2. Introduction to Statistics for the Social Sciences SBS200, COMM200, GEOG200, PA200, POL200, or SOC200 Lecture Section 001, Spring 2015 Room 150 Harvill Building 8:00 - 8:50 Mondays, Wednesdays & Fridays.

4. Schedule of readings Before next exam (April 10 th ) 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

5. By the end of lecture today 3/27/15 Use this as your study guide Logic of hypothesis testing Steps for hypothesis testing Levels of significance (Levels of alpha) what does p < 0.05 mean? what does p < 0.01 mean? One-sample z-tests and t-tests

6. Homework due Assignment 15 One-sample z and t hypothesis tests Due: Friday, March 30 th

7. Labs continue this week Project 2

8. Standard deviation and Variance For Sample and Population These would be helpful to know by heart – please memorize these formula Review

9. Five steps to hypothesis testing Step 1: Identify the research problem (hypothesis) Describe the null and alternative hypotheses Step 2: Decision rule Alpha level? ( α =.05 or.01)? 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 Step 5: Conclusion - tie findings back in to research problem One or two tailed test? Balance between Type I versus Type II error Critical statistic (e.g. z or t or F or r) value?

10. Five steps to hypothesis testing Step 1: Identify the research problem (hypothesis) Describe the null and alternative hypotheses Step 2: Decision rule Alpha level? ( α =.05 or.01)? 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 Step 5: Conclusion - tie findings back in to research problem Critical statistic (e.g. z or t) value? How is a t score same as a z score? Population versus sample standard deviation How is a t score different than a z score?

11. .. A note on z scores, and t score: Difference between means Variability of curve(s) Difference between means Numerator is always distance between means (how far away the distributions are or “effect size”) Denominator is always measure of variability (how wide or much overlap there is between distributions) Variability of curve(s) (within group variability)

12. . A note on variability versus effect size Difference between means Variability of curve(s) Variability of curve(s) (within group variability) Difference between means

13. .. Variability of curve(s) Variability of curve(s) (within group variability) Difference between means A note on variability versus effect size

14. . Effect size is considered relative to variability of distributions 1. Larger variance harder to find significant difference Treatment Effect Treatment Effect 2. Smaller variance easier to find significant difference x x

15. . Effect size is considered relative to variability of distributions Treatment Effect Treatment Effect x x Variability of curve(s) (within group variability) Difference between means

16. Preview of what to expect with the homework assignment

17. Hypothesis testing: one sample t-test Is the mean of my observed sample consistent with the known population mean or did it come from some other distribution? We are given the following problem: 800 students took a chemistry exam. Accidentally, 25 students got an additional ten minutes. Did this extra time make a significant difference in the scores? The average number correct by the large class was 74. The scores for the sample of 25 was 76, 72, 78, 80, 73 70, 81, 75, 79, 76 77, 79, 81, 74, 62 95, 81, 69, 84, 76 75, 77, 74, 72, 75 Please note: In this example we are comparing our sample mean with the population mean (One-sample t-test) L e t ’ s j u m p r i g h t i n a n d d o a t - t e s t

18. Hypothesis testing Step 1: Identify the research problem / hypothesis Did the extra time given to this sample of students affect their chemistry test scores? Null: The extra time did not affect their chemistry test scores Alternative: The extra time did affect their chemistry test scores Independent Variable? Dependent Variable? IV: Nominal Ordinal Interval or Ratio? DV: Nominal Ordinal Interval or Ratio? Extra time vs. no extra time Test Scores IV: Nominal DV: Ratio One tail or two tail test? Two-tailed test

19. Hypothesis testing Step 2: Decision rule =.05 Degrees of freedom (df) = ( n - 1) = (25 - 1) = 24 two tail test n = 25 What is formula for degres of freedom? n – 1

20. α =.05 (df) = 24 Critical t (24) = 2.064 two tail test

21. . Hypothesis testing Step 3: Calculations µ = 74 = 76.44 76 72 78 80 73 70 81 75 79 76 77 79 81 74 62 95 81 69 84 76 75 77 74 72 75 76 – 76.44 72 – 76.44 78 – 76.44 80 – 76.44 73 – 76.44 70 – 76.44 81 – 76.44 75 – 76.44 79 – 76.44 76 – 76.44 77 – 76.44 79 – 76.44 81 – 76.44 74 – 76.44 62 – 76.44 95 – 76.44 81 – 76.44 69 – 76.44 84 – 76.44 76 – 76.44 75 – 76.44 77 – 76.44 74 – 76.44 72 – 76.44 75 – 76.44 = -0.44 = -4.44 = +1.56 = + 3.56 = -3.44 = -6.44 = +4.56 = -1.44 = +2.56 = -0.44 = +0.56 = +2.56 = +4.56 = -2.44 = -14.44 = +18.56 = +4.56 = -7.44 = +7.56 = -0.44 = -1.44 = +0.56 = -2.44 = -4.44 = -1.44 0.1936 19.7136 2.4336 12.6736 11.8336 41.4736 20.7936 2.0736 6.5536 0.1936 0.3136 6.5536 20.7936 5.9536 208.5136 344.4736 20.7936 55.3536 57.1536 0.1936 2.0736 0.3136 5.9536 19.7136 2.0736 x (x - x) (x - x) 2 868.16 24 = 6.01 N = 25 Σx = 1911 Σ(x- x) = 0 Σ(x- x) 2 = 868.16 ΣxΣx = N 1911 25 =

22. . s = 6.01 = 76.44 - 74 1.20 = 2.03 76.44 - 74 6.01 25 critical t Observed t (24) = 2.03 Step 3: Calculations µ = 74 = 76.44 N = 25 ΣxΣx = N 1911 25 = Hypothesis testing

23. . Step 4: Make decision whether or not to reject null hypothesis Step 6: Conclusion: The extra time did not have a significant effect on the scores 2.03 is not farther out on the curve than 2.064, so, we do not reject the null hypothesis Observed t (24) = 2.03 Critical t (24) = 2.064

24. . Hypothesis testing: Did the extra time given to these 25 students affect their average test score? notice we are comparing a sample mean with a population mean: single sample t-test The mean score for those students who where given extra time was 76.44 percent correct, while the mean score for the rest of the class was only 74 percent correct. A t-test was completed and there appears to be no significant difference in the test scores for these two groups t(24) = 2.03; n.s. Start summary with two means (based on DV) for two levels of the IV Describe type of test (t-test versus z-test) with brief overview of results Finish with statistical summary t(24) = 2.03; ns Or if it had been different results that *were* significant: t(24) = -5.71; p < 0.05 Type of test with degrees of freedom n.s. = “not significant” p<0.05 = “significant” n.s. = “not significant” p<0.05 = “significant” Value of observed statistic

25. . What if we had chosen a one-tail test? Step 2: Decision rule =.05 Degrees of freedom (df) = (n - 1) Critical t (24) = = (25 - 1) = 24 1.711 Step 1: Identify the research problem / hypothesis Did the extra time given to this sample of students improve their chemistry test scores? Null: The extra time did not improve their chemistry test scores Alternative: The extra time did improve their chemistry test scores

26. . α =.05 (df) = 24 Critical t (24) = 1.711 one tail test

27. . Hypothesis testing Step 3: Calculations µ = 74 = 76.44 76 72 78 80 73 70 81 75 79 76 77 79 81 74 62 95 81 69 84 76 75 77 74 72 75 76 – 76.44 72 – 76.44 78 – 76.44 80 – 76.44 73 – 76.44 70 – 76.44 81 – 76.44 75 – 76.44 79 – 76.44 76 – 76.44 77 – 76.44 79 – 76.44 81 – 76.44 74 – 76.44 62 – 76.44 95 – 76.44 81 – 76.44 69 – 76.44 84 – 76.44 76 – 76.44 75 – 76.44 77 – 76.44 74 – 76.44 72 – 76.44 75 – 76.44 = -0.44 = -4.44 = +1.56 = + 3.56 = -3.44 = -6.44 = +4.56 = -1.44 = +2.56 = -0.44 = +0.56 = +2.56 = +4.56 = -2.44 = -14.44 = +18.56 = +4.56 = -7.44 = +7.56 = -0.44 = -1.44 = +0.56 = -2.44 = -4.44 = -1.44 0.1936 19.7136 2.4336 12.6736 11.8336 41.4736 20.7936 2.0736 6.5536 0.1936 0.3136 6.5536 20.7936 5.9536 208.5136 344.4736 20.7936 55.3536 57.1536 0.1936 2.0736 0.3136 5.9536 19.7136 2.0736 x (x - x) (x - x) 2 868.16 24 = 6.01 N = 25 Σx = 1911 Σ(x- x) = 0 Σ(x- x) 2 = 868.16 ΣxΣx = N 1911 25 = (exactly same as two-tail test)

28. . s = 6.01 = 76.44 - 74 1.20 = 2.03 76.44 - 74 6.01 25 critical t Step 3: Calculations µ = 74 = 76.44 N = 25 ΣxΣx = N 1911 25 = Hypothesis testing One-tailed test has no effect on calculations stage (exactly same as two-tail test)

29. . Hypothesis testing Step 4: Make decision whether or not to reject null hypothesis Step 6: Conclusion: The extra time did have a significant effect on the scores 2.03 is not farther out on the curve than 1.711, so, we do not reject the null hypothesis Observed t (24) = 2.03 Critical t (24) = 1.711

30. . Hypothesis testing: Did the extra time given to these 25 students affect their average test score? notice we are comparing a sample mean with a population mean: single sample t-test The mean score for those students who where given extra time was 76.44 percent correct, while the mean score for the rest of the class was only 74 percent correct. A one-tailed t-test was completed and there appears to be significant difference in the test scores for these two groups t(24) = 2.03; p < 0.05. Start summary with two means (based on DV) for two levels of the IV Describe type of test (t-test versus z-test) with brief overview of results Finish with statistical summary t(24) = 2.03; ns Or if it had been different results that *were* significant: t(24) = -5.71; p < 0.05 Type of test with degrees of freedom Value of observed statistic n.s. = “not significant” p<0.05 = “significant”

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

32. Melvin Mark Melvin 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

33. Ho: µ = 5 Ha: µ ≠ 5 Bags of potatoes from that plant are not different from other plants Bags of potatoes from that plant are different from other plants Two tailed test (α =.05) 1.96 6 – 5.25 = 4.0 1 16 √ =.25 4.0 1.96 -1.96 1 4 = z- score : because we know the population standard deviation

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

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

36. Yes These three will always match Probability of Type I error is always equal to alpha.05 Because the observed z (2.67) is bigger than critical z (2.13) 1.753 No Because observed t (2.67) is still bigger than critical t (1.753) 2.947 consultant did improve morale Yes Because observed t (2.67) is not bigger than critical t(2.947) consultant did not improve morale she did not she did 2.131 2.947 No These three will always match

37. 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) 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 Finish with statistical summary z = 4.0; p < 0.05 Or if it *were not* significant: z = 1.2 ; n.s. Value of observed statistic n.s. = “not significant” p<0.05 = “significant”