Paired-Samples t Test
Research Scenario
Question
Woody wants to understand if workers prioritize job security or pay. He asks 30 individuals who work in different settings and to rate his or her concern about salary level and job security on a scale from 1 to 10.
Instrument & Scoring
Woody’s data file contains 30 cases, one for each employee, and two variables: ratings identifying level of concern for salary level and job security each.
Null Hypothesis
Woody is interested in understanding whether employees are more concerned about pay or job security, or if there is no difference in importance, when evaluating the two. Therefore, the null hypothesis for this study would be that employees do not report a significantly higher degree of concern for pay over job security (or vice versa). In other words, no difference between the ratings.
Assumptions
- Difference scores—which are created by subtracting one variable from another (in this case job security scores—column 1 in the data set—were subtracted from pay scores—column 2 in the data set)—are normally distributed in the population.
- The cases represent a random sample from the population, and the scores on the test variable are independent of each other.
Check
Of the two major assumptions for the paired-samples t test, only the first can be evaluated using data alone. The second assumption is something that is best addressed through implementing proper research/sampling practices–simply put, if this assumption is found to be violated in the data analysis phase there is little the researcher can do to address the issue.
Difference Scores are Normally Distributed
To evaluate the first assumption of normality, develop a histogram of the computed difference scores. Furthermore, and if desired, conduct skewness and kurtosis analyses.
skewness: 0.46
kurtosis: 1.28
Conclusion
In reviewing the histogram along with the skewness and kurtosis values above, the data appears to approximate a normal distribution. The first assumption is confirmed.
Paired-Samples t Test Overview
The Paired-Samples t Test compares two means that are from the same individual, object, or related units. The two means typically represent two different times (e.g., pre-test and post-test with an intervention between the two time points) or two different but related conditions or units (e.g., left and right ears, twins).
The purpose of the test is to determine whether there is statistical evidence that the mean difference between paired observations on a particular outcome is significantly different from zero. With the dependent variable, or test variable, measured at two different times or for two related conditions or units.
Other names for the Paired-Samples t Test include:
- Dependent t Test
- Paired t Test
- Repeated Measures t Test
Paired-Samples t Test Overall Equation
\[ t = \frac{\frac{\Sigma d}{N}}{\sqrt\frac{\Sigma d^2 - \frac{(\Sigma d)^2}{N}}{N(N-1)}} \]
Simplified Equation
\[ t = \frac{\frac{\text{Sum of Differences}}{N}}{\sqrt\frac{\text{Sum of Squared Differences} - \frac{(\text{Sum of Differences})^2}{N}}{N(N-1)}} \]
Data Needed to Complete the Paired-Samples t Test
To conduct a paired-samples t test three pieces of information are required, which Woody will need to compute.
- sum of differences = ?
- sum of squared differences = ?
- sample size (N) = ?
| Pay | Job Security | Difference | Squared Difference | |
|---|---|---|---|---|
| 1 | 4 | 3 | 1 | 1 |
| 2 | 9 | 1 | 8 | 64 |
| 3 | 8 | 6 | 2 | 4 |
| 4 | 5 | 7 | -2 | 4 |
| 5 | 4 | 6 | -2 | 4 |
| 6 | 6 | 5 | 1 | 1 |
| 7 | 6 | 3 | 3 | 9 |
| 8 | 4 | 4 | 0 | 0 |
| 9 | 7 | 8 | -1 | 1 |
| 30 | 6 | 2 | 4 | 16 |
| Note: | ||||
| This table is for illustrative purposes only and is not representative of all 30 cases in the dataset. |
In the table above, Woody has gathered the outstanding data needed to conduct the paired-samples t test:
- sum of differences = 35
- sum of squared differences = 189
- sample size (N) = 30
Step 1: Input All of the Data
\[ t = \frac{\frac{35}{30}}{\sqrt\frac{189 - \frac{1225}{30}}{30(29)}} \]
Step 2: Simplify the Numerator and Denominator
\[ t = \frac{1.17}{0.41} \]
Step 3: Round the Resulting t Value
\[ t = 2.83 \]
Step 4: Determine Significance
To determine if the computed t value is significant and therefore, reject the null hypothesis, consult a table of critical t values. A critical value is the value that a test statistic (in that case the t statistic) must exceed in order for the null hypothesis to be rejected. In this instance, using degrees of freedom equal to 29 with a p value of 0.05, the critical value for this t test is 2.042.
Paired-Samples t Test Conclusion
The observed difference appears to be significant since the computed t value of 2.83 is larger than the critical t value of 2.042.
Effect Size
An effect size is a way of quantifying the size of the difference between two groups; or in this case, a measure of magnitude of difference of concern regarding pay and job security for Woody’s participants. To evaluate effect size, use Cohen’s d effect size.
Cohen’s d Overall Equation
\[ d = \frac{t}{\sqrt{n}} \]
Input t and sample size \[d = \frac{2.83}{\sqrt{30}} \]
Cohen’s d Outcome
\[d = 0.52 \]
The resulting Cohen’s d is considered to be moderate.
A work by Alex Aguilar
aaguilar@thechicagoschool.edu