Ed 602 - Lesson 12 - The dependent t-test

Lesson 12 will consist of the following topics

• Text Assignment for Lesson 12

• The dependent t-test

• Example problem using the dependent t-test

• Using the Excel Spreadsheet program to calculate the dependent t-test

• Additional problem using the dependent t-test

• Lesson 12 Assignment

• Lesson 12 Quiz

Text Assignment for Lesson 12

For lesson 12, read pages 129-131 in Practical Statistics for Educators, Third Edition by Ruth Ravid (2005, University Press of America)
or read pages 309-316 in Basic Statistics for Behavioral Science Research 2nd ed by Mary B. Harris (1998, Allyn and Bacon)
or read pages 195-197 in Practical Statistics for Educators, 2nd Edition by Ruth Ravid (2000, University Press of America)
or read pages 186-189 in Practical Statistics for Educators by Ruth Ravid (1994, University Press of America).

The Dependent t-test

In our last lesson we looked at the process for making inferences about research involving two samples. We discussed the independent t-test used when the two samples were independent of one another. In this lesson we are going to discuss the case in which the two samples are not independent of one another or are dependent on one another. There are two situations in which the two samples are not independent:

1. Where the subjects making up the two samples are matched on some variable before being put in the two groups.
   For example we might wish to conduct an experiment measuring the effect of a new method of teaching reading on reading comprehension test scores. However we are concerned about the effect of intelligence on reading comprehension test scores, so we control for intelligence by matching the students in the study on intelligence. We give all the subjects to participate in the study an intelligence test. Then we take the two students with the highest IQ and randomly put one of the two in the experimental group and the other in the control group. We then take the two children with the next highest intelligence and do the same thing until we have selected our two groups but they are matched on the factor of tested intelligence.
2. The situation where the two groups are the same subjects administered a pre-test and a post-test.
   For example we want to study the effect of a new teaching method on reading comprehension scores so we administer a reading comprehension test to the group (pretest), then we apply the experimental teaching method to the group of students, and follow this by administering a reading comprehension test to the students again (the post test). We then see if there is a significant difference betwwen the pre-test scores and the post-test scores.

When subjects are connected to one another by either of these methods the variance is constrained, so when we use a statistical test to measure the significance of a difference between the means we must use a test that takes into consideration these constrained variances. This is what the dependent t-test does.

The formula for the dependent t is:

Where D is the difference between pairs of scores,

Notice that we subtract the score for the first X from the paired second X. This is probably so that when we are finding the difference between the pre-test and post-test, that we subtract the pre-test (X₁) from the post-test (X₂).

and the degrees of freedom for the dependent-t test is

df = n - 1

and

n

Example problem using the dependent t-test

Research Problem: The Beck Depression Scale (pre-test) was administered to ten adolescents undergoing anger management thereapy. After four weeks of therapy the same scale was administered again (post-test). Does the anger management therapy significantly reduce the scores on the depression scale?

In this problem we are comparing pre-test and post-test scores for a group of subjects. This would be an appropriate situation for the dependent t-test.

The pre-test and post-test scores, as well as D and D² are shown in the following table.

For our problem:

and the degrees of freedom for this problem is:

We now have the information we need to complete the six step process for testing statistical hypotheses for our research problem.

1. State the null hypothesis and the alternative hypothesis based on your research question.
   
   
   Note: Our problem stated that the therapy would decrease the depression score. Therefore our alternative hypothesis states that mu1 (the pre-test score) will be greater than mu2 (the post-test score).
2. Set the alpha level.
   
   Note: As usual we will set our alpha level at .05, we have 5 chances in 100 of making a type I error.
3. Calculate the value of the appropriate statistic. Also indicate the degrees of freedom for the statistical test if necessary.
   t = -2.623
   df = n - 1 = 10 - 1 = 9
   Note: We have calculated the t-value and will also need to know the degrees of freedom when we go to look up the critical value of t.
4. Write the decision rule for rejecting the null hypothesis.
   Reject H₀ if t is <= -1.833
   Note: To write the decision rule we need to know the critical value for t, with an alpha level of .05 and a one-tailed test. We can do this by looking at Appendix C (Distribution of t) on page 318 of the text book. Look for the column of the table under .05 for Level of significance for one-tailed tests, read down the column until you are level with 9 in the df column, and you will find the critical value of t which is 1.833. That means our result is significant if the calculated t value is less than or equal to -1.833.
   
   Note: Why are we looking for a negative value of t? This is a little tricky, but we are looking at the posttest being less than the pretest. Now the dependent t is calculated by subtracting the pretest from the posttest so if the posttest is actually less than the pretest, posttest minus pretest will be a negative quantity. I hope this makes sense to you. If not send me email (wasson@mnstate.edu) and I will try to clarify it further.
5. Write a summary statement based on the decision.
   Reject H₀, p < .05, one-tailed
   Note: Since our calculated value of t (-2.623) is less than or equal to -1.833, we reject the null hypothesis and accept the alternative hypothesis.
6. Write a statement of results in standard English.
   The management therapy did significantlly reduce the depression scores for the adolescents.

Please send electronic mail to the course instructor if you have any questions about this lesson or other concerns.