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Psychology 10

Psychology 10. Analysis of Psychological Data April 9, 2014. The plan for today. Another example of a repeated-measures t test . An illustration of confidence interval coverage . An example of comparisons of several independent means (including confidence intervals and effect sizes).

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Psychology 10

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  1. Psychology 10 Analysis of Psychological Data April 9, 2014

  2. The plan for today • Another example of a repeated-measures t test. • An illustration of confidence interval coverage. • An example of comparisons of several independent means (including confidence intervals and effect sizes).

  3. (Fake) example: Does advertising work? • The manager of a convenience store is interested in whether advertising affects sales. • Selects 13 items from store, records number sold in a one-week period before and after running an ad on the local TV station.

  4. Are we ready to gather data? • Two-tailed hypothesis • Alpha = .05 • H0: mD = 0.

  5. The Data

  6. The Data

  7. The Data

  8. Calculating the test statistic • First, the mean difference: • The differences sum to 32. • 32/13 = 2.461538. • Next, the variance of the differences: • The squared deviations sum to 200. • SS = 200 – 322/13 = 121.2308. • s2= 121.2308/12 = 10.10256

  9. Calculating the test statistic (cont.) • The standard error of the mean difference: • The repeated-measures t statistic: • t = 2.461538 / 0.8815441 = 2.792.

  10. Evaluating the test statistic • N = 13, so df = 12. • From the t table, the critical value is 2.179. • 2.792 > 2.179, so we reject the null hypothesis. • We conclude that advertising does increase the likelihood that people will buy things.

  11. Assumptions • Differences are independent. • Differences are normally distributed.

  12. Independent? • Independent: • The various objects probably have different prices and are different types of things. • Maybe advertising works better for, say, higher priced objects. • Maybe advertising works better for food items than for durable goods. • If so, that could create clusters of dependent differences.

  13. Normally distributed? Differences: -2 | 00 -0 | 0 0 | 00 2 | 00 4 | 0000 6 | 00 • It is difficult to evaluate normality in small data sets, but no major issues.

  14. What is the effect size? • d = 2.461538 / √10.10256 = 0.77. • Alternatively, we could argue that “number of objects” is a metric that is intrinsically meaningful, and describe the effect as 2.46 objects.

  15. A confidence interval • A 95% confidence interval for the mean difference is given by 2.461538 ± 2.179 × 0.8815441 = (0.54, 4.38). • Interpret?

  16. A demonstration about confidence intervals • Digression using statistical software.

  17. A new situation: comparisons among five means • The Eysenck memory experiment: • Memorizing word lists • Level of processing • Five different strategy groups: • Counting • Rhyming • Adjective • Imagery • Intentional

  18. Interest in differences among means • Means are 7.0, 6.9, 11.0, 13.4, and 12.0. • Standard deviations are 1.83, 2.13, 2.49, 4.50, and 3.74. • We could use two-sample independent groups t tests to compare these means.

  19. Counting vs Rhyming • Standard deviations are 1.83 and 2.13, so the pooled variance estimate is (9×1.832+9×2.132)/18 = 3.9429. • The standard error of the difference between means is √(3.9429/10+3.9429/10) = 0.8880203. • t = (7.0-6.9)/0.8880203 = 0.11. • df = 10+10-2, tcrit = 2.101, NS.

  20. Counting vs Adjective • Standard deviations are 1.83 and 2.49, so the pooled variance estimate is (9×1.832+9×2.492)/18 = 4.7745. • The standard error of the difference between means is √(4.7745/10+4.7745/10) = 0.9771898. • t = (7.0-11)/0.9771898 = -4.09. • df = 10+10-2, tcrit = 2.101; reject.

  21. Counting vs Imagery • Standard deviations are 1.83 and 4.5, so the pooled variance estimate is (9×1.832+9×4.52)/18 = 11.79945. • The standard error of the difference between means is √(11.79945 /10+ 11.79945 /10) = 1.536193. • t = (7.0-13.4)/1.536193 = -4.17. • df = 10+10-2, tcrit = 2.101; reject.

  22. Counting vs Intentional • Standard deviations are 1.83 and 3.74, so the pooled variance estimate is (9×1.832+9×3.742)/18 = 8.66825. • The standard error of the difference between means is √(8.66825 /10+ 8.66825 /10) = 1.316681. • t = (7.0-12)/1.316681 = -3.80. • df = 10+10-2, tcrit = 2.101; reject.

  23. This is getting boring.

  24. That’s a lot of tests. • We’ve just done 10 tests, all at an alpha level of .05. • The chance that we’ve committed a Type I error somewhere among those tests is certainly greater than .05. • This is known as the “multiple testing problem.”

  25. The multiple testing problem • One solution to the multiple testing problem: the Bonferroni adjustment. • Divide our alpha equally across all of the tests. • With 10 tests, to have an overall alpha of .05, we would need to use .05/10 = .005 for each test. • The critical value of t for alpha of .005, df = 9, is 3.69.

  26. Bonferroni adjusted results.

  27. Bonferroni adjusted results. • Nothing changed. But the rhyming-adjective and rhyming-intentional comparisons were pretty close to NS. • Imagine that we’d had 10 groups instead of 5. That would involve 45 comparisons, so our corrected alpha would be .05/45 = .00111. • The critical t would be 4.71.

  28. Next time • This problem is what motivates the analysis of variance (ANOVA). • We will introduce ANOVA next time, working with this same example.

  29. Exercise • A confidence interval for hunger.

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