, you need to define what your null and alternative hypotheses are. The choices are two-tailed, left-tailed or right-tailed.
• Step 2: Then, you need to specify your significance level. Typically, you will choose α = 0.05. This is the tolerance you accept to making a type I error
• Step 3: Based on the significance level you chose and the type of tail, you find the critical t-statistics by either looking at a t-distribution table or use a calculator or Excel. Then, you clearly state your rejection region
• Step 4: You compute the t-statistic using the formula specified above t = Dbar/(sd/√n)
• Step 5: Based on the the calculated t-statistic and if it falls in the rejection region or not, you determine whether you reject the null hypothesis or not
• Step 6: Use the conclusion of the t-test to give an interpretation in the context of the setting of the specific problem.

Paired t test example

Question : Assume that you have the following sample of paired data.

Can the null hypothesis that the population mean difference is zero be rejected at the .05 significance level.

From the sample data, it is found that the corresponding sample means are:

Also, the provided sample standard deviations are:

and the sample size is n = 7. For the score differences we have

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

This corresponds to a two-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is \(\alpha = 0.05\), and the critical value for a two-tailed test is \(t_c = 2.447\).

The rejection region for this two-tailed test is \(R = \{t: |t| > 2.447\}\)

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that \(|t| = 1.109 \le t_c = 2.447\), it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is \(p = 0.31\), and since \(p = 0.31 \ge 0.05\), it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean difference \(\mu_D = \mu_1 - \mu_2\) is different than 0, at the \(\alpha = 0.05\) significance level.

Confidence Interval

The 95% confidence interval is \(-0.862 < \mu_D < 2.291\).

What is the non-parametric alternative of the paired t-test?

This is a parametric test that should be used only if the normality assumption is met. If it fails, you should use instead this Wilcoxon Signed Ranks test . This paired t-test calculator deals with mean and standard deviation of pairs.

Other t-test applications

Often times you have two samples that are not paired, in which case you would use a t-test for two independent samples calculator . Notice that in that case the samples don't have to necessarily have the same size.

Related Calculators

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8.3.2 - hypothesis testing.

Below are the procedures for conducting a hypothesis test for two paired means. This is often referred to as a "paired means \(t\) test," "dependent means \(t\) test," or "matched pairs \(t\) test." 

Data must be paired. The difference between the two groups must be normally distributed in the population or the sample size must be at least 30.

The possible combinations of null and alternative hypotheses are:

Where \( \mu_d \) is the hypothesized difference in the population.

The calculation of the test statistic for dependent samples is similar to the calculation you performed earlier in this lesson for a single sample mean. In this formula, \(\overline{x}_d\) is used in place of \(\overline{x}\) and \(s_d\) is used in place of \(s\):

\(t=\frac{\bar{x}_d-\mu_0}{\dfrac{s_d}{\sqrt{n}}}\)

\(\overline{x}_d\) = observed sample mean difference \(\mu_0\) = mean difference specified in the null hypothesis \(s_d\) = standard deviation of the differences \(n\) = sample size (i.e., number of unique individuals)

When testing hypotheses about a mean difference, a \(t\) distribution is used to find the \(p\) value. The degrees of freedom are equal to \(n-1\) where \(n\) is the number of pairs. 

 If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis.

Based on your decision in Step 4, write a conclusion in terms of the original research question.