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Psychology 10

Psychology 10. Analysis of Psychological Data April 7, 2014. The plan for today. One more example of the two-sample t test. More on confidence intervals. Introducing the repeated-measures t test. Problem 23 on page 349.

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Psychology 10

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  1. Psychology 10 Analysis of Psychological Data April 7, 2014

  2. The plan for today • One more example of the two-sample t test. • More on confidence intervals. • Introducing the repeated-measures t test.

  3. Problem 23 on page 349 • Cheating behavior: participants reported number of puzzle problem correctly solved. • Paid $.50 for each correct answer. • Easier to lie in dim room. • H0 : mdim–mbright= 0. • a two-tailed = .01.

  4. The data

  5. Summary statistics • Mean for the light room = 68 / 9 = 7.5555556. • Mean for the dim room = 102 / 9 = 11.3333333. • SS for the light room = 556 – 682 / 9 = 42.222222. • SS for the dim room = 1194 – 1022 / 9 = 38.0.

  6. The t statistic • Total SS = 42.222222 + 38 = 80.22222. • So s2P = 80.22222 / (9 + 9 – 2) = 5.0138889. • The standard error of the difference between means is √(5.0138889 / 9 + 5.0138889 / 9) = 1.0555556. • t = (11.33333 – 7.5555556) / 1.0555556 = 3.579.

  7. Evaluating the t statistic • df = 9 + 9 – 2 = 16. • tcrit = 2.921. • Our value of 3.579 is larger than that, so we reject the null hypothesis. • We conclude that people are more likely to be dishonest in a dim environment.

  8. Assumptions • Independence between groups: If assignment was random, OK. • Independence within groups: insufficient information. • Equal population variances: compare standard deviations in the samples. • 2.17 and 2.30 are very similar. • Normality?

  9. Normality Dim Light | 4 | 00 | 6 | 000 00 | 8 | 00 0000 | 10 | 0 0 | 12 | 0 00 | 14 |

  10. What is the effect size? • The pooled variance was 5.0138889, so the pooled sd is 2.239171. • d = (11.33333 – 7.55556) / 2.239171 = 1.69.

  11. What about a confidence interval? • Since we used a = .01, let’s do a 99% confidence interval. • (11.33333 – 7.55556) ± 2.921 × 1.0555556 = (0.69, 6.86). • Interpretation?

  12. Another kind of t test • Sometimes, it is more efficient to use repeated observations on the same participants. • For example, rather than treating one group of patients for headache, and not treating a second group, I might assess the same patients’ headaches before and after treatment.

  13. Repeated measures t tests • Such designs have potential problems (e.g., headaches might simply get better over time). • However, all else being equal, they are often more powerful. • Your book refers to this test as the t test for related groups.

  14. The t test for related groups • That language is employed because sometimes researchers use designs that involving matching participants on various characteristics, and treating them as if they were the same person. • That can work really well in large investigations with highly reliable measurement for the matching variables.

  15. Related groups (cont.) • However, in the context of psychological research, samples are often smaller and matching variables may be unreliable. • It has been shown that matched t tests are underpowered under those circumstances. • We will confine our attention to using the procedure for true repeated-measures designs.

  16. Repeated measures t tests • You already know how to do this test. • The repeated-measures t test is simply a one-sample t test applied to differences (measure at time 2 – measure at time 1). • The null hypothesis is that mD = 0.

  17. Example of a repeated-measures t test • A 1983 study of the relation between stress and beta-endorphin levels in the blood measured patients’ levels 12 hours before surgery, and again 10 minutes prior to surgery. • The null hypothesis is mD = 0, and the research hypothesis is mD ≠ 0. • Let’s use a two-tailed alpha level of .01.

  18. Beta-endorphins • The sum of the differences is 148.3. • The sum of the squared differences is 4436.33. • M = 148.3 / 19 = 7.805263158. • SS = 4436.33 – 148.32 / 19 = 3278.809474.

  19. Beta-endorphins (cont.) • s2 = 3278.809474 / 18 = 182.1560819. • s = √ 182.1560819 = 13.4965211. • sM = 13.4965211 / √19 = 3.096314293. • t = (7.805263158 – 0) / 3.096314293 = 2.521. • df = 19 – 1 = 18, so the critical value is 2.878. • We fail to reject the null hypothesis.

  20. Assumptions • We assume that the differences are independent of one another. • We assume that the differences are normally distributed.

  21. Stem and leaf plot of differences -0 | 44322 0 | 000366678 1 | 03 2 | 0 3 | 9 4 | 5

  22. Effect size? • d = 7.805263158 / 13.4965211 = 0.58. • How about a 95% confidence interval? • The critical value of t for a test at a = .05 would be 2.101, so the 95% CI is given by 7.805263158 ± 2.101 × 3.096314293 = (1.30, 14.31). • Interpret? • Note that we would have rejected the null at a = .05.

  23. Next time • Introducing the analysis of variance.

  24. Exercise • Before and after pulse rate.

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