2. Lecturer’s desk INTEGRATED LEARNING CENTER ILC 120 Screen 11 10 2 1 98 7 6 5 13 12 15 14 17 16 19 18 4 3 Row A Row B Row C Row D Row E Row F Row G Row H Row I Row J Row K Row L Computer Storage Cabinet Cabinet Table 20 11 10 2 1 9 8 7 6 5 21 20 23 22 25 24 27 26 4 3 28 13 12 14 16 15 17 18 19 11 10 2 1 9 8 7 6 5 21 20 23 22 25 24 26 4 3 13 12 14 16 15 17 18 19 11 10 2 1 9 8 7 6 5 21 20 23 22 25 24 26 4 3 13 12 14 16 15 17 18 19 11 10 2 1 9 8 7 6 5 21 20 23 22 25 24 27 26 4 3 28 13 12 14 16 15 17 18 19 29 11 10 2 1 9 8 7 6 5 21 20 23 22 25 24 27 26 4 3 28 13 12 14 16 15 17 18 19 11 10 2 1 9 8 7 6 5 21 20 23 22 25 24 27 26 4 3 13 12 14 16 15 17 18 19 11 10 2 1 9 8 7 6 5 21 20 23 22 25 24 26 4 3 13 12 14 16 15 17 18 19 11 10 2 1 9 8 7 6 5 21 20 23 22 25 24 4 3 13 12 14 16 15 17 18 19 11 10 2 1 9 8 7 6 5 21 20 23 22 24 4 3 13 12 14 16 15 17 18 19 11 10 2 1 9 8 7 6 5 21 20 23 22 4 3 13 12 14 16 15 17 18 19 11 10 9 8 7 6 5 4 3 13 12 14 16 15 17 18 19 broken desk

3. Introduction to Statistics for the Social Sciences SBS200, COMM200, GEOG200, PA200, POL200, or SOC200 Lecture Section 001, Fall, 2014 Room 120 Integrated Learning Center (ILC) 10:00 - 10:50 Mondays, Wednesdays & Fridays. http://www.youtube.com/watch?v=oSQJP40PcGI

4. Reminder A note on doodling

5. Schedule of readings Before next exam (November 21 st ) 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. No homework due – Friday (November 14 th )

7. Labs continue this week with Project 2

9. By the end of lecture today 11/12/14 Use this as your study guide Hypothesis testing with Analysis of Variance (ANOVA) Constructing brief, complete summary statements

10. One way analysis of variance Variance is divided Total variability Within group variability (error variance) Between group variability (only one factor) Remember, 1 factor = 1 independent variable (this will be our numerator – like difference between means) Remember, error variance = random error (this will be our denominator – like within group variability Remember, one-way = one IV

11. 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 t (or F) is bigger then critical t (or F) then reject null Step 5: Conclusion - tie findings back in to research problem Critical statistic (e.g. z or t or F or r) value? MS Within MS Between F = Still, difference between means Still, variability of curve(s)

12. . Difference between means Variability of curve(s) “Between Groups” Variability “Within Groups” Variability

13. Sum of squares (SS): The sum of squared deviations of some set of scores about their mean Mean squares (MS): The sum of squares divided by its degrees of freedom Mean square within groups: sum of squares within groups divided by its degrees of freedom Mean square between groups: sum of squares between groups divided by its degrees of freedom Mean square total: sum of squares total divided by its degrees of freedom MS Within MS Between F =

14. ANOVA Variability within groups Variability between groups F = Variability Between Groups Variability Within Groups “Between” variability bigger than “within” variability so should get a big (significant) F Variability Between Groups Variability Within Groups “Between” variability getting smaller “within” variability staying same so, should get a smaller F Variability Between Groups “Between” variability getting very small “within” variability staying same so, should get a very small F Variability Within Groups

15. ANOVA Variability within groups Variability between groups F = “Between” variability bigger than “within” variability so should get a big (significant) F “Between” variability getting smaller “within” variability staying same so, should get a smaller F “Between” variability getting very small “within” variability staying same so, should get a very small F (equal to 1) Variability Within Groups Variability Between Groups Variability Within Groups Variability Between Groups

16. . Effect size is considered relative to variability of distributions Treatment Effect Treatment Effect x x Variability within groups Variability between groups

17. Homework

20. Type of major in school 4 (accounting, finance, hr, marketing) Grade Point Average 0.05 2.83 3.02 3.24 3.37

21. Homework 0.3937 0.1119 0.3937 / 0.1119 = 3.517 3.517 3.009 3 24 0.03 If observed F is bigger than critical F: Reject null & Significant! If p value is less than 0.05: Reject null & Significant! # groups - 1 # scores - number of groups # scores - 1 4-1=3 28 - 4=24 28 - 1=27

22. Homework Yes F (3, 24) = 3.517;p < 0.05 The GPA for four majors was compared. The average GPA was 2.83 for accounting, 3.02 for finance, 3.24 for HR, and 3.37 for marketing. An ANOVA was conducted and there is a significant difference in GPA for these four groups (F (3,24) = 3.52; p < 0.05).

23. Number of observations in each group Average for each group (We REALLY care about this one)

24. “SS” = “Sum of Squares” - will be given for exams Number of groups minus one (k – 1)  4-1=3 Number of people minus number of groups (n – k)  28-4=24

25. MS between MS within SS between df between SS within df within

28. Type of executive 3 (banking, retail, insurance) Hours spent at computer 0.05 10.8 8 8.4

29. 11.46 2 11.46 / 2 = 5.733 5.733 3.88 2 12 0.0179 If observed F is bigger than critical F: Reject null & Significant! If p value is less than 0.05: Reject null & Significant!

30. Yes F (2, 12)= 5.73; p < 0.05 The number of hours spent at the computer was compared for three types of executives. The average hours spent was 10.8 for banking executives, 8 for retail executives, and 8.4 for insurance executives. An ANOVA was conducted and we found a significant difference in the average number of hours spent at the computer for these three groups, (F (2,12) = 5.73; p < 0.05).

31. Number of observations in each group Just add up all scores Average for each group

32. “SS” = “Sum of Squares” - will be given for exams Number of groups minus one (k – 1)  3-1=2 Number of people minus number of groups (n – k)  15-3=12

33. MS between MS within SS between df between SS within df within

35. Let’s try one In a one-way ANOVA we have three types of variability. Which picture best depicts the random error variability (also known as the within variability)? a. Figure 1 b. Figure 2 c. Figure 3 d. All of the above 1. 2. 3.

36. Let’s try one Which figure would depict the largest F ratio a. Figure 1 b. Figure 2 c. Figure 3 d. All of the above Variability within groups Variability between groups F = 1. 2. 3. “F ratio” is referring to "observed F”

37. Let’s try one Winnie found an observed z of.74, what should she conclude? (Hint: notice that.74 is less than 1) a. Reject the null hypothesis b. Do not reject the null hypothesis c. Not enough info is given small observed z score x x If your observed z is within one standard deviation of the mean, you will never reject the null

38. Let’s try one Winnie found an observed t of.04, what should she conclude? (Hint: notice that.04 is less than 1) a. Reject the null hypothesis b. Do not reject the null hypothesis c. Not enough info is given small observed t score x

39. Let’s try one Winnie found an observed F ratio of.9, what should she conclude? a. Reject the null hypothesis b. Do not reject the null hypothesis c. Not enough info is given 1. 2. 3.

40. Let’s try one An ANOVA was conducted comparing different types of solar cells and there appears to be a significant difference in output of each (watts) F(4, 25) = 3.12; p < 0.05. In this study there were __ types of solar cells and __ total observations in the whole study? a. 4; 25 b. 5; 30 c. 4; 30 d. 5; 25 # groups - 1 # scores - # of groups # scores - 1 F(4, 25) = 3.12; p < 0.05 How many observations within each group?

41. Let’s try one An ANOVA was conducted comparing different types of solar cells and there appears to be significant difference in output of each (watts) F(4, 25) = 3.12; p < 0.05. In this study ___ a. we rejected the null hypothesis b. we did not reject the null hypothesis p <.05 F(4, 25) = 3.12; p < 0.05 Observed F bigger than Critical F

42. Let’s try one An ANOVA was conducted comparing different types of solar cells. The analysis was completed using an alpha of 0.05. But Julia now wants to know if she can reject the null with an alpha of at 0.01. In this study ___ a. we rejected the null hypothesis b. we did not reject the null hypothesis p <.05 F(4, 25) = 3.12; p < 0.05 Comparison of the Observed F and Critical F Is no longer are helpful because the critical F is no longer correct. We must use the p value p >.01

43. Let’s try one An ANOVA was conducted comparing home prices in four neighborhoods (Southpark, Northpark, Westpark, Eastpark). For each neighborhood we measured the price of four homes. Please complete this ANOVA table. Degrees of freedom between is _____; degrees of freedom within is ____ a. 16; 4 b. 4; 16 c. 12; 3 d. 3; 12.

44. Let’s try one An ANOVA was conducted comparing home prices in four neighborhoods (Southpark, Northpark, Westpark, Eastpark). For each neighborhood we measured the price of four homes. Please complete this ANOVA table. Mean Square between is _____; Mean Square within is ____ a. 300, 300 b. 100, 100 c. 100, 25 d. 25, 100.

45. Let’s try one An ANOVA was conducted comparing home prices in four neighborhoods (Southpark, Northpark, Westpark, Eastpark). For each neighborhood we measured the price of four homes. Please complete this ANOVA table. The F ratio is: a..25 b. 1 c. 4 d. 25.

46. An ANOVA was conducted comparing home prices in four neighborhoods (Southpark, Northpark, Westpark, Eastpark). For each neighborhood we measured the price of four homes. Please complete this ANOVA table. We should: a. reject the null hypothesis b. not reject the null hypothesis Let’s try one p <.05 Observed F bigger than Critical F

47. An ANOVA was conducted comparing home prices in four neighborhoods (Southpark, Northpark, Westpark, Eastpark). For each neighborhood we measured the price of four homes. The most expensive neighborhood was the ____ neighborhood a. Southpark b. Northpark c. Westpark d. Eastpark Let’s try one

48. An ANOVA was conducted comparing home prices in four neighborhoods (Southpark, Northpark, Westpark, Eastpark). For each neighborhood we measured the price of four homes. Please complete this ANOVA table. The best summary statement is: a. F(3, 12) = 4.0; n.s. b. F(3, 12) = 4.0; p < 0.05 c. F(3, 12) = 3.49; n.s. d. F(3, 12) = 3.49; p < 0.05

49. A t-test was conducted to see whether “Bankers” or “Retailers” spend more time in front of their computer. Which best summarizes the results from this excel output: a. Bankers spent significantly more time in front of their computer screens than Retailers, t(3.5) = 8; p < 0.05 b. Bankers spent significantly more time in front of their computer screens than Retailers, t(8) = 3.5; p < 0.05 c. Retailers spent significantly more time in front of their computer screens than Bankers, t(3.5) = 8; p < 0.05 d. Retailers spent significantly more time in front of their computer screens than Bankers, t(8) = 3.5; p < 0.05 e. There was no difference between the groups

50. A t-test was conducted to see whether “Bankers” or “Retailers” spend more time in front of their computer. Which critical t would be the best to use a. 3.5 b. 1.859 c. 2.306 d..004 e..008 Let’s try one

51. An ANOVA was conducted and there appears to be a significant difference in the number of cookies sold as a result of the different levels of incentive F(2, 27) = ___; p < 0.05. Please fill in the blank a. 3.3541 b..00635 c. 6.1363 d. 27.00

52. Let’s try one An ANOVA was conducted and we found the following results: F(3,12) = 3.73 ____. Which is the best summary a. The critical F is 3.89; we should reject the null b. The critical F is 3.89; we should not reject the null c. The critical F is 3.49; we should reject the null d. The critical F is 3.49; we should not reject the null