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2. Introduction to Statistics for the Social SciencesSBS200, COMM200, GEOG200, PA200, POL200, or SOC200Lecture Section 001, Summer Session II, 20139:00 - 11:20am Monday - FridayRoom 312 Social Sciences (Monday – Thursdays)Room 480 Marshall Building (Fridays) Welcome http://www.youtube.com/watch?v=oSQJP40PcGI

3. Please click in My last name starts with a letter somewhere between A. A – D B. E – L C. M – R D. S – Z Please double check All cell phones other electronic devices are turned off and stowed away

4. Homework due – Thursday (July 25th) • On class website: • Please print and complete homework worksheet #11 • Hypothesis Testing with t-tests and ANOVAs

5. Schedule of readings • Before Friday (July 26th) • Please read chapters 7 – 11 in Ha & Ha • Please read Chapters 2, 3, and 4 in Plous • Chapter 2: Cognitive Dissonance • Chapter 3: Memory and Hindsight Bias • Chapter 4: Context Dependence

6. Use this as your study guide By the end of lecture today7/24/13 Logic of hypothesis testing Steps for hypothesis testing Levels of significance (Levels of alpha) Interpreting t-tests and ANOVAs

7. Questions from the homework assignment?

8. 95% 26.08 < µ< 33.92 mean + z σ = 30 ± (1.96)(2) 99% 24.84 < µ< 35.16 mean + z σ = 30 ± (2.58)(2)

9. Melvin Melvin Mark Difference not due sample size because both samples same size Difference not due population variability because same population Yes! Difference is due to sloppiness and random error in Melvin’s sample Melvin

10. √ z- score : because we know the population standard deviation Ho: µ = 5 Bags of potatoes from that plant are not different from other plants Ha: µ ≠ 5 Bags of potatoes from that plant are different from other plants Two tailed test (α = .05) 1.96 1 1 = .25 = 6 – 5 4 16 = 4.0 .25 4.0 -1.96 1.96

11. Because theobserved z (4.0 ) is bigger than critical z (1.96) These three will always match Yes Probability of Type I error is always equal to alpha Yes Yes .05 1.64 No Because observed z (4.0) is still bigger than critical z (1.64) 2.58 No Because observed z (4.0) is still bigger than critical z(2.58) there is not there is a difference there is there is no difference 1.96 2.58

12. t- score : because we don’t know the population standard deviation Two tailed test (α = .05) n – 1 =16 – 1 = 15

13. two tail test α= .05 (df) = 15 Critical t(15) = 2.131

14. -2.13 2.13 √ t- score : because we don’t know the population standard deviation Two tailed test (α = .05) n – 1 =16 – 1 = 15 Critical t(15) = 2.131 89 - 85 2.667 6 16

15. Because theobserved z (2.67) is bigger than critical z (2.13) These three will always match Yes Probability of Type I error is always equal to alpha Yes Yes .05

16. one tail test α= .05 (df) = 15 Critical t(15) = 1.753

17. Yes Yes Yes .05 1.753 No Because observed t (2.67) is still bigger than critical t (1.753)

18. two tail test α= .01 (df) = 15 Critical t(15) = 2.947

19. Yes Yes Yes .05 1.753 No Because observed t (2.67) is still bigger than critical t (1.753) 2.947 Yes Because observed t (2.67) is not bigger than critical t(2.947) No These three will always match No No she did not consultant did improve morale she did consultant did not improve morale 2.131 2.947

20. Finish with statistical summaryz = 4.0; p < 0.05 Or if it *were not* significant: z = 1.2 ; n.s. Start summary with two means (based on DV) for two levels of the IV Describe type of test (z-test versus t-test) with brief overview of results n.s. = “not significant” p<0.05 = “significant” The average weight of bags of potatoes from this particular plant is 6 pounds, while the average weight for population is 5 pounds. A z-test was completed and this difference was found to be statistically significant. We should fix the plant. (z = 4.0; p<0.05) Value of observed statistic

21. Finish with statistical summaryt(15) = 2.67; p < 0.05 Or if it *were not* significant: t(15) = 1.07; n.s. Start summary with two means (based on DV) for two levels of the IV Describe type of test (z-test versus t-test) with brief overview of results n.s. = “not significant” p<0.05 = “significant” The average job-satisfaction score was 89 for the employees who went On the retreat, while the average score for population is 85. A t-test was completed and this difference was found to be statistically significant. We should hire the consultant. (t(15) = 2.67; p<0.05) Value of observed statistic df

22. . . A note on z scores, and t score: • Numerator is always distance between means • (how far away the distributions are) • Denominator is always measure of variability • (how wide or much overlap there is between distributions) Difference between means Difference between means Difference between means Variabilityof curve(s) Variabilityof curve(s) Variabilityof curve(s)

23. Five steps to hypothesis testing Step 1: Identify the research problem (hypothesis) How is a single sample t-test different than two sample t-test? Describe the null and alternative hypotheses How is a single sample t-test most similar to the two sample t-test? Step 2: Decision rule • Alpha level? (α= .05 or .01)? • Critical statistic (e.g. z or t) value? Step 3: Calculations Step 4: Make decision whether or not to reject null hypothesis If observed z (or t) is bigger then critical z (or t) then reject null Single sample standard deviation versus average standard deviation for two samples Single sample has one “n” while two samples will have an “n” for each sample Step 5: Conclusion - tie findings back in to research problem

24. Independent samples t-test Are the two means significantly different from each other, or is the difference just due to chance? 24 – 21 t = variability Donald is a consultant and leads training sessions. As part of his training sessions, he provides the students with breakfast. He has noticed that when he provides a full breakfast people seem to learn better than when he provides just a small meal (donuts and muffins). So, he put his hunch to the test. He had two classes, both with three people enrolled. The one group was given a big meal and the other group was given only a small meal. He then compared their test performance at the end of the day. Please test with an alpha = .05 Small meal 19 23 21 Big Meal 22 25 25 Mean= 21 Mean= 24 Got to figure this part out: We want to average from 2 samples - Call it “pooled” x1 – x2 t = variability

25. Hypothesis testing Step 1: Identify the research problem Did the size of the meal affect the learning / test scores? Step 2: Describe the null and alternative hypotheses Ho: The size of the meal has no effect on test scores H1: The size of the meal does have an effect on test scores Step 3: Decision rule α= .05 One tail or two tail test?

26. Hypothesis testing Step 3: Decision rule α= .05 n1 = 3; n2 = 3 Degrees of freedom total (df total) = (n1 - 1) + (n2 – 1) = (3 - 1) + (3 – 1) = 4 Degrees of freedom total (df total) = (n total - 2) two tailed test Notice: Two different ways to think about it Critical t(4) = 2.776

27. two tail test α= .05 (df) = 4 Critical t(4) = 2.776

28. Mean= 21 Mean= 24 Big Meal Deviation From mean -2 1 1 Small Meal Deviation From mean -2 2 0 SquaredDeviation 4 4 0 Squared deviation 4 1 1 Big Meal 22 25 25 Small meal 19 23 21 Σ = 6 Σ = 8 = 1.732 6 Notice: s2 = 3.0 1 2 1 Notice: Simple Average = 3.5 = 2.0 8 Notice: s2 = 4.0 2 2 2 (n1 – 1) s12 + (n2 – 1) s22 S2pooled = n1 + n2 - 2 (3 – 1) (1.732) 2 + (3 – 1) (2)2 = 3.5 S2pooled = 31+ 32- 2

29. 24 – 21 = 1.5275 Mean= 21 Mean= 24 Big Meal Deviation From mean -2 1 1 Small Meal Deviation From mean -2 2 0 SquaredDeviation 4 4 0 Squared deviation 4 1 1 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3 Σ = 6 Σ = 8 S2p= 3.5 24 - 21 = 1.964 3.5 3.5 3 3 Observed t

30. Hypothesis testing Step 5: Make decision whether or not to reject null hypothesis Observed t = 1.964 Critical t = 2.776 1.964 is not farther out on the curve than 2.776 so, we do not reject the null hypothesis t(4) = 1.964; n.s. Step 6: Conclusion: There appears to be no difference in test scores between the two groups

31. How to report the findingsfor a t-test Mean of small meal was 21 Mean of big meal was 24 One paragraph summary of this study. Describe the IV & DV. Present the two means, which type of test was conducted, and the statistical results. Finish with statistical summaryt(4) = 1.96; ns Observed t = 1.964 Start summary with two means (based on DV) for two levels of the IV df = 4 Or if it *were* significant: t(9) = 3.93; p < 0.05 Describe type of test (t-test versus anova) with brief overview of results We compared test scores for large and small meals. The mean test scores for the big meal was 24, and was 21 for the small meal. A t-test was calculated and there appears to be no significant difference in test scores between the two types of meals t(4) = 1.964; n.s. n.s. = “not significant” p<0.05 = “significant” n.s. = “not significant” p<0.05 = “significant” Type of test with degrees of freedom Value of observed statistic

32. Mean= 21 Figuring out the two means using Excel Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3

33. Mean= 21 Figuring out the two means Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3

34. Mean= 21 Figuring out the two means Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3 Mean= 21 Mean= 24

35. Mean= 21 Complete a t-test Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3 May want to remove means so don’t include in the t-test

36. Mean= 21 Complete a t-test Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3 If checked you’ll want to include the labels in your variable range If checked, you’ll want to include the labels in your variable range If checked you’ll want to include the labels in your variable range

37. Finding Means Finding Means

38. Finding degrees of freedom

39. Finding Observed t

40. Finding Critical t

41. Finding p value (Is it less than .05?)

43. We compared test scores for large and small meals. The mean test scores for the big meal was 24, and was 21 for the small meal. A t-test was calculated and there appears to be no significant difference in test scores between the two types of meals, t(4) = 1.964; n.s. Type of test with degrees of freedom n.s. = “not significant” p<0.05 = “significant” Value of observed statistic Start summary with two means (based on DV) for two levels of the IV Finish with statistical summaryt(4) = 1.96; ns Describe type of test (t-test versus anova) with brief overview of results Or if it *were* significant: t(9) = 3.93; p < 0.05

44. Independent samples t-test What if we ran more subjects? Donald is a consultant and leads training sessions. As part of his training sessions, he provides the students with breakfast. He has noticed that when he provides a full breakfast people seem to learn better than when he provides just a small meal (donuts and muffins). So, he put his hunch to the test. He had two classes, both with three people enrolled. The one group was given a big meal and the other group was given only a small meal. He then compared their test performance at the end of the day. Please test with an alpha = .05 Small meal 19 23 21 19 23 21 19 23 21 Big Meal 22 25 25 22 25 25 22 25 25 Mean= 21 Mean= 24

45. Hypothesis testing Notice: Additional participants don’t affect this part of the problem Step 1: Identify the research problem Did the size of the meal affect the test scores? Step 2: Describe the null and alternative hypotheses Ho: The size of the meal has no effect on test scores H1: The size of the meal does have an effect on test scores One tail or two tail test?

46. Hypothesis testing Step 3: Decision rule α= .05 n1 = 9; n2 = 9 Degrees of freedom total (df total) = (n1 - 1) + (n2 – 1) = (9 - 1) + (9 – 1) = 16 Degrees of freedom total (dftotal) = (n total - 2) = 18 – 2 = 16 two tailed test Notice: Two different ways to think about it Critical t(16) = 2.12

47. two tail test α= .05 (df) = 16 Critical t(16) = 2.12

48. Mean= 21 Mean= 24 Big Meal Deviation From mean 2 -1 -1 2 -1 -1 2 -1 -1 Small Meal Deviation From mean 2 -2 0 2 -2 0 2 -2 0 SquaredDeviation 4 4 0 4 4 0 4 4 0 Squared deviation 4 1 1 4 1 1 4 1 1 Big Meal 22 25 25 22 25 25 22 25 25 Small meal 19 23 21 19 23 21 19 23 21 Σ = 18 Σ = 24 = 1.50 18 Notice: s2 = 2.25 1 8 1 Notice: Simple Average = 2.625 = 1.732 24 Notice: s2 = 3.0 2 8 2

49. Mean= 21 Mean= 24 Big Meal 22 25 25 22 25 25 22 25 25 Small meal 19 23 21 19 23 21 19 23 21 Sp = 2.625 S22 = 3.00 S21 = 2.25 S2 = 1.732 S1 = 1.5 (n1 – 1) s12 + (n2 – 1) s22 S2pooled = n1 + n2 - 2 (9 – 1) (1.50)2 + (9 – 1) (1.732)2 = 2.625 S2pooled = 9 + 9 - 2

50. 24 – 21 = 0.7638 Mean= 21 Mean= 24 Big Meal 22 25 25 22 25 25 22 25 25 Small meal 19 23 21 19 23 21 19 23 21 Sp = 2.625 S2 = 1.732 S1 = 1.5 24 - 21 = 3.928 2.625 2.625 9 9