Module problem set 5
ecl569
College of Doctoral Studies
PSY 845: Module 5 Problem Set Solutions
Problem 1:
You are designing your dissertation. You want to see if humor affects systolic blood pressure (BP). You plan to have one condition where a lecture is presented that includes humorous remarks and humorous pictures in the accompanying PowerPoint that is used in the lecture. You plan to have a second condition where a lecture (on a different topic but of equal interest and difficulty) is presented that includes NO humorous remarks or humorous pictures in the accompanying PowerPoint.
You plan to measure each participant’s BP at the end of the lecture that is presented.
a. How would you assign participants if you plan to use an independent t-test to compare systolic BP after the humorous versus the nonhumorous lecture?
b. How would you assign participants if you plan to use a repeated measures design and the t-test for related samples to compare systolic BPs in the two conditions?
c. How would you assign participants if you plan to use a matched pairs design and the t-test for related samples to compare systolic BPs in the two conditions?
Solution:
a. Randomly assign half of the participants to condition A and half to condition B. Each participant would only hear ONE of the two lectures.
b. All of my participants would hear both lectures. However, to control for carryover effects, I would have half of these participants hear the humorous lecture first, and the other half hear the nonhumorous lecture first.
c. I would need to pretest all of my participants on some variable that I think could affect their systolic BP during the experiment. For example, I might want to know what their resting systolic BP is prior to including them in the experiment. I could premeasure on this. Then, I would pair up individuals who have similar resting systolic BPs. I would randomly assign one member of each pair to Condition A and the other member of the pair would go into Condition B. Thus, I would believe that the members of each pair were similar in systolic BP at the beginning of the lecture, which should predict that the condition they were in in the experiment would cause any differences in the systolic BPs of the matched individuals because otherwise they’d be the same.
Problem 2:
Why is a difference score used in computing the t-value for a related t-test?
Solution:
When you have the same person (or people who have been prematched on relevant characteristics) in both conditions, it is assumed that any differences in scores on the DV from one condition to the other is due to the differences in the treatment effect, rather than to other types of variability. Thus, only this difference for the two scores of the same person (or members of the matched pair) is needed. In these repeated-measures designs, the scores in both conditions are assumed to be positively correlated. That is, those who are lower on the dependent variable in one condition will also be lower on the dependent variable (DV) in the other condition, and those who are higher on the DV in one condition will also be higher on the DV in the other condition. Differences in the scores will be due to the effects of treatment, but while the mean score in each condition may vary, the distributions of the scores for the two groups should be similar.
Problem 3:
Compute the t-value for a repeated measures design:
Null hypothesis: There is no difference in outcome scores (the DV measure) between condition A and Condition B.
Alternative hypothesis: There is a difference in outcome scores (the DV measure) between condition A and Condition B.
You have set alpha at .05. Based on the t-value you compute, do you accept or reject the null hypothesis?
|
Condition A |
Condition B |
S1 |
15 |
10 |
S2 |
15 |
12 |
S3 |
12 |
8 |
S4 |
19 |
18 |
S5 |
17 |
15 |
Solution:
|
Condition A |
Condition B |
Difference Score (D) |
D2 |
S1 |
15 |
10 |
5 |
25 |
S2 |
15 |
12 |
3 |
9 |
S3 |
12 |
8 |
4 |
16 |
S4 |
19 |
18 |
1 |
1 |
S5 |
17 |
15 |
2 |
4 |
D = 15
Mean D, MD = 3 D2 = 55
D2/n = 11
(D)2
SS = D2 - n = 55 – [(15)2/5] = 55 – (225/5) = 55 – 45 = 10
Sample Variance = s2 = SS/n-1 = 10/4 = 2.5
Estimated standard error = sm = s2/n =
MD 3
t = sm = .71 = 4.22 with a df = 4 (number of paired data – 1). Go to Table B.2. The alternative hypothesis here is nondirectional, meaning that the 5% is divided between the two tails. I find in the Table that the critical t-value for rejecting the null hypothesis with alpha = .05 and a two-tailed test to be 2.776. My computed t-value is 4.22. Thus, my value is greater than the critical value and I can reject the null hypothesis.
Computing the t-Statistic for Repeated Measures
First, take each pair of scores (either for the same participant or for a participant and his/her matched pair mate) and subtract the score on the DV that was obtained in the treatment condition (or post, if a pre-post design) from the one that was obtained in the no treatment condition (or pre, if a pre-post design). The resulting value for each participant or pair is the D score. The D scores are then used for all further computations of t.
Next, compute the Mean of the D scores, MD. Then, compute the SS by squaring each of the D scores, adding them together for the first term; then square the sum of the Ds and divide this answer by n. Subtract the second term from the first term to get SS.
(D)2
SS = D2 - n .
The degrees of freedom remain n-1.
The sample variance and estimated standard error are then calculated. The sample variance is found by dividing the SS by the degree of freedom (s2 = SS/n-1). The estimated standard error is found by dividing the sample variance by the sample size and then calculating the square root of that quotient (sm = s2/n).
Compute the t-statistic by dividing the mean of the D scores, MD, by the estimated standard error (MD/sm). Then compare the computed t-value with the criterion value in a t-value table that matches the alpha level and the direction/nondirectionality of the hypothesis. (When possible, use a t-value calculator to get the actual value of p.)
Problem 4:
Compute the effect size, Cohen’s d, for the problem in Question 3. (See p. 361 in textbook, Chap. 11). Interpret the effect size (what does it tell you?).
Solution:
Sample mean difference MD
Estimated d = Sample Standard Deviation = s
s is the square root of the sample variance. So, going back to our data and computations in problem 3, we find that MD = 3. The variance is 2.5, so s is the square root of this = 1.58.
3
d = 1.58 = 1.9. This is a large effect size for the difference between the scores in Condition A and in Condition B.
Problem 5:
What does 95% confidence interval mean? What would it be for the study described in Question 3? Why do we compute it? (See pp. 361-362. This is also discussed in previous chapters).
Solution:
A confidence interval is a range in which we expect the population mean to fall. 95% confidence interval means that we are 95% sure that the mean difference between the mean in Condition A and the mean in Condition B will fall in the range we compute. You can also compute 99% confidence intervals, etc. We compute it because it gives us another estimate of effect size by estimating the population mean difference between the two treatment conditions.
For a repeated measures design with one IV and two conditions, as with the study in question 2, we use the mean difference, MD, and estimated standard error of the mean difference, sM, to construct the interval.
Our MD = 3; the sM = .71.
Go to Table B.2. Look at what the critical t is for df = 4 with .05 combined in both tails. You will see the critical value (which we looked up before) is 2.776.
Estimated population mean = MD + t(SM)
= 3 – 2.776(.71) (lower limit of the range for 95% CI)
= 3 + 2.776(.71) (upper limit of the range for 95% CI)
95% Confidence Interval for the size of the population mean difference between
condition A and condition B = 1.03 to 4.97.
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