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Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Spring 2018 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays.

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Presentation on theme: "Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Spring 2018 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays."— Presentation transcript:

1 Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Spring 2018 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays & Fridays. Welcome 3/28/18

2 Lecturer’s desk Projection Booth Screen Screen Harvill 150 renumbered
Row A 15 14 Row A 13 3 2 1 Row A Row B 23 20 Row B 19 5 4 3 2 1 Row B Row C 25 21 Row C 20 6 5 1 Row C Row D 29 23 Row D 22 8 7 1 Row D Row E 31 23 Row E 23 9 8 1 Row E Row F 35 26 Row F 25 11 10 1 Row F Row G 35 26 Row G 25 11 10 1 Row G Row H 37 28 27 13 Row H 12 1 Row H 41 29 28 14 Row J 13 1 Row J 41 29 Row K 28 14 13 1 Row K Row L 33 25 Row L 24 10 9 1 Row L Row M 21 20 19 Row M 18 4 3 2 1 Row M Row N 15 1 Row P 15 1 Harvill 150 renumbered table 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Projection Booth Left handed desk

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4 Before next exam (April 6th)
Schedule of readings Before next exam (April 6th) Please read chapters in OpenStax textbook Please read Chapters 2, 3, and 4 in Plous Chapter 2: Cognitive Dissonance Chapter 3: Memory and Hindsight Bias Chapter 4: Context Dependence

5 Labs continue this week Project 3
Lab sessions Labs continue this week Project 3

6 Project 3: Analysis of Variance (ANOVA)

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8 Writing Assignment – ANOVA – Another Example
1. Write formula for standard deviation of sample 2. Write formula for variance of sample 3. Re-write formula for variance of sample using the nicknames for the numerator and denominator SS df = MS ANOVA table 4. Complete this ANOVA table Source SS df MS F 5. Given a critical F(4,45) = 2.58 Write a summary statement of findings (all three parts) Between 32 ? ? ? Within 450 ? ? Total 482 ?

9 Writing Assignment – ANOVA – Another Example
1. Write formula for standard deviation of sample 2. Write formula for variance of sample 3. Re-write formula for variance of sample using the nicknames for the numerator and denominator SS df = MS ANOVA table 4. Complete this ANOVA table Source SS df MS F 5. Given a critical F(4,45) = 2.58 Write a summary statement of findings (all three parts) Between 32 4 8 0.80 Within 450 45 10 Total 482 49

10 Make decision whether or not to reject null hypothesis
Observed F(4,45) = 0.80 Critical F(4,45) = 2.58 0.80 is not larger than our critical of 2.58 so, we do not reject the null hypothesis Observed F(4,45) = 0.80 Conclusion: There appears to be no main effect of type of major on number of units taken in school The average number of units taken by Geography, Sociology, Nutrition, Communication and Criminal Justice majors are 11.6, 11.1, 9.3, 11.3 and 11 respectively. An ANOVA was conducted and there appears to be no main effect of the number of units taken as a result of the different majors. F(4,45) = 0.80; n.s.

11 Hand in your writing assignment

12 Let’s do same problem Using MS Excel
A girlscout troop leader wondered whether providing an incentive to whomever sold the most girlscout cookies would have an effect on the number cookies sold. She provided a big incentive to one troop (trip to Hawaii), a lesser incentive to a second troop (bicycle), and no incentive to a third group, and then looked to see who sold more cookies. Troop 1 (Nada) 10 8 12 7 13 Troop 2 (bicycle) 12 14 10 11 13 Troop 3 (Hawaii) 14 9 19 13 15 n = 5 x = 10 n = 5 x = 12 n = 5 x = 14

13 Let’s do one Replication of study (new data)

14 Let’s do same problem Using MS Excel

15 Let’s do same problem Using MS Excel

16 SSbetween dfbetween 40 2 =20 “Sum of Squares” MSbetween MSwithin # groups - 1 20 7.33 =2.73 3-1=2 88 12 =7.33 # scores - # of groups SSwithin dfwithin 15-3=12 # scores - 1 15- 1=14

17 No, so it is not significant Do not reject null
F critical (is observed F greater than critical F?) P-value (is it less than .05?) “Sum of Squares”

18 Make decision whether or not to reject null hypothesis
Observed F = 2.73 Critical F(2,12) = 3.89 2.7 is not farther out on the curve than 3.89 so, we do not reject the null hypothesis Also p-value is not smaller than 0.05 so we do not reject the null hypothesis Step 6: Conclusion: There appears to be no effect of type of incentive on number of girl scout cookies sold

19 Make decision whether or not to reject null hypothesis
Observed F = F(2,12) = 2.73; n.s. Critical F(2,12) = 2.7 is not farther out on the curve than 3.89 so, we do not reject the null hypothesis Conclusion: There appears to be no effect of type of incentive on number of girl scout cookies sold The average number of cookies sold for three different incentives were compared. The mean number of cookie boxes sold for the “Hawaii” incentive was 14 , the mean number of cookies boxes sold for the “Bicycle” incentive was 12, and the mean number of cookies sold for the “No” incentive was 10. An ANOVA was conducted and there appears to be no significant difference in the number of cookies sold as a result of the different levels of incentive F(2, 12) = 2.73; n.s.

20 incentive then the means are significantly different from each other
Main effect of incentive: Will offering an incentive result in more girl scout cookies being sold? If we have a “effect” of incentive then the means are significantly different from each other we reject the null we have a significant F p < 0.05 To get an effect we want: Large “F” - big effect and small variability Small “p” - less than 0.05 (whatever our alpha is) We don’t know which means are different from which …. just that they are not all the same 20

21 Homework

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

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

26 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).

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

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

29 SS between df between MS between MS within SS within df within

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

33 2 12 11.46 2 If observed F is bigger than critical F: Reject null & Significant! If observed F is bigger than critical F: Reject null & Significant! 11.46 / 2 = 5.733 5.733 3.88 If p value is less than 0.05: Reject null & Significant! 0.0179

34 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).

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

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

37 MS between MS within SS between df between SS within df within

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39 Thank you! See you next time!!


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