PSY 201 Ch 11 - Repeated Measures
1. What assumptions must be satisfied for repeated-measures t tests to be valid?
1. The observations within a treatment are independent. The population distribution of D scores is assumed to be normal.
1. For a research study comparing two treatment conditions, what characteristic differentiates a repeated-measures design from an independent-measures design?
. For a repeated-measures design, the same group of individuals is tested in both of the treatments. An independent-measures design uses a separate group for each treatment
1. A researcher is investigating the effectiveness of acupuncture treatment for chronic back pain. A sample of n = 4 participants is obtained from a pain clinic. Each individual ranks the current level of pain and then begins a 6-week program of acupuncture treatment. At the end of the program, the pain level is rated again and the researcher records the amount of difference between the two ratings. For this sample, pain level decreased by an average of M = 4.5 points with SS = 27. a. Are the data sufficient to conclude that acupuncture has a significant effect on back pain? Use a two-tailed test with a = .05. b. Can you conclude that acupuncture significantly reduces back pain? Use a one-tailed test with a = .05.
1. a. For these data, the sample variance is 9, the standard error is 1.50, and t = 3.00. With df = 3, the critical values are t = ±3.182. Fail to reject the null hypothesis. b. For a one-tailed test, the critical value is t = 2.353. Reject the null hypothesis and conclude that acupuncture treatment significantly reduces pain
1. For the following studies, indicate whether a repeatedmeasures t test is the appropriate analysis. Explain your answers. a. A researcher is comparing the amount of time spent playing video games each week for college males versus college females. b. A researcher is comparing two new designs for cell phones by having a group of high school students send a scripted text message on each model and measuring the difference in speed for each student. c. A researcher is evaluating the effects of fatigue by testing people in the morning when they are well rested and testing again at midnight when they have been awake for at least 14 hours.
1. a. This is an independent-measures experiment with two separate samples. b. This is repeated-measures. The same individuals are measured twice. c. This is repeated-measures. The same individuals are measured twice.
10. Research has shown that losing even one night's sleep can have a significant effect on performance of complex tasks such as problem solving (Linde & Bergstroem, 1992). To demonstrate this phenomenon, a sample of n = 25 college students was given a problem-solving task at noon on one day and again at noon on the following day. The students were not permitted any sleep between the two tests. For each student, the difference between the first and second score was recorded. For this sample, the students averaged MD = 4.7 points better on the first test with a variance of s2 = 64 for the difference scores. a. Do the data indicate a significant change in problemsolving ability? Use a two-tailed test with a = .05. b. Compute an estimated Cohen's d to measure the size of the effect
10. a. The null hypothesis says that losing one night's sleep will have no effect on performance. For these data, the estimated standard error is 1.60, and t(24) = 2.94. With df = 24, the critical boundaries are ±2.064. Reject the null hypothesis. b. Cohen's d = 4.7/64 = 0.5875
11. Strack, Martin, and Stepper (1988) reported that people rate cartoons as funnier when holding a pen in their teeth (which forced them to smile) than when holding a pen in their lips (which forced them to frown). A researcher attempted to replicate this result using a sample of n = 25 adults between the ages of 40 and 45. For each person, the researcher recorded the difference between the rating obtained while smiling and the rating obtained while frowning. On average the cartoons were rated as funnier when the participants were smiling, with an average difference of MD = 1.6 with SS = 150. a. Do the data indicate that the cartoons are rated significantly funnier when the participants are smiling? Use a one-tailed test with a = .01. b. Compute r2 to measure the size of the treatment effect. c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report
11. a. The null hypothesis says that there is no difference in judgments for smiling versus frowning. For these data, the sample variance is 6.25, the estimated standard error is 0.5, and t = 1.6/0.5 = 3.20. For a one-tailed test with df = 24, the critical value is 2.492. Reject the null hypothesis. b. r2 = 10.24/34.24 = 0.299 (29.9%) c. The cartoons were rated significantly funnier when people held a pen in their teeth compared to holding a pen in their lips, t(24) = 3.20, p < .01, one tailed, r2 = 0.299.
12. How would you react to doing much worse on an exam than you expected? There is some evidence to suggest that most individuals believe that they can cope with this kind of problem better than their fellow students (Igou, 2008). In the study, participants read a scenario of a negative event and were asked to use a 10-point scale to rate how it would affect their immediate well-being (-5 strongly worsen to +5 strongly improve). Then they were asked to imagine the event from the perspective of an ordinary fellow student and rate how it would affect that person. The difference between the two ratings was recorded. Suppose that a sample of n = 25 participants produced a mean difference ofMD = 1.28 points (self rated higher) with a standard deviation of s = 1.50 for the difference scores. a. Is this result sufficient to conclude that there is a significant difference in the ratings for self versus others? Use a two-tailed test with a = .05. b. Compute r2 and estimate Cohen's d to measure the size of the treatment effect. c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.
12. a. The null hypothesis states that self-ratings are not different from ratings of others. For these data the estimated standard error is 0.30 and t = 4.27. With df = 24, the critical value is 2.064. Reject the null hypothesis and conclude that self-ratings are significantly different from ratings of others. b. r2 = 4.272/(4.272 + 24) = 0.432 and Cohen's d = 1.28/1.50 = 0.853. c. Participants rated their own ability to cope significantly higher than the ability of others, t(24) = 4.27, p < .01, d = 0.853
13. Research results indicate that physically attractive people are also perceived as being more intelligent (Eagly, Ashmore, Makhijani, & Longo, 1991). As a demonstration of this phenomenon, a researcher obtained a set of 10 photographs, 5 showing men who were judged to be attractive and 5 showing men who were judged to be unattractive. The photographs were shown to a sample of n = 25 college students and the students were asked to rate the intelligence of the person in the photo on a scale from 1 to 10. For each student, the researcher determined the average rating for the 5 attractive photos and the average for the 5 unattractive photos, and then computed the difference between the two scores. For the entire sample, the average difference was MD = 2.7 (attractive photos rated higher) with s = 2.00. Are the data sufficient to conclude that there was a significant difference in perceived intelligence for the two sets of photos? Use a two-tailed test at the .05 level of significance
13. The null hypothesis states that there is no difference in the perceived intelligence between attractive and unattractive photos. For these data, the estimated standard error is 0.4 and t = 2.7/0.4 = 6.75. With df = 24, the critical value is 2.064. Reject the null hypothesis.
14. Researchers have noted a decline in cognitive functioning as people age (Bartus, 1990). However, the results from other research suggest that the antioxidants in foods such as blueberries may reduce and even reverse these age-related declines (Joseph et al., 1999). To examine this phenomenon, suppose that a researcher obtains a sample of n = 16 adults who are between the ages of 65 and 75. The researcher uses a standardized test to measure cognitive performance for each individual. The participants then begin a 2-month program in which they receive daily doses of a blueberry supplement. At the end of the 2-month period, the researcher again measures cognitive performance for each participant. The results show an average increase in performance of MD = 7.4 with SS = 1215. a. Does this result support the conclusion that the antioxidant supplement has a significant effect on cognitive performance? Use a two-tailed test with a = .05. b. Construct a 95% confidence interval to estimate the average cognitive performance improvement for the population of older adults.
14. a. The null hypothesis states that the antioxidant has no effect on cognitive performance. For these data, the estimated standard error is 2.25, and t(15) = 3.29. With df = 15 the critical value is 2.131. Reject the null hypothesis, the antioxidant has a significant effect. b. For 95% confidence use t = 2.131 and the interval extends from 2.61 to 12.19.
15. The following data are from a repeated-measures study examining the effect of a treatment by measuring a group of n = 4 participants before and after they receive the treatment. a. Calculate the difference scores and MD. b. Compute SS, sample variance, and estimated standard error. c. Is there a significant treatment effect? Use a = .05, two tails. Participant Before Treatment After Treatment A 7 10 B 6 13 C 9 12 D 5 8
15. a. The difference scores are 3, 7, 3, and 3. MD = 4. b. SS = 12, sample variance is 4, and the estimated standard error is 1. c. With df = 3 and α = .05, the critical values are t = ±3.182. For these data, t = 4.00. Reject H0. There is a significant treatment effect.
16. A researcher for a cereal company wanted to demonstrate the health benefits of eating oatmeal. A sample of 9 volunteers was obtained and each participant ate a fixed diet without any oatmeal for 30 days. At the end of the 30-day period, cholesterol was measured for each individual. Then the participants began a second 30-day period in which they repeated exactly the same diet except that they added 2 cups of oatmeal each day. After the second 30-day period, cholesterol levels were measured again and the researcher recorded the difference between the two scores for each participant. For this sample, cholesterol scores averaged MD = 16 points lower with the oatmeal diet with SS = 538 for the difference scores. a. Are the data sufficient to indicate a significant change in cholesterol level? Use a two-tailed test with a = .01. b. Compute r2, the percentage of variance accounted for by the treatment, to measure the size of the treatment effect. c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.
16. a. The null hypothesis states that the oatmeal has no effect on cholesterol level. For these data, the sample variance is 67.25, the estimated standard error is 2.73, and t(8) = 5.86. With df = 8 and α = .01, the critical values are t = ±3.355. Reject the null hypothesis. b. r2 = 5.862/(5.862 + 8) = 0.811 c. The results show that the oatmeal has a significant effect on cholesterol levels, t(8) = 5.86, p < .01, r2 = 0.811.
17. A variety of research results suggest that visual images interfere with visual perception. In one study, Segal and Fusella (1970) had participants watch a screen, looking for brief presentations of a small blue arrow. On some trials, the participants were also asked to form a mental image (for example, imagine a volcano). The results for a sample ofn = 6, show that participants made an average of MD = 4.3 more errors while forming images than while not forming images. The difference scores had SS = 63. Do the data indicate a significant difference between the two conditions? Use a two-tailed test with a = .05
17. The null hypothesis states that the images have no effect on performance. For these data, the sample variance is 12.6, the estimated standard error is 1.45, and t(5) = 2.97. With df = 5 and α = .05, the critical values are t = ±2.571. Reject the null hypothesis, the images have a significant effect.
18. One of the primary advantages of a repeated-measures design, compared to independent-measures, is that it reduces the overall variability by removing variance caused by individual differences. The following data are from a research study comparing two treatment conditions.
18. a. The pooled variance is 6.4 and the estimated standard error is 1.46. b. For the difference scores the variance is 2 and the estimated standard error is 0.58.
. Assume that the data are from an independentmeasures study using two separate samples, each with n = 6 participants. Compute the pooled variance and the estimated standard error for the mean difference. b. Now assume that the data are from a repeatedmeasures study using the same sample of n = 6 participants in both treatment conditions. Compute the variance for the sample of difference scores and the estimated standard error for the mean difference. (You should find that the repeatedmeasures design substantially reduces the variance and the standard error.) Treatment 1 Treatment 2 Difference 10 13 3 12 12 0 8 10 2 6 10 4 5 6 1 7 9 2 M = 8 SS = 34 M = 10 SS = 30 MD = 2 SS = 10
19. a. The pooled variance is 6.4 and the estimated standard error is 1.46. b. For the difference scores the variance is 12.8, the estimated standard error is 1.46.
2. Participants enter a research study with unique characteristics that produce different scores from one person to another. For an independent-measures study, these individual differences can cause problems. Briefly explain how these problems are eliminated or reduced with a repeated-measures study
2. One problem with individual differences is that the participants in one group may be noticeably different (smarter, older, etc.) than those in another group and these differences may explain why the groups have different means. This problem is eliminated with a repeated-measures design because the same individuals are in both groups. The second problem is that individual differences can increase variance. In a repeated-measures design, the individual differences are subtracted out of the variance.
2. Describe some situations for which a repeated-measures design is well suited.
2. The repeated-measures design is suited to situations in which a particular type of subject is not readily available for study. This design is helpful because it uses fewer subjects (only one sample is needed). Certain questions are addressed more adequately by a repeatedmeasures design—for example, any time one would like to study changes across time in the same individuals. Also, when individual differences are large, a repeated-measures design is helpful because it reduces the amount of this type of error in the statistical analysis.
2. Describe the data used to compute the sample mean and the sample variance for the repeated-measures t statistic.
2. The two scores obtained for each individual are used to compute a difference score. The sample of difference scores is used to compute the mean and variance.
2. Compute the effect size using both Cohen's d and r2 acupuncture study in the previous question.
2. d = 4.5/3 = 1.50 and r2 = 9/12 = 0.75
20. A researcher uses a matched-subjects design to investigate whether single people who own pets are generally happier than singles without pets. A mood inventory questionnaire is administered to a group of 20- to 29-year-old non—pet owners and a similar age group of pet owners. The pet owners are matched one to one with the non—pet owners for income, number of close friendships, and general health. The data are as follows: Matched Non-Pet Pet Pair Owner Owner A 12 14 B 8 7 C 10 13 D 9 9 E 7 13 F 10 12 a. Is there a significant difference in the mood scores for non—pet owners versus pet owners? Test with a = .05 for two tails. b. Construct the 95% confidence interval to estimate the size of the mean difference in mood between the population of pet owners and the population of non—pet owners. (You should find that a mean difference of IID = 0 is an acceptable value, which is consistent with the conclusion from the hypothesis test.) 19. The previous problem demonstrates that removing individual differences can substantially reduce variance and lower the standard error. However, this benefit only occurs if the individual differences are consistent across treatment conditions. In problem 18, for example, the first two participants (top two rows) consistently had the highest scores in both treatment conditions. Similarly, the last two participants consistently had the lowest scores in both treatments. To construct the following data, we started with the scores in problem 18 and scrambled the scores in treatment 1 to eliminate the consistency of the individual differences. a. Assume that the data are from an independentmeasures study using two separate samples, each with n = 6 participants. Compute the pooled variance and the estimated standard error for the mean difference. b. Now assume that the data are from a repeatedmeasures study using the same sample of n = 6 participants in both treatment conditions. Compute the variance for the sample of difference scores and the estimated standard error for the mean difference. (This time you should find that removing the individual differences does not reduce the variance or the standard error.) reatment 1 Treatment 2 Difference 6 13 7 7 12 5 8 10 2 10 10 0 5 6 1 12 9 -3
20. a. The null hypothesis says that pets do not have an effect on mood, H 0: μD = 0. With df = 5 and α = .05, the critical values are t = ±2.571. For these data, MD = 2, SS = 30, the estimated standard error is 1.00, and t(5) = 2.00. Fail to reject H0. b. For 95% confidence, use t = ±2.571. The interval extends from 0.571 to 4.571. (Note that a value of zero is contained in the interval.)
21. There is some evidence suggesting that you are likely to improve your test score if you rethink and change answers on a multiple-choice exam (Johnston, 1975). To examine this phenomenon, a teacher gave the same final exam to two sections of a psychology course. The students in one section were told to turn in their exams immediately after finishing, without changing any of their answers. In the other section, students were encouraged to reconsider each question and to change answers whenever they felt it was appropriate. Before the final exam, the teacher had matched 9 students in the first section with 9 students in the second section based on their midterm grades. For example, a student in the no-change section with an 89 on the midterm exam was matched with student in the change section who also had an 89 on the midterm. The final exam grades for the 9 matched pairs of students are presented in the following table. a. Do the data indicate a significant difference between the two conditions? Use a two-tailed test with a = .05. b. Construct a 95% confidence interval to estimate the size of the population mean difference. c. Write a sentence demonstrating how the results of the hypothesis test and the confidence interval would appear in a research report. Matched Pair No-Change Section Change Section #1 71 86 #2 68 80 #3 91 88 #4 65 74 #5 73 82 #6 81 89 #7 85 85 #8 86 88 #9 65 76
21. a. The null hypothesis says that changing answers has no effect, H0: μD = 0. With df = 8 and α = .05, the critical values are t = ±2.306. For these data, MD = 7, SS = 288, the standard error is 2, and t(8) = 3.50. Reject H0 and conclude that changing answers has a significant effect on exam performance. b. For 95% confidence use t = ±2.306. The interval extends from 2.388 to 11.612. c. Changing answers resulted in significantly higher exam scores, t(8) = 3.50, p < .05, 95% CI [2.388, 11.612].
22. The teacher from the previous problem also tried a different approach to answering the question of whether changing answers helps or hurts exam grades. In a separate class, students were encouraged to review their final exams and change any answers they wanted to before turning in their papers. However, the students had to indicate both the original answer and the changed answer for each question. The teacher then graded each exam twice, one using the set of original answers and once with the changes. In the class of n = 22 students, the average exam score improved by an average of MD = 2.5 points with the changed answers. The standard deviation for the difference scores was s = 3.1. Are the data sufficient to conclude that rethinking and changing answers can significantly improve exam scores? Use a one-tailed test at the .01 level of significance
22. The null hypothesis says that changing answers should not affect the student's scores. With df = 21 and α = .01, the one-tailed critical value is t = 2.518. For these data, the estimated standard error is 0.66, and t(21) = 3.79. Reject H0.
23. At the Olympic level of competition, even the smallest factors can make the difference between winning and losing. For example, Pelton (1983) has shown that Olympic marksmen shoot much better if they fire between heartbeats, rather than squeezing the trigger during a heartbeat. The small vibration caused by a heartbeat seems to be sufficient to affect the marksman's aim. The following hypothetical data demonstrate this phenomenon. A sample of n = 8 Olympic marksmen fires a series of rounds while a researcher records heartbeats. For each marksman, a score is recorded for shots fired during heartbeats and for shots fired between heartbeats. Do these data indicate a significant difference? Test with a = .05. Participant During Heartbeats Between Heartbeats A 93 98 B 90 94 C 95 96 D 92 91 E 95 97 F 91 97 G 92 95 H 93 9
23. The null hypothesis says that there is no difference between shots fired during versus between heart beats, H0: μD = 0. With α = .05, the critical region consists of t values beyond ±2.365. For these data, MD = 3, SS = 36, s2 = 5.14, the standard error is 0.80, and t(7) = 3.75. Reject H0 and conclude that the timing of the shot has a significant effect on the marksmen's scores.
24. The Preview section of this chapter presented a repeated-measures research study demonstrating that swearing can help reduce pain (Stephens, Atkins, & Kingston, 2009). In the study, each participant was asked to plunge a hand into icy water and keep it there as long as the pain would allow. In one condition, the participants repeated their favorite curse words while their hands were in the water. In the other condition, the participants repeated a neutral word. Data similar to the results obtained in the study are shown in the following table. a. Do these data indicate a significant difference in pain tolerance between the two conditions? Use a two-tailed test with a = .05. b. Compute ,-2, the percentage of variance accounted for, to measure the size of the treatment effect. c. Write a sentence demonstrating how the results of the hypothesis test and the measure of effect size would appear in a research report. Amount of Time (in Seconds) Participant Swear Words Neutral Words 1 94 59 2 70 61 3 52 47 4 83 60 5 46 35 6 117 92 7 69 53 8 39 30 9 51 56 10 73 61
24. a. For these data, MD = 14, s2 = 128, the standard error is 3.58, and t(9) = -3.91. For a two- tailed test with α = .05 the critical boundary is t = 2.262. Reject H0 and conclude that swearing significantly increases pain tolerance. b. r2 = 15.288/24.288 = 0.629. c. Swearing significantly increased the amount of time that participants could tolerate the icy water, t(9) = -3.91, p < .05, r2 = 0.629.
3. Explain the difference between a matched-subjects design and a repeated-measures design.
3. For a repeated-measures design the same subjects are used in both treatment conditions. In a matched-subjects design, two different sets of subjects are used. However, in a matched-subjects design, each subject in one condition is matched with respect to a specific variable with a subject in the second condition so that the two separate samples are equivalent with respect to the matching variable.
3. In words and in symbols, what is the null hypothesis for a repeated-measures t test?
3. The null hypothesis states that, for the general population, the average difference between the two conditions is zero. In symbols, pD = 0.
3. How is a matched-subjects design similar to a repeated-measures design? How do they differ?
3. They are similar in that the role of individual differences in the experiment is reduced. They differ in that there are two samples in a matched-subjects design and only one in a repeatedmeasures study.
3. A computer printout for a repeated-measures t test reports a p value of p = .021. a. Can the researcher claim a significant effect with a =.01? b. Is the effect significant with a =.05?
3. a. The exact p value, p = .021, is not less than a = .01. Therefore, the effect is not significant for a = .01 (p > .01). b. The p value is less than .05, so the effect is significant with a = .05
4. A researcher conducts an experiment comparing two treatment conditions and obtains data with 10 scores for each treatment condition. a. If the researcher used an independent-measures design, how many subjects participated in the experiment? b. If the researcher used a repeated-measures design, how many subjects participated in the experiment? c. If the researcher used a matched-subjects design, how many subjects participated in the experiment?
4. a. An independent-measures design would require two separate samples, each with 10 subjects, for a total of 20 subjects. b. A repeated-measures design would use the same sample of n = 10 subjects in both treatment conditions. c. A matched-subjects design would require two separate samples with n = 10 in each, for a total of 20 subjects.
4. The data from a research study consist of 10 scores in each of two different treatment conditions. How many individual subjects would be needed to produce these data a. For an independent-measures design? b. For a repeated-measures design? c. For a matched-subjects design?
4. a. The independent-measures design would require 20 subjects (two separate samples with n = 10 in each). b. The repeated-measures design would require 10 subjects (the same 10 individuals are measured in both treatments). c. The matched-subjects design would require 20 subjects (10 matched pairs).
5. A sample of n = 9 individuals participates in a repeatedmeasures study that produces a sample mean difference of MD = 6.5 with SS = 200 for the difference scores. a. Calculate the standard deviation for the sample of difference scores. Briefly explain what is measured by the standard deviation. b. Calculate the estimated standard error for the sample mean difference. Briefly explain what is measured by the estimated standard error.
5. a. The standard deviation is 5 points and measures the average distance between an individual score and the sample mean. b. The estimated standard error is 1.67 points and measures the average distance between a sample mean and the population mean.
6. a. A repeated-measures study with a sample of n = 25 participants produces a mean difference ofMD = 3 with a standard deviation of s = 4. Based on the mean and standard deviation, you should be able to visualize (or sketch) the sample distribution. Use a two-tailed hypothesis test with a = .05 to determine whether it is likely that this sample came from a population with p, = 0. b. Now assume that the sample standard deviation is s = 12, and once again visualize the sample distribution. Use a two-tailed hypothesis test with a = .05 to determine whether it is likely that this sample came from a population with pp = 0. Explain how the size of the sample standard deviation influences the likelihood of finding a significant mean difference
6. a. The estimated standard error is 0.8 points and t(8) = 3.75. With a critical boundary of ±2.064, reject the null hypothesis. b. The estimated standard error is 2.4 point and t(8) = 1.25. With a critical boundary of ±2.064, fail to reject the null hypothesis. c. A larger standard deviation (or variance) reduces the likelihood of finding a significant difference.
7. a. A repeated-measures study with a sample of n = 9 participants produces a mean difference of MD = 3 with a standard deviation of s = 6. Based on the mean and standard deviation, you should be able to visualize (or sketch) the sample distribution. Use a two-tailed hypothesis test with a = .05 to determine whether it is likely that this sample came from a population with No = 0. b. Now assume that the sample mean difference is MD = 12, and once again visualize the sample distribution. Use a two-tailed hypothesis test with a = .05 to determine whether it is likely that this sample came from a population with 1.tp = 0. c. Explain how the size of the sample mean difference influences the likelihood of finding a significant mean difference.
7. a. The estimated standard error is 2 points and t(8) = 1.50. With a critical boundary of ±2.306, fail to reject the null hypothesis. b. With MD = 12, t(8) = 6.00. With a critical boundary of ±2.306, reject the null hypothesis. c. The larger the mean difference, the greater the likelihood of finding a significant difference.
8. A sample of difference scores from a repeated-measures experiment has a mean ofMD = 4 with a standard deviation of s = 6. a. If n = 4, is this sample sufficient to reject the null hypothesis using a two-tailed test with a = .05? b. Would you reject H0 if n = 16? Again, assume a two-tailed test with a = .05. c. Explain how the size of the sample influences the likelihood of finding a significant mean difference.
8. a. With n = 4, the estimated standard error is 3 and t(3) = 4/3 = 1.33. With critical boundaries of ±3.182, fail to reject H0. b. With n = 16, the estimated standard error is 1.5 and t(15) = 4/1.5 = 2.67. With critical boundaries of ±2.131, reject H0 c. If other factors are held constant, a larger sample increases the likelihood of finding a significant mean difference.
9. As mentioned in Chapters 2 and 3 (pp. 38 and 81), Steven Schmidt (1994) reported a series of studies examining the effect of humor on memory. In one part of the study, participants were presented with a list containing a mix of humorous and nonhumorous sentences, and were then asked to recall as many sentences as possible. Schmidt recorded the number of humorous and the number of nonhumorous sentences recalled by each individual. Notice that the data consist of two memory scores for each participant. Suppose that a difference score is computed for each individual in a sample of n = 16 and the resulting data show that participants recalled an average of = 3.25 more humorous sentences than nonhumorous, with SS = 135. Are these results sufficient to conclude that humor has a significant effect on memory? Use a two-tailed test with a = .05.
9. The sample variance is 9, the estimated standard error is 0.75, and t(15) = 4.33. With critical boundaries of ±2.131, reject H0
In a matched-subjects design, each individual in one sample is
matched with an individual in the other sample. The matching is done so that the two individuals are equivalent (or nearly equivalent) with respect to a specific variable that the researcher would like to control.
A repeated-measures design, or a within-subject design, is one in which the dependent variable is
measured two or more times for each individual in a single sample. The same group of subjects is used in all of the treatment conditions.
