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Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Fall 2017 Room 150 Harvill Building 10:00 - 10:50 Mondays, Wednesdays.

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Presentation on theme: "Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Fall 2017 Room 150 Harvill Building 10:00 - 10:50 Mondays, Wednesdays."— Presentation transcript:

1 Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Fall 2017 Room 150 Harvill Building 10: :50 Mondays, Wednesdays & Fridays. Welcome

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

3 A note on doodling

4 Start Project 3 This Week
Lab sessions Everyone will want to be enrolled in one of the lab sessions Start Project 3 This Week

5 Schedule of readings Before next exam (November 17th)
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

6

7 Prep Project 3 Study Type 2: t-test
We are looking to compare two means Study Type 2: t-test Study Type 3: One-way Analysis of Variance (ANOVA) Comparing more than two means Prep Project 3

8 Comparing ANOVAs with t-tests
Prep Project 3 Comparing ANOVAs with t-tests Similarities still include: Using distributions to make decisions about common and rare events Using distributions to make inferences about whether to reject the null hypothesis or not The same 5 steps for testing an hypothesis Tells us generally about number of participants / observations Tells us generally about number of groups / levels of IV The three primary differences between t-tests and ANOVAS are: 1. ANOVAs can test more than two means 2. We are comparing sample means indirectly by comparing sample variances 3. We now will have two types of degrees of freedom t(16) = 3.0; p < F(2, 15) = 3.0; p < 0.05 Tells us generally about number of participants / observations

9 Prep Project 3 A girl scout troop leader wondered whether providing an
incentive to whomever sold the most girl scout 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. How many levels of the Independent Variable? What is Independent Variable? Troop 1 (nada) 10 8 12 7 13 Troop 2 (bicycle) 12 14 10 11 13 Troop 3 (Hawaii) 14 9 19 13 15 What is Dependent Variable? How many groups? n = 5 x = 10 n = 5 x = 12 n = 5 x = 14 Prep Project 3

10 ANOVA: Using MS Excel Prep Project 3
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 Prep Project 3 n = 5 x = 10 n = 5 x = 12 n = 5 x = 14

11 Let’s do one Replication of study (new data)
Prep Project 3

12 Let’s do same problem Using MS Excel
Prep Project 3

13 Let’s do same problem Using MS Excel
Prep Project 3

14 # scores - number of groups
SSbetween dfbetween 40 2 40 2 =20 3-1=2 # groups - 1 MSbetween MSwithin # scores - number of groups 15-3=12 Prep Project 3 SSwithin dfwithin 88 12 =7.33 20 7.33 =2.73 88 12 # scores - 1 15- 1=14

15 Prep Project 3 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?) Prep Project 3

16 Prep Project 3 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 Prep Project 3

17 Prep Project 3 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. Prep Project 3

18 Study Type 3: One-way Analysis of Variance (ANOVA)
We are looking to compare two means Study Type 2: t-test Study Type 3: One-way Analysis of Variance (ANOVA) Comparing more than two means

19 Review of the homework assignment

20 6 – 5 = 4.0 .25 Two tailed test 1.96 (α = .05) 1 1 = = .25 16 4 √ 4.0
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 1.96 (α = .05) 1 1 = = .25 6 – 5 4 16 = 4.0 .25 4.0 -1.96 1.96

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

22 89 - 85 Two tailed test (α = .05) n – 1 =16 – 1 = 15
-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 2.667 6 16

23 Because the observed z (2.67) is bigger than critical z (2.13)
These three will always match Yes Yes Probability of Type I error is always equal to alpha 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 consultant did improve morale she did not consultant did not improve morale she did 2.131 2.947

24 Value of observed statistic
Finish with statistical summary z = 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

25 Value of observed statistic
Finish with statistical summary t(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

26 Thank you! See you next time!!


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