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Fridayt-test.pptxWeek Four
Paired t-tests and independent sample t-tests
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Statistical Tests:
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Differences in two groups of data
In many instances, it is necessary to compare numeric data from two groups to see which one is greater than the other. How do we know that random chance did not cause the apparent difference between groups if one group had a mean score of 79 and the other produced an average score of 81? In this instance, I am thinking about continuous data, means, and normal curves, whether skewed or not. Such a situation calls for a t-test.
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Connection to your dissertation
Page 23 of the Dissertation and Case Study Handbook contains an example of how to report a t-test.
(t [df] = t statistic, p less than or greater than .05).
In chapter four and your abstract (where you report test results) also include means of the groups.
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What is the difference in types of t-tests?
Before we go further, it is important to point out the situational differences between a paired samples and independent samples t-test. An independent samples t-test is used to compare data from two groups when there is no direct connection between the two groups (i.e., males and females). A paired samples t-test is used in a pre- and post-test scenario. Another example would be data on freshmen infractions as compared to the same students’ infractions tallied during their sophomore year. If you were comparing different groups of freshmen to sophomores, then it would require an independent samples t-test.
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Types of t- tests
Independent samples also known as:
Two-Sample Assuming Equal Variance
Paired samples also known as:
Repeated Measures
Matched
Natural
Correlated samples t-tests
Paired sample for means
T distribution on table D in Appendix C (Spatz, 10th edition)
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An example of a paired samples t-test from a dissertation
A paired sample t-test was administered to compare data generated between Transactional Leadership MBE: Active and Laissez-Faire Leadership MBE: Passive. Data in Table 8 show a significant difference between the samples Transactional Leadership characteristic of Management by Exception: Active (M = 1.44) and the Laissez-Faire Leadership characteristic of Management by Exception: Passive (M = 1.01), (t [66] = 4.39, p < .01).
| MBE: Active | MBE: Passive | |
| Mean | 1.440298507 | 1.013432836 |
| Variance | 0.576684758 | 0.60936228 |
| Observations | 67 | 67 |
| Pearson Correlation | 0.468086249 | |
| Hypothesized Mean Difference | 0 | |
| Df | 66 | |
| t Stat | 4.398296847 | |
| P(T<=t) one-tail | 2.03352E-05 | |
| t Critical one-tail | 1.668270515 | |
| P(T<=t) two-tail | 4.06705E-05 | |
| t Critical two-tail | 1.996564396 |
Table 8: Paired sample t-test Transactional Leadership MBE: Active and Laissez-Faire Leadership MBE: Passive
Let’s try a t-test together
A teacher wanted to know an intervention in her math class made a difference in student scores. She gave the students a pre-test before the intervention, and a post-test after the intervention.
How should she proceed?
Since she is looking at the same students, before and after the intervention, she selects a pair two sample t-test.
She loads the students scores in excel (row one contains labels, row two is student Aa__ , row three is student Ab___,…)
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Using excel:
1. Select the Data Tab
2. Select Data
Analysis
3. Scroll down to t-test, Paired Two Sample for Means
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Pre and Post Test Scores:
4. Click Variable 1 Range Box, Highlight Column A (Rows 1-21).
5. Click Variable 2 Range, Highlight Column B (Rows 1-21).
6. Click the Box labels (since row one does not contain numbers, but identifies our variables)
7. Click the output range box, and select an open cell (d18, for example).
8. Click OK.
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Your screen should look like this now.
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The results
| t-Test: Paired Two Sample for Means | ||
| Pre-Test | Post-Test | |
| Mean | 71 | 75.75 |
| Variance | 180.5263158 | 77.03947368 |
| Observations | 20 | 20 |
| Pearson Correlation | 0.238766431 | |
| Hypothesized Mean Difference | 0 | |
| df | 19 | |
| t Stat | -1.497409772 | |
| P(T<=t) one-tail | 0.075357904 | |
| t Critical one-tail | 1.729132812 | |
| P(T<=t) two-tail | 0.150715809 | |
| t Critical two-tail | 2.093024054 |
The pre-test scores (M = 71) were not significantly different than the post-test scores (M = 75.75), (t [19] = -1.498, p > .05).
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Look on the table, at 19 df, and for the alpha level under .05 in the two-tailed test column.
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An independent samples t-test example from a dissertation
Comparison of Post Legislation Implementation on Energy Use in Energy Star Districts and Districts that Employed Energy Managers. As can be seen in Table 7, the independent samples t-test for post legislation Energy Star school districts and post legislation school districts which employed energy managers indicated a significant post-legislation reduction in the mean energy use among districts that were Energy Star districts (M = 55.72), compared to districts that employed energy managers (M = 61.83); (t [40] = -1.972, p <.05). The results were significant at the .05 level, in a one-tail test, which ruled out random chance. Districts that were designated Energy Star had a greater reduction in energy consumption than those districts that only employed an energy manager.
| Energy Star Post-Legislation (2013) | Energy Manager Post-Legislation (2013) | |
| Mean | 55.71905 | 61.82857 |
| Variance | 66.82162 | 134.5731 |
| Observations | 21 | 21 |
| Pooled Variance | 100.6974 | |
| Hypothesized Mean Difference | 0 | |
| df | 40 | |
| t Stat | -1.97284 | |
| P(T<=t) one-tail | 0.027726 | |
| t Critical one-tail | 1.683852 | |
| P(T<=t) two-tail | 0.055452 | |
| t Critical two-tail | 2.021075 |
Let’s try another t-test
A state administrator was assigned two counties to evaluate their ACT performance. He wanted to know if there was a difference between ACT scores in the districts even though they use the same common core.
How should he proceed?
He selected an independent sample t-test to determine if there was a difference in the two district’s scores.
Columns were labeled for the two counties. He randomly listed all the ACT scores for each student in County A, in column A. He randomly listed all ACT scores for each student in County B, in column B.
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Click data tab, select data analysis again. Scroll down to t-test two sample assuming equal variances.
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Next:
Click Variable 1 Range, highlight Column A (Rows 1-21.)
Click Variable 2 Range, highlight Column B (Rows 1-25).
Click Labels box.
Click output range, and select a cell (d1, for example).
Click OK.
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The results:
| t-Test: Two-Sample Assuming Equal Variances | ||
| County A | County B | |
| Mean | 23.2 | 22.41666667 |
| Variance | 35.32631579 | 33.64492754 |
| Observations | 20 | 24 |
| Pooled Variance | 34.40555556 | |
| Hypothesized Mean Difference | 0 | |
| df | 42 | |
| t Stat | 0.441089379 | |
| P(T<=t) one-tail | 0.330705325 | |
| t Critical one-tail | 1.681952357 | |
| P(T<=t) two-tail | 0.66141065 | |
| t Critical two-tail | 2.018081703 |
County B ‘s ACT scores (M = 22.4) were not significantly different than County A’s ACT scores (M = 23.2), (t [42] = .44, p > .05). There is no significant difference in County A and County B’s ACT performance.
Always look at the two-tail first. Is .66 greater than .05?
Next, look at the one-tail, is .33 greater than .05?
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Next…
Practice problems using excel to solve both paired and independent sample t-tests.
Understand the difference in paired and independent sample. What do the results mean?
Note: results that look like: 4.1203E-08 indicate a very small number. If you get this result for the two tailed test, then we report that p<.001
Review: descriptive stats, skew, parametric vs. non-parametric, quantitative vs. qualitative paradigms, null/research hypothesis, chi square tests, Spearman’s rs, paired samples t-tests, and independent sample t-tests.
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