1. Psych 230 Psychological Measurement and Statistics Pedro Wolf November 18, 2009

2. This Time…. Analysis of Variance (ANOVA) Concepts of variability Why bother with ANOVA? Conducting a test

3. Statistical Testing 1.Decide which test to use 2.State the hypotheses (H 0 and H A ) 3.Calculate the obtained value 4.Calculate the critical value (size of  ) 5.Make our conclusion

4. Statistical Testing 1.Decide which test to use 2.State the hypotheses (H 0 and H A ) 3.Calculate the obtained value 4.Calculate the critical value (size of  ) 5.Make our conclusion 6.Conduct post-hoc tests

5. Analysis of Variance (ANOVA)

6. Analysis of Variance In this statistical test, we are interested in seeing if there are significant differences between more than two groups In an experiment involving only two conditions of the independent variable, you can use either a t- test or the ANOVA

7. Analysis of Variance We will look at the variance within each group and compare that to the variance found between the groups Remember: Variance is the degree to which scores are dispersed in our data

8. Analysis of Variance Remember: H 0 is that all the group means in our experiment are the same – any difference between them is due to random chance H 1 is that there is a difference between our group means – a difference that is so unlikely to have happened by chance that we conclude it is due to the independent variable

9. Analysis of Variance There is always variability in our data This variability can be due to two factors: 1.The independent variable Systematic factors 2.Error variance Random factors

10. Analysis of Variance So, to draw a conclusion about whether the independent variable makes a difference to the dependent variable, we need to know what kind of variance there is in our data – variance due to the independent variable (systematic) – variance due to random factors (error)

11. Analysis of Variance To assess this, we need to look at the variance both within each of our experimental conditions and also at the variance between each of our experimental conditions Assume H 0 is true - there is no effect of our independent variable What type of variance might we expect to see?

12. Analysis of Variance Example: we want to see if people differ in their shoe size by where they sit in the class – three groups of students: front row, middle row and back row – expect to find significant differences in shoe size? Front row: vary in their shoe size: 6,8,5,4,7 Middle row: vary in their shoe size : 9,9,4,6,7 Back row: vary in their shoe size : 4,6,12,10,8

13. Analysis of Variance BackMiddleFront 694 896 5412 4610 778 678Mean=7

14. Analysis of Variance BackMiddleFront 694 896 5412 4610 778 678Mean=7 Total variance

15. Analysis of Variance BackMiddleFront 694 896 5412 4610 778 678Mean=7 Variance within the groups Total variance

16. Analysis of Variance Variance between the groups BackMiddleFront 694 896 5412 4610 778 678Mean=7 Variance within the groups Total variance

17. Analysis of Variance BackMiddleFront 694 896 5412 4610 778 678Mean=7 Total Variance = Variance between + Variance within Variance within the groups Total variance Variance between the groups

18. Analysis of Variance So, when H 0 is actually true, we should expect the same amount of variance both within each group and between the groups If we divide Variance between by Variance within, we should get? If H 0 is true, this ratio should be close to 1

19. Analysis of Variance How about if H 1 is actually true? In this case, we know the independent variable is having some effect So, we should expect more variance between each group than there is within each group

20. Analysis of Variance Example: we want to see if people differ in their attendance by where they sit in the class – front row, middle row and back row – expect to find significant differences in attendance? Front row: vary in their attendance: 6,5,7,6,6 Middle row: vary in their attendance: 7,8,6,7,7 Back row: vary in their attendance: 8,8,7,9,8

21. Analysis of Variance Variance between the groups BackMiddleFront 678 588 767 679 678 678Mean=7 Total Variance = Variance between + Variance within Variance within the groups Total variance

22. Analysis of Variance When H A is true, we have more variance between each group than there is within each group If we divide Variance between by Variance within, we should get? If H A is true, this ratio should be more than 1 – the F-ratio

23. ANOVA A one-way ANOVA is performed when only one independent variable is tested in the experiment Example: we are interested in the differences between freshmen, sophomores and juniors on tests of socialization – dependent variable: socialization scores – independent variable: class standing

24. ANOVA A two-way ANOVA is performed when two independent variables are tested in the experiment Example: we are interested in the differences between male and female freshmen, sophomores and juniors on tests of socialization – dependent variable: socialization scores – independent variable 1: class standing – independent variable 2: gender

25. ANOVA When an independent variable is studied using independent samples in all conditions, it is called a between-subjects factor A between-subjects factor involves using the formulas for a between-subjects ANOVA

26. ANOVA When a factor is studied using related (dependent) samples in all levels, it is called a within-subjects factor This involves a set of formulas called a within- subjects ANOVA

27. ANOVA

29. ANOVA - assumptions 1.All conditions contain independent samples 2.The dependent scores are normally distributed, interval or ratio score 3.The variances of the populations are homogeneous

30. ANOVA - why bother? We want to see if there are differences between our three groups: – Freshmen – Sophomores – Juniors Why not just do a bunch of t-tests? – Freshmen vs. Sophomores – Freshmen vs. Juniors – Sophomores vs. Juniors

31. ANOVA - experiment-wise error The overall probability of making a Type I error somewhere in an experiment is call the experiment-wise error rate When we use a t-test to compare only two means in an experiment, the experiment-wise error rate equals 

32. ANOVA - experiment-wise error When there are more than two means in an experiment, the multiple t-tests result in an experiment-wise error rate that is much larger than the  we have selected – Freshmen vs. Sophomores:  =0.05 – Freshmen vs. Juniors:  =0.05 – Sophomores vs. Juniors:  =0.05 – experiment-wise error rate = 0.05+0.05+0.05=approx 0.15 Using the ANOVA allows us to compare the means from all levels of the factor and keep the experiment- wise error rate equal to 

33. Conducting an ANOVA

34. 1. Decide which test to use Are we comparing a sample to a population? – Yes: Z-test if we know the population standard deviation – Yes: One-sample T-test if we do not know the population std dev – No: Keep looking Are we looking for the difference between samples? – Yes: How many samples are we comparing? Two: Use the Two-sample T-test – Are the samples independent or related? » Independent: Use Independent Samples T-test » Related: Use Related Samples T-test More than Two: Use Anova test

35. 2. State the Hypotheses H 0 :  1 =  2 = ……. =  k – there is no difference in the means H A : not all  s are equal – there is a difference between some of the means Only conduct two-tailed tests using ANOVA

36. 3. Calculate the obtained value (F obt ) The statistic for the ANOVA is F When F obt is significant, this indicates only that somewhere among the means at least two of them differ significantly It does not indicate which groups differ significantly When the F-test is significant, we perform post hoc comparisons (step 6)

37. 3. Calculate the obtained value (F obt ) Remember, we are trying to compare the between group variance to the within group variance We use the mean squares to calculate this The mean square within groups is an estimate of the variability in scores as measured by differences within the conditions The mean square between groups is an estimate of the differences in scores that occurs between the levels in a factor

38. 3. Calculate the obtained value (F obt ) The F-ratio is therefore the mean square between groups divided by the mean square within groups

39. 3. Calculate the obtained value (F obt ) The F-ratio is therefore the mean square between groups divided by the mean square within groups

40. 3. Calculate the obtained value (F obt ) The F-ratio is therefore the mean square between groups divided by the mean square within groups When H 0 is true, F obt should be close to 1 When H A is true, F obt should be greater than 1

41. 3. Calculate the obtained value (F obt ) The ANOVA table: SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot

42. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot

43. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot

44. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot df tot = N - 1 df wn = N - k df bn = k - 1

45. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot df tot = N - 1 df wn = N - k df bn = k - 1

46. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 df tot = N - 1 df wn = N - k df bn = k - 1

47. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 df tot = N - 1 df wn = N - k df bn = k - 1

48. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 df tot = N - 1 df wn = N - k df bn = k - 1

49. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 df tot = N - 1 df wn = N - k df bn = k - 1

50. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 df tot = N - 1 df wn = N - k df bn = k - 1

51. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot

52. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot

53. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot = [(30 2 /5) + (35 2 /5) + (40 2 /5)] - (105 2 /15) = [(900/5) + (1225/5) + 1600/5)] - (11025/15) = [180 + 245 + 320] - (735) = 745 - 735 = 10

54. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between10df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot = [(30 2 /5) + (35 2 /5) + (40 2 /5)] - (105 2 /15) = [(900/5) + (1225/5) + 1600/5)] - (11025/15) = [180 + 245 + 320] - (735) = 745 - 735 = 10

55. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between10df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot

56. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between10df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot = (751) - (105 2 /15) = 751 - 735 = 751 - 735 = 16

57. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between10df bn MS bn F obt WithinSS wn df wn MS wn Total16df tot = (751) - (105 2 /15) = 751 - 735 = 751 - 735 = 16

58. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between10df bn MS bn F obt WithinSS wn df wn MS wn Total16df tot = 16 - 10 = 6

59. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between10df bn MS bn F obt Within6df wn MS wn Total16df tot

60. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between10df bn MS bn F obt Within6df wn MS wn Total16df tot = 3 - 1 = 2 df bn = k - 1

61. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between102MS bn F obt Within6df wn MS wn Total16df tot = 3 - 1 = 2 df bn = k - 1

62. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between102MS bn F obt Within6df wn MS wn Total16df tot = 15 - 3 = 12 df wn = N - k

63. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between102MS bn F obt Within612MS wn Total16df tot = 15 - 3 = 12 df wn = N - k

64. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between102MS bn F obt Within612MS wn Total16df tot = 15 - 1 = 14 df tot = N - 1

65. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between102MS bn F obt Within612MS wn Total1614 = 15 - 1 = 14 df tot = N - 1

66. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between102MS bn F obt Within612MS wn Total1614

67. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between102MS bn F obt Within612MS wn Total1614 = 10 / 2 = 5

68. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between1025F obt Within612MS wn Total1614 = 10 / 2 = 5

69. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between1025F obt Within612MS wn Total1614 = 6 / 12 = 0.5

70. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between1025F obt Within6120.5 Total1614 = 6 / 12 = 0.5

71. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between1025F obt Within6120.5 Total1614

72. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between1025F obt Within6120.5 Total1614 = 5 / 0.5 = 10

73. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between1025 Within6120.5 Total1614 = 5 / 0.5 = 10

74. Analysis of Variance BackMiddleFront 678 588 767 679 678 678Mean=7 ∑X=30∑X=35∑X=40∑X tot =105 ∑X 2 =182∑X 2 =247∑X 2 =322∑X 2 tot =751 n 1 =5n 2 =5n 3 =5N = 15 k=3 SourceSum of SquaresdfMean SquaresF Between1025 Within6120.5 Total1614

75. 4. Calculate the critical value Assume  =0.05 Always a two-tail test with ANOVA df between = k - 1 df within = N - k k is the number of levels, or groups N is the total number of subjects df between df within  12345 12 0.05 4.75 3.88 3.49 3.26 3.11

76. F crit and F obt F crit = 3.88 F obt = 10

77. 5. Make our Conclusion F crit = 3.88 F obt = 10 If F obt is outside the rejection region, we retain H 0 If F obt is inside the rejection region, we reject H 0 and accept H A We conclude that there is a significant difference between some of our groups – but which groups?

78. 6. Conduct post-hoc tests To determine which means differ from each other significantly, we conduct post-hoc tests Post hoc comparisons are like t-tests –we compare all possible pairs of means from a factor, one pair at a time There are many possible post-hoc tests to use: –Tukey’s HSD –Scheffe –Bonferroni

79. 6. Conduct post-hoc tests When the n’s in all levels of the factor are equal, we can use Tukey’s HSD test –“honestly significant difference” Calculate the difference between each pair of means Compare each difference between the means to the HSD If the absolute difference between two means is greater than the HSD, then these means differ significantly

80. 6. Conduct post-hoc tests Calculating the HSD: where q k is found using the appropriate table

81. 6. Conduct post-hoc tests HSD = (q k ) [√(MS wn /n)] q k : Look up Table L pg. 562 - Need: –k (number of means being compared) –df wn = (3.77) [√(0.5/5)] = (3.77) [√(0.1)] = (3.77) (0.316) = 1.19

82. 6. Conduct post-hoc tests HSD = 1.19 Difference between means: ‘Back’ (mean=6) and ‘Middle’ (mean=7) : 1 ‘Middle’ (mean=7) and ‘Front’ (mean=8): 1 ‘Back’ (mean=6) and ‘Front’ (mean=8): 2 Because 2 is greater than 1.19, there is a significant difference between the ‘front’ and ‘back’ groups in terms of their attendance

83. ANOVA - example

84. Analysis of Variance A clinical psychologist has noted that autistic children seem to respond to treatment better if they are in a familiar environment. To evaluate the influence of environment, the psychologist selects a group of 18 children who are currently in treatment and randomly divides them into three groups. One group continues to receive treatment in the clinic as usual. For the second group, treatment sessions are conducted entirely in the child’s home. The third group gets half of the treatment in the clinic and half at home. After six weeks the data look as follows (high scores are good): –Clinic: 14, 13, 14, 13, 10, 15 –Home: 10, 12, 11, 13, 11, 10 –Both: 11, 14, 14, 15, 13, 15 Do the data indicate any significant differences between the three settings? Use  =0.05

85. 1. Decide which test to use Are we comparing a sample to a population? – Yes: Z-test if we know the population standard deviation – Yes: One-sample T-test if we do not know the population std dev – No: Keep looking Are we looking for the difference between samples? – Yes: How many samples are we comparing? Two: Use the Two-sample T-test – Are the samples independent or related? » Independent: Use Independent Samples T-test » Related: Use Related Samples T-test More than Two: Use Anova test

86. 2. State the Hypotheses H 0 :  1 =  2 = ……. =  k – there is no difference in the means H A : not all  s are equal – there is a difference between some of the means

87. 3. Calculate the obtained value (F obt ) ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 df tot = N - 1 df wn = N - k df bn = k - 1

88. 3. Calculate the obtained value (F obt ) ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 df tot = N - 1 df wn = N - k df bn = k - 1

89. 3. Calculate the obtained value (F obt ) ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 df tot = N - 1 df wn = N - k df bn = k - 1

90. 3. Calculate the obtained value (F obt ) ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 df tot = N - 1 df wn = N - k df bn = k - 1

91. 3. Calculate the obtained value (F obt ) ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 df tot = N - 1 df wn = N - k df bn = k - 1

92. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3

93. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = [(80 2 /6) + (67 2 /6) + (82 2 /6)] - (229 2 /18) = [(6400/6) + (4489/6) + 6724/6)] - (52441/18) = [1066.67 + 748.17 + 1120.67] - (2913.39) = 2935.51 - 2913.39 = 22.12

94. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = [(80 2 /6) + (67 2 /6) + (82 2 /6)] - (229 2 /18) = [(6400/6) + (4489/6) + 6724/6)] - (52441/18) = [1066.67 + 748.17 + 1120.67] - (2913.39) = 2935.51 - 2913.39 = 22.11

95. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = (2969) - (229 2 /18) = 2969 - 2913.39 = 55.61

96. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11df bn MS bn F obt WithinSS wn df wn MS wn Total55.61df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = (2969) - (229 2 /18) = 2969 - 2913.39 = 55.61

97. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11df bn MS bn F obt WithinSS wn df wn MS wn Total55.61df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 55.61 - 22.12 = 33.49

98. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11df bn MS bn F obt Within33.49df wn MS wn Total55.61df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 55.61 - 22.12 = 33.49

99. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11df bn MS bn F obt Within33.49df wn MS wn Total55.61df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 3 - 1 = 2 df bn = k - 1

100. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.112MS bn F obt Within33.49df wn MS wn Total55.61df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 3 - 1 = 2 df bn = k - 1

101. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.112MS bn F obt Within33.49df wn MS wn Total55.61df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 18 - 3 = 15 df wn = N - k

102. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.112MS bn F obt Within33.4915MS wn Total55.61df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 18 - 3 = 15 df wn = N - k

103. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.112MS bn F obt Within33.4915MS wn Total55.61df tot ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 18 - 1 = 17 df tot = N - 1

104. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.112MS bn F obt Within33.4915MS wn Total55.6117 ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 18 - 1 = 17 df tot = N - 1

105. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.112MS bn F obt Within33.4915MS wn Total55.6117 ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 22.12 / 2 = 11.06

106. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11211.055F obt Within33.4915MS wn Total55.6117 ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 22.12 / 2 = 11.06

107. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11211.055F obt Within33.4915MS wn Total55.6117 ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 33.49 / 15 = 2.23

108. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11211.055F obt Within33.49152.23 Total55.6117 ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 33.49 / 15 = 2.23

109. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11211.055F obt Within33.49152.23 Total55.6117 ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 11.06 / 2.23 = 4.96

110. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11211.0554.957 Within33.49152.23 Total55.6117 ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3 = 11.06 / 2.23 = 4.96

111. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between22.11211.054.96 Within33.49152.23 Total55.6117 ClinicHomeBoth 141011 131214 1114 1315 101113 151015 13.3311.1713.67Mean=12.72 ∑X=80∑X=67∑X=82∑X tot =229 ∑X 2 =1082∑X 2 =755∑X 2 =1322∑X 2 tot =2969 n 1 =6n 2 =6n 3 =6N = 18 k=3

112. 4. Calculate the critical value Assume  =0.05 Always a two-tail test with ANOVA df between = k - 1 = 3 - 1 = 2 df within = N - k = 18 - 3 = 15 F crit for 2 and 15 degrees of freedom and  = 0.05 is 3.68

113. F crit and F obt F crit = 3.68 F obt = 4.96

114. 5. Make our Conclusion F crit = 3.68 F obt = 4.96 F obt is inside the rejection region, we reject H 0 and accept H A We conclude that there is a significant difference between some of our groups – which groups?

115. 6. Conduct post-hoc tests HSD = (q k ) [√(MS wn /n)] q k : Look up Table L - Need: –k (number of means being compared) –df wn = (3.675) [√(2.233/6)] = (3.675) [√(0.372)] = (3.675) (0.61) = 2.242

116. 6. Conduct post-hoc tests HSD = 2.242 Difference between means: Clinic - Home = 2.166 Clinic - Both = -0.334 Home - Both = -2.5 Therefore, there is a significant difference in scores between treatment at home, and treatment at both home and at the clinic

117. ANOVA - your turn

118. Problem A psychologist is interested in the effects of room temperature on learning. 15 subjects are recruited for an experiment to answer this questions - 5 subjects undergo a learning task in a room with a temperature of 50 degrees, another group of 5 learn the same material in room with a temperature of 70 degrees and the final group learn in a temperature of 90 degrees. The data are below, with higher numbers indicating greater learning. Does room temperature affect rates of learning? Use  =0.05 –50 degrees: 0, 1, 3, 1, 0 –70 degrees : 4, 3, 6, 3, 4 –90 degrees : 1, 2, 2, 0, 0

119. Analysis of Variance 50°70°90° 041 132 362 130 040 df tot = N - 1 df wn = N - k df bn = k - 1

120. 1. Decide which test to use Are we comparing a sample to a population? – Yes: Z-test if we know the population standard deviation – Yes: One-sample T-test if we do not know the population std dev – No: Keep looking Are we looking for the difference between samples? – Yes: How many samples are we comparing? Two: Use the Two-sample T-test – Are the samples independent or related? » Independent: Use Independent Samples T-test » Related: Use Related Samples T-test More than Two: Use Anova test – No: Keep looking Are we looking for the relationship between variables? – Yes: Use the Correlation test

121. 2. State the Hypotheses H 0 :  1 =  2 = ……. =  k – there is no difference in the means H A : not all  s are equal – there is a difference between some of the means

122. Analysis of Variance 50°70°90° 041 132 362 130 040

123. Analysis of Variance 50°70°90° 041 132 362 130 040 141Mean=2

124. Analysis of Variance 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30

125. Analysis of Variance 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106

126. Analysis of Variance 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15

127. Analysis of Variance 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

128. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

129. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF BetweenSS bn df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot = [(5 2 /5) + (20 2 /5) + (5 2 /5)] - (30 2 /15) = [(25/5) + (400/5) + 25/5)] - (900/15) = [5 + 80 + 5] - (60) = 90 - 60 = 30 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

130. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3 = [(5 2 /5) + (20 2 /5) + (5 2 /5)] - (30 2 /15) = [(25/5) + (400/5) + 25/5)] - (900/15) = [5 + 80 + 5] - (60) = 90 - 60 = 30

131. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30df bn MS bn F obt WithinSS wn df wn MS wn TotalSS tot df tot = (106) - (30 2 /15) = 106 - 60 = 46 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

132. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30df bn MS bn F obt WithinSS wn df wn MS wn Total46df tot = (106) - (30 2 /15) = 106 - 60 = 46 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

133. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30df bn MS bn F obt WithinSS wn df wn MS wn Total46df tot = 46 - 30 = 16 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

134. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30df bn MS bn F obt Within16df wn MS wn Total46df tot = 46 - 30 = 16 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

135. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30df bn MS bn F obt Within16df wn MS wn Total46df tot = 3 - 1 = 2 df bn = k - 1 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

136. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between302MS bn F obt Within16df wn MS wn Total46df tot = 3 - 1 = 2 df bn = k - 1 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

137. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between302MS bn F obt Within16df wn MS wn Total46df tot = 15 - 3 = 12 df wn = N - k 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

138. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between302MS bn F obt Within1612MS wn Total46df tot = 15 - 3 = 12 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3 df tot = N - 1

139. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between302MS bn F obt Within1612MS wn Total46df tot = 15 - 1 = 14 df tot = N - 1 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

140. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between302MS bn F obt Within1612MS wn Total4614 = 15 - 1 = 14 df tot = N - 1 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

141. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between302MS bn F obt Within1612MS wn Total4614 = 30 / 2 = 15 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

142. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30215F obt Within1612MS wn Total4614 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3 = 30 / 2 = 15

143. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30215F obt Within1612MS wn Total4614 = 16 / 12 = 1.33 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

144. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30215F obt Within16121.33 Total4614 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3 = 16 / 12 = 1.33

145. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between30215F obt Within16121.33 Total4614 = 15 / 1.33 = 11.28 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

146. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between3021511.28 Within16121.33 Total4614 = 15 / 1.33 = 11.28 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

147. 3. Calculate the obtained value (F obt ) SourceSum of SquaresdfMean SquaresF Between3021511.28 Within16121.33 Total4614 50°70°90° 041 132 362 130 040 141Mean=2 ∑X=5∑X=20∑X=5∑X tot =30 ∑X 2 =11∑X 2 =86∑X 2 =9∑X 2 tot =106 n 1 =5n 2 =5n 3 =5N = 15 k=3

148. 4. Calculate the critical value Assume  =0.05 Always a two-tail test with ANOVA df between = k - 1 = 3 - 1 = 2 df within = N - k = 15 - 3 = 12 F crit for 2 and 15 degrees of freedom and  = 0.05 is 3.88

149. F crit and F obt F crit = 3.88 F obt = 11.28

150. 5. Make our Conclusion F crit = 3.88 F obt = 11.28 F obt is inside the rejection region, we reject H 0 and accept H A We conclude that there is a significant difference between some of our groups – which groups?

151. 6. Conduct post-hoc tests HSD = (q k ) [√(MS wn /n)] q k : Look up Table L - Need: –k (number of means being compared) –df wn = (3.77) [√(1.33/5)] = (3.77) [√(0.266)] = (3.77) (0.52) = 1.96

152. 6. Conduct post-hoc tests HSD = 1.96 Difference between means: 50° group - 70° group = -3 50° group - 90° group = 0 70° group - 90° group = 3 Therefore, there is a significant increase in learning when the temperature is 70° as compared to either 50° or 90°