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MGMT 276: Statistical Inference in Management Winter, 2013

MGMT 276: Statistical Inference in Management Winter, 2013. Welcome. http://www.youtube.com/watch?v=Ahg6qcgoay4&watch_response. Please start portfolios. Homework Assignment #11 due – Thursday (January 9 th ). On class website:

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MGMT 276: Statistical Inference in Management Winter, 2013

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  1. MGMT 276: Statistical Inference in ManagementWinter, 2013 Welcome http://www.youtube.com/watch?v=Ahg6qcgoay4&watch_response

  2. Please start portfolios

  3. Homework Assignment #11 due – Thursday (January 9th) On class website: Please print and complete homework worksheet Hypothesis testing with ANOVA 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. Readings: (Before the next exam – January 13th) Lind Chapter 10: One sample Tests of Hypothesis Chapter 11: Two sample Tests of Hypothesis Chapter 12: Analysis of Variance Chapter 13: Linear Regression and Correlation Chapter 14: Multiple Regression Chapter 15: Chi-Square Plous Chapter 2: Cognitive Dissonance Chapter 3: Memory and Hindsight Bias Chapter 4: Context Dependence Chapter 17: Social Influences Chapter 18: Group Judgments and Decisions

  5. Homework review

  6. Is there a difference in mpg between these two cars There is no difference in mpg between these two cars There is a difference in mpg between these two cars

  7. 0.05 0.05 2-tail 18 2.101

  8. two tail test α= .05 (df) = 18 Critical t(18) = 2.101

  9. S12 .64 S22 1.0 .82 3.704 = .82 (10 – 1) (.80)2 + (10 – 1) (1)2 17 – 18.5 -1.5 t = S2pooled = = .4049691 101+ 102- 2 .82/10 + .82/10 = -3.704

  10. Yes Yes Yes There is a difference there is no difference there is a difference There is no difference

  11. The average mpg is 18.5 for the Ford Explorer and 17.0 for the Expedition. A t-test was conducted and found this difference to be significantly different , t(18) = 3.70; p < 0.05 18.6 18.3 18.0 17.7 17.4 17.1 16.8 0 Miles per gallon Expedition Explorer Type of Car

  12. . Homework Is there an increase in foot size from 1960 to 1980 Is there no difference (or a decrease) in foot size from 1960 to 1980 There is an increase in foot size from 1960 to 1980 0.05 1-tail 22 1.717

  13. one tail test α= .05 (df) = 22 Critical t(22) = 1.717

  14. S12 .3845 S22 .2027 = .2936 .2936 = 2.26 2.26 (12 – 1) (.6201)2 + (12 – 1) (4502)2 8.208 – 7.708 0.5 S2pooled = t = = .2212 121+ 122- 2 .2936/12 + .2936/12

  15. . Homework Yes Yes Yes there is not there is a difference there is there is no difference The average foot size for women in 1960 is 7.7, while the average foot size for women in 1980 is 8.2. A t-test was conducted and found that the increase in foot size is statistically significant, t(22) = 2.26; p < 0.05

  16. . Homework 8.6 8.3 8.0 7.7 7.4 0 Shoe Size 1980 1960 Year of birth

  17. . Homework – same problem using excel Year of Birth Shoe Size 8.2 7.7 1-tail 0.05 2.26 1.7170 22 .017014309 Yes The average foot size for women in 1960 is 7.7, while the average foot size for women in 1980 is 8.2. A t-test was conducted and found that the increase in foot size is statistically significant, t(22) = 2.26; p < 0.05

  18. . Homework

  19. . Homework Type of instruction Exam score 50 40 2-tail 0.05 CAUTION This is significant with alpha of 0.05 BUT NOT WITH alpha of 0.01 2.66 2.02 38 p = 0.0113 yes The average exam score for those with instruction was 50, while the average exam score for those with no instruction was 40. A t-test was conducted and found that instruction significantly improved exam scores, t(38) = 2.66; p < 0.05

  20. . Homework Type of Staff Travel Expenses 142.5 130.29 2-tail 0.05 1.53679 2.2 11 p = 0.153 no The average expenses for sales staff is 142.5, while the average expenses for the audit staff was 130.29. A t-test was conducted and no significant difference was found, t(11) = 1.54; n.s.

  21. Use this as your study guide By the end of lecture today1/8/14 Logic of hypothesis testing Steps for hypothesis testing Hypothesis testing with analysis of variance (ANOVA) Interpreting excel output of hypothesis tests Constructing brief, complete summary statements

  22. Study Type 3: One-way ANOVA Single Independent Variable comparing more than twogroups Single Dependent Variable (numerical/continuous) Used to test the effect of the IV on the DV Ian was interested in the effect of incentives for girl scouts on the number of cookies sold. He randomly assigned girl scouts into one of three groups. The three groups were given one of three incentives and looked to see who sold more cookies. The 3 incentives were 1) Trip to Hawaii, 2) New Bike or 3) Nothing. This is an example of a true experiment How could we make this a quasi-experiment? Independent Variable: Type of incentive Levels of Independent Variable: None, Bike, Trip to Hawaii Dependent Variable: Number of cookies sold Levels of Dependent Variable: 1, 2, 3 up to max sold Between participant design Causal relationship: Incentive had an effect – it increased sales

  23. Study Type 3: One-way ANOVA Single Independent Variable comparing more than two groups Single Dependent Variable (numerical/continuous) Used to test the effect of the IV on the DV Ian was interested in the effect of incentives for girl scouts on the number of cookies sold. He randomly assigned girl scouts into one of three groups. The three groups were given one of three incentives and looked to see who sold more cookies. The 3 incentives were 1) Trip to Hawaii, 2) New Bike or 3) Nothing. This is an example of a true experiment Dependent variable is always quantitative Sales per Girl scout Sales per Girl scout New Bike None Trip Hawaii New Bike None Trip Hawaii In an ANOVA, independent variable is qualitative (& more than two groups)

  24. One-way ANOVA versus Chi Square Be careful you are not designing a Chi Square If this is just frequency you may have a problem This is an Chi Square Total Number of Boxes Sold Sales per Girl scout This is an ANOVA New Bike None Trip Hawaii New Bike None Trip Hawaii These are just frequencies These are just frequencies These are just frequencies These are means These are means These are means

  25. “Between Groups”Variability . Difference between means Difference between means Difference between means Variabilityof curve(s) “Within Groups”Variability Variabilityof curve(s) Variabilityof curve(s)

  26. 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)? Still, difference between means • Critical statistic (e.g. z or t or F or r) value? Step 3: Calculations MSBetween F = MSWithin Still, variabilityof curve(s) 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

  27. One-way ANOVA Number of cookies sold Bike None Hawaii trip Incentives • One-way ANOVAs test only one independent variable • - although there may be many levels • “Factor” = one independent variable • “Level” = levels of the independent variable • treatment • condition • groups • “Main Effect” of independent variable = difference between levels • Note: doesn’t tell you which specific levels (means) differ from each other A multi-factor experiment would be a multi-independent variables experiment

  28. 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 < 0.05 F(2, 15) = 3.0; p < 0.05 Tells us generally about number of participants / observations

  29. One way analysis of varianceVariance is divided Remember, one-way = one IV Total variability Between group variability (only one factor) Within group variability (error variance) 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

  30. Three different types of variance Between groups Within groups Total Between Groups Variability Total Variability Variability between groups F = Within Groups Variability Variability within groups

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

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

  33. . Effect size is considered relativeto variability of distributions Treatment Effect x Variability between groups Treatment Effect x Variabilitywithin groups

  34. x = 10 x = 14 x = 12 What if we want to compare 3 means? One independent variable with 3 means 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 (Hawaii) 14 9 19 13 15 Troop 2 (bicycle) 12 14 10 11 13 Troop 3 (nada) 10 8 12 7 13 Note: 5 girls in each troop

  35. 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 3 (Hawaii) 14 9 19 13 15 Troop 1 (nada) 10 8 12 7 13 Troop 2 (bicycle) 12 14 10 11 13 What is Dependent Variable? How many groups? n = 5 x = 10 n = 5 x = 12 n = 5 x = 14

  36. 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

  37. Hypothesis testing: Step 1: Identify the research problem Is there a significant difference in the number of cookie boxes sold between the girlscout troops that were given the different levels of incentive? Describe the null and alternative hypotheses

  38. Hypothesis testing: = .05 Decision rule Degrees of freedom (between) = number of groups - 1 = 3 - 1 = 2 Degrees of freedom (within) = # of scores - # of groups = (15-3) = 12* Critical F(2,12) = 3.98 *or = (5-1) + (5-1) + (5-1) = 12.

  39. Appendix B.4 (pg.518) F (2,12) α= .05 Critical F(2,12) = 3.89

  40. “SS” = “Sum of Squares”- will be given for exams- you can think of this as the numerator in a standard deviation formula ANOVA table F Source df MS SS Between ? ? ? ? Within ? ? ? Total ? ?

  41. “SS” = “Sum of Squares”- will be given for exams ANOVA table F Source df MS SS 3-1=2 # groups - 1 Between 40 ? 2 ? ? ? 15-3=12 Within ? 88 ? 12 # scores - number of groups ? Total ? 128 ? 14 # scores - 1 15- 1=14

  42. ANOVA table MSbetween MSwithin 40 88 SSbetween 12 2 ANOVA table dfbetween F Source df MS SS ? Between 40 2 ? 2.73 20 Within 88 12 ? 7.33 Total 128 14 SSwithin dfwithin 88 20 =2.73 =7.33 40 7.33 12 =20 2

  43. Make decision whether or not to reject null hypothesis Observed F = 2.73 Critical F(2,12) = 3.89 2.73 is not farther out on the curve than 3.89 so, we do not reject the null hypothesis F(2,12) = 2.73; n.s. 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.

  44. Let’s do same problemUsing 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

  45. Let’s do same problemUsing MS Excel

  46. Let’s do same problemUsing MS Excel

  47. Let’s do oneReplication of study(new data)

  48. Let’s do same problemUsing MS Excel

  49. Let’s do same problemUsing MS Excel

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

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