A paired t-test (Paired T Distribution, Paired T Test, Paired Comparison test, Paired Sample Test) is a statistical method that compares the mean and standard deviation of two matched groups to determine if there is a significant difference between the two groups. In other words, It tests whether the average difference between the two measurements is statistically significant from zero.
Student’s T distribution is used for finding confidence intervals for the population mean when the sample size is less than 30, and the population standard deviation is unknown. Further, Students’ t-test is divided into paired and unpaired t-test.
Also, See Student’s T test – for when samples are <30 in size.
When the sample groups are not independent, the appropriate method to test for differences between the groups is known as a paired comparison test (or paired t-test or paired sample test).
Furthermore, use the Paired T Distribution to identify if a change has made a significant impact on a process. Paired t-test is similar to 1-sample t-test. A 1-sample t-test compares one sample mean. Whereas, a paired t-test calculates the difference between paired values and then performs a 1-sample t-test on the differences.
When to use Paired t-test
The Paired T Distribution, Paired T Test, Paired Comparison test, Paired Sample Test is a parametric procedure. Paired samples t-test are used when same group tested twice. It is often used in “before and after” designs where the same individuals are measured both before and after a treatment or improvement to see if changes occurred over time.
- Two repeated or matched samples, In other words, must have a before and after design or matched pairs.
- Paired samples t-test can have only two groups. Use ANOVA for more than two measures.
- The non-negotiable assumption for the paired samples is dependent variable must be quantitative.
- Paired t-test assumes no extreme outliers.
- The sampling distribution of the dependent variable should be normally distributed. If it is a non-normal data, use Wilcoxon test.
- Dependent variable is a continuous variable measure at the interval or ratio.
- Unlike the independent samples t-test, there is no assumption for homogeneity of variance with a paired sample t-test.
What are the hypothesis of Paired t-test
- The null hypothesis for a paired t-test is that the average difference between the two population means is zero (0). In other words, there is no significant difference between two population means.
- The alternative hypothesis for a paired t-test – there is a significant difference between the two population means.
How to conduct Paired t-test
- Establish the null hypothesis and alternative hypothesis
- Determine the significance level
- Calculate the difference of each observation in two groups
- Then, compute the mean difference (x̅ – µ)
- Calculate the standard deviation of differences (s) and then calculate the standard error i.e s/√n (where n is sample size)
- Compute the t-statistic, t= (x̅ – µ)/ s/√n
- Determine t critical value with n-1 degrees of freedom
- Finally, interpret the result. If the test statistic falls in critical region, in other words, t statistic > t critical, reject the null hypothesis.
Example of Paired t-test
Example: Two operators are checking the same dimension on the same sample of 10 parts. Below are the results. Is there a significant operator measurement error? Test at the 5% significance level.
We need to calculate t using, t= (x̅ – µ)/ s/√n and then compare to a table value t critical.
- H0 =There is no significant measurement error between the two operators.
- H1 =There is a significant measurement error between two operators.
- DF (degrees of freedom) = n-1 ; 10-1 =9
- Significance level =5%
Calculate the difference of each observation in two groups
Compute the mean difference (x̅ – µ)
- µ (mean of first test) = 59.5 ; take the average of the 10 data points in Op1
- x̅ (mean of new test) = 60.2 ; take the average of the 10 data points in Op2
- x̅-µ = 60.2-59.5 =0.7
Calculate the standard deviation of differences (s)
- D= -7 ; d2=73
- Use sd = sqrt [ (Σ(di – d)2 / (n – 1) ] where di is the difference for pair i, d is the sample mean of the differences, or s=(√ (n*d2 -(D2/df)) / √(n)
- (√ (10*73 -(-7*-7/9)) / √(10) =2.7508
calculate the standard error
standard error = s/√n =2.7508/√10 =0.869
Compute the t-statistic
t= (x̅ – µ)/ s/√n =(0.7)/0.869 =0.805
Determine t critical value with n-1 degrees of freedom
Since this is a two-tailed test at an alpha of 5% t critical = 2.262
Interpret the results
Ccompare t statistic to t critical 0.805 < 2.262. In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. If the t calculated was smaller than the critical value, then the data did fit the model. Hence, we fail to reject the null hypothesis and say that there is no difference between the two mean values.
ASQ Six Sigma Black Belt Exam Paired T Test Questions
A black belt plans to test the performance of workers before and after training. Which of the following hypothesis tests should be used to determine whether the training actually improved the workers’ performance?
(A) 2-sample z test
(B) 2-sample t test
(C) Paired t test
(D) F test