Full Lecture Presentation on ANOVA

a full lecture presentation on ANOVA .
areas covered include;
a. definition and purpose of anova
b. one-way anova
c. factorial anova
d. mutiple anova
e MANOVA
f. POST-HOC TESTS - types
f. easy step by step process of calculating post hoc test.

1. Masinde Muliro UNIVERSITY LECTURE PRESENTATION ON ANOVA

2. Presenter Kololi Gellause

3. INTRODUCTION What is Anova? It is a statistical procedure used to test the degree to which the means of 2 or more groups differ in an experiment In Anova we look at how he group means vary from each other that they exceed individual differences within the groups. Developed by Sir Ronald A. Fisher in 1920’s.

4. purpose The purpose of ANOVA is much the same as the t tests, the goal is to determine whether the mean differences that are obtained for sample data are sufficiently large to justify a conclusion that there are mean differences between the populations from which the samples were obtained .

5. Why Anova? Two sample t-test are problematic thus increasing the risk of type 1 error. The more t test you run the greater the risk of type one error - at 0.5 level of significance with hundred comparisons 5 will show some difference when none exists -the difference between ANOVA and the t tests is that ANOVA can be used in situations where there are two or more means being compared, whereas the t tests are limited to situations where only two means are involved. - Anova avoids type 1 error in situations where a study is comparing more than two population means Anova allows to see if there are differences between the means with an OMNIBUS test - they test whether the explained variance in a set of data is significantly greater than unexplained variance

6. CONDITIONS FOR ANOVA Data must be experimental With many experimental designs the sample sizes must be equal for various factor level combinations

7. Assumptions Normality: The values in each group are normally distributed. Homogeneity of variances: The variance within each group should be equal for all groups. Independence of error: The error(variation of each value around its own group mean) should be independent for each value.

8. Types of Anova One way Anova - One way repeated measure factorial Anova Multiple Anova Manova

9. Independent variables 1 2+ Dependent variables 1 One-way Anova Factorial Anova 2+ Multiple Anova MANOVA

10. Variance Why do scores vary? What contributes to differences in scores individual differences Which group you are in There is variation anytime that all of the data values are not identical and this variation can come from different sources such as the model or the factor -There is always the left over variation that cant be explained by any other sources. This source is called the error

11. Variation can be: Within group variation variability or differences in particular groups between groups differences depending what groups one is in (what treatment you received) - We are actually examining the ratio of differences (variances) from treatment to variances from individual differences if the ratio is large there is significant impact from treatment

12. . We can know this variation by evaluating the ratio of the MST to MSE and conducting F test We can be able to compare multiple means Between group variances reflect differences in the way the groups were treated within group variances reflects individual differences Null hypothesis – there is no significant differences in means Alternative hypothesis – there is significant differences between the means We are comparing variance estimates Variance = SS/DF

13. Degrees of freedom This are the number of values that are free to vary once certain parameters have been established Usually this is one less than the sample size but in general its the number of values minus the number of parameters being estimated abbreviated as (df)

14. Variance (MS) The sample variances is the average squared deviation from the mean found by dividing the variation by the degree of freedom Variance (MS) = variation ∕ df It is abbreviated as MS for mean of squares MS = SS/df

15. F F is the f statistic F is the ratio of two sample variances There will be an f statistic for each source except for the error and the total The MS column contains variances The F test statistic for each source is the MS for that row divided by the MS of the error row

16. F F requires a pair of degrees of freedom, one for the numerator and another for the denominator The numerator df is the degree of freedom for the source The numerator df is the df for the error row F is always a right tail test

17. Notes on Anova The MS(Total) isn’t actually part of the ANOVA table, but it represents the sample variance of the response variable, so it’s useful to find The total df is one less than the sample size You would either need to find a Critical F value or the p-value to finish the hypothesis test

18. One way Anova Measures the outcome of intervention in subjects over time There is one independent variable like time on therapy, an independent variable that has multiple levels e.g time on therapy then a dependent variable

19. one way Anova Determines means of 2 or more independent groups significantly different from one another. Only 1 independent variable (factor/grouping variable) with ≥3 levels Only 1 dependent variable Grouping variable- nominal Outcome variable- interval or ratio

20. Steps.. State null and alternative hypothesis State alpha Calculate degree of freedom State decision rule Calculate test statistic Calculate variance between samples Calculate variance within samples Calculate f statistic If F’ is significant, perform post hoc test State results and conclusion

21. Steps.. State the null and alternative hypothesis H0 = U1 = U2------ U j Ho = all sample means are equal H a = not all of the U j are equal at least one of the sample has different means State alpha 0.05 Calculate degrees of freedom (n-k) and (n-1) k - No of samples n – Total number of observation State decision rule if calculated f value > table value of F, reject H0 Calculate test statistic

22. Calculating variance between samples Calculate the mean of each sample. Calculate the Grand average Take the difference between means of various samples & grand average. Square these deviations & obtain total which will give sum of squares between samples (SSC) Divide the total obtained in step 4 by the degrees of freedom to calculate the mean sum of square between samples (MSC ).

23. Calculating Variance within Samples Calculate mean value of each sample Take the deviations of the various items in a sample from the mean values of the respective samples. Square these deviations & obtain total which gives the sum of square within the samples (SSE) Divide the total obtained in 3 rd step by the degrees of freedom to calculate the mean sum of squares within samples (MSE).

24. Mean sum of squares 1. MSC ( Between samples) MSC = SSC/k-1 MSE (within samples) MSE = SSE / n-k when k= No of samples n = total number of observations

25. Calculation of F statistic f = variability between groups/ variability within groups F statistic = MSC/MSE compare the F statistic value with F (critical) value which is obtained by looking for it in the F distribution tables against degrees of freedom The calculated value of F if is > table value Ho is rejected

26. Within-Group Variance Between-Group Variance Between-group variance is large relative to the within-group variance, so F statistic will be larger & > critical value, therefore statistically significant . Conclusion – At least one of group means is significantly different from other group means

27. Within-Group Variance Between-Group Variance W ithin-group variance is larger, and the between-group variance smaller, so F will be smaller ( reflecting the likely-hood of no significant differences between these 3 sample means )

28. Example: 3 sample population of preschool children obtained from normal populations with equal variances. were grouped in accordance to their visual acuity as 6/24, 6/36 and 6/60 obtained during a vision screening and their scores for ease of performing daily activities were assessed and recorded using a scale of 1-10 (1 coded for being unable to perform a particular task and 10 for most ease in performing a task ) test the hypothesis that sample means of their scores are equal

29. 9 + 6.8 + 4.2 3 Grand average = = 6.67 Visual acuity 6/24 6/36 6/60 Ease of doing daily activity 8 5 3 10 5 3 8 8 5 9 7 4 10 9 6 total 45 34 21 M1=9 m2-=6.8 M3=4.2

30. Variance BETWEEN samples Sum of squares between samples (SSC) = n 1 ( M1 – Grand avg) 2 + n2 ( M2 – Grand avg) 2 + n3 ( M3 – Grand avg) 2 5 ( 9 - 6.67) 2 + 5 ( 6.8 - 6.67) 2 + 5 ( 4.42 - 6.67) 2 = 52.55 Calculation of M ean sum of S quares between samples ( MSC) k= No of Samples, n = Total No of observations

31. Variance WITH IN samples X1 (X1 – M1) 2 X2 (X2– M2) 2 X3 (X3– M3) 2 8 1 5 3.24 3 1.44 10 1 5 3.24 3 1.44 8 1 8 1.44 5 0.64 9 7 0.04 4 0.04 10 1 9 4.84 6 3.24 4 12.8 6.8 Sum of squares within samples (SSE) = 4 + 12.8 + 6.8=23.6 Calculation of Mean Sum Of Squares within samples ( MSE)

32. Calculation of ratio F sss F = 26.27/ 1.97 = 13.34 The Table value of F at 5% level of significance for d.f 2 & 12 is 3.88 The calculated value of F > table value H0 is rejected. Hence there is significant difference in sample means

33. Post hoc test Uses: Post hoc tests are designed for situations in which the researcher has already obtained a significant omnibus F- test . Thus he/she analyses the differences among the means to find which means are significantly different from each other

34. Types of post-hoc Tests.. Depending upon research design & research question: Bonferroni (more flexible) Only some pairs of sample means are to be tested Desired alpha level is divided by no. of comparisons Tukey’s HSD Procedure when all pairs of sample means are to be tested Scheffe’s Procedure (when sample sizes are unequal)

35. Tukey honest sig difference Tukey tests calculates a new critical value that can be used to evaluate whether the differences between any two pairs of means is significant . The critical value is a little different because it involves the mean difference that has to be exceeded to achieve significance. Simply one calculates one critical value and then the differences between all possible pairs of means, each critical value is then compared to the tukey critical value. If the difference is larger than the tukey value, the comparison is significant

36. Tukey HSD The formula for the critical value is t.hsd = M 1 -M 2 √Ms w (1/n) 1. M is the treatment group mean 2. Ms w is the mean square error degree of freedom from the overall F-test 3. n is the sample size for each treatment group

37. Post hoc for our data t.hsd = M 1 -M 2 √Ms w (1/n) M1=9 M2=6.8 M3= 4.2 n = 5 Ms w = 12 M1 vs M2 = 9-6.8 = 1.42009 √ 12 × (1/5) 2. M1 vs M3 = 9-4.2 = 3.098 √ 12 × (1/5)

38. Cont.. M2 vs M3 = 6.8- 4.2 = 1.678 √ 12 × (1/5) M1 vs M2 = 1.42009 M1 vs M3 = 3.0983 M2 vs M3 = 1.67829 Using Tukey's sig/ probability table , taking into account ( df w =12 n =3) the mean comparison between the three pairwise means M1 and M2 and M3 are statistically significant.

39. THE ANOVA OUTPUT TABLE The ANOVA table is composed of rows, each row represents one source of variation For each source of variation … The variation is in the SS column The degrees of freedom is in the df column The variance is in the MS column The MS value is found by dividing the SS by the df

40. . The ANOVA OUTPUT Table Source SS (variation) df MS (variance) F Explained* Error Total

41. LETS FILL OUR OUTPUT TABLE WITH THE VALUES WE CALCULATED . Source SS df MS F Explained (Between groups) 52.55 2 26.27 Error (within groups) 23.6 12 1.97 Total 76.15 14 Divide SS by df to get MS.

42. Find F . Source SS df MS F Explained 52.55 2 26.27 13.34 Error 23.6 12 1.97 Total 76.15 14 F = / =

43. Violations of Assumptions Normality Choose the non-parametric Kruskal-Wallis H Test which does not require the assumption of normality. Homogeneity of variances Welch test or Brown and Forsythe test or Kruskal-Wallis H Test

44. Factorial Anova

45. Factorial Anova Introduction Often, we wish to study 2 (or more) factors in a single experiment, for example.. Compare two or more treatment protocols Compare scores of people who are young, middle-aged, and elderly The baseline experiment will therefore have two factors as Independent Variables Treatment type Age groups

46. Two-Way ANOVA Factorial ANOVA has One dependent variable interval or ratio with a normal distribution Two independent variables nominal (define groups), and independent of each other Three hypothesis tests: Test effect of each independent variable controlling for the effects of the other independent variable One : H : factor A has no impact on Outcome (no relationship) Two : H : factor B has no impact on Outcome Three : Test interaction effect for combinations of categories H : there is no relationship between the interaction effect of factors A and B on the Outcome

47. Structure of the Two-Factor Analysis of Variance

48. Assumptions for the Two-Factor ANOVA The validity of the ANOVA presented in this chapter depends on three assumptions common to other hypothesis tests The observations within each sample must be independent of each other The populations from which the samples are selected must be normally distributed The populations from which the samples are selected must have equal variances (homogeneity of variance)

49. Factorial ANOVA asks three questions and tests three null hypotheses First: Does Factor 1 have any impact on the Outcome? Null : The groups defined by Factor 1 will have the same Mean Outcome. Second: Does Factor 2 have any impact on the Outcome? Null : The groups defined by Factor 2 will have the same Mean Outcome. Third: Do Factor 1 and Factor 2 interact in influencing Outcome? Null : No combination of Factor 1 and Factor 2 produces unusually high or unusually low mean Outcome scores.

50. Plotting the Main Effects Plot the means of each group (defined as a combination of Factor 1 and Factor 2) If all the null hypotheses are true , all the points will have about the same Mean Outcome level.

51. The Null Result: No Effects The two row means are the same The two column means are the same All groups have the same mean score Neither factor had any effect

52. Main effects Row means: the same Column means: differ Row means: differ Column means: the same

53. Degrees of freedom for Two-Factor ANOVA df total = N – 1 df within treatments = Σ df inside each treatment df between treatments = k – 1 df A = number of groups – 1 df B = number of groups – 1 df AxB = df between treatments df A df B

54. total sums of squares can be partitioned into “explained” and “unexplained” components … i.e Explained Sums of Squares component (variation explained by differences between groups) Unexplained Sums of Squares component (variation explained by differences within groups)

55. EXAMPLE Researcher wants to test a new anti-anxiety medication. They measure the anxiety of 36 participants on three different dosages of the medication 0mg, 50 mg, and 100 mg. The participants are also divided based on what school they are attending , which researchers hypothesize will also affect the anxiety levels . Anxiety is rated on a scale of 1-10, with 10 being high intensity and 1 being low intensity Use alpha = 0.05 to conduct your analysis

56. example 0 mg 50 mg 100 mg High school students 3 5 7 4 4 8 5 3 7 3 5 8 4 5 7 3 5 7 College students 1 5 9 2 4 8 1 3 9 1 5 8 1 5 7 2 4 9

57. Factorial ANOVA will produce an F-ratio for each main effect and for each interaction. Main effect: school attended – F ratio . Main effect: dosage – F ratio . Interaction effect: school attended by dosage – F ratio

58. We will consider the effect of multiple independent variables on a single dependent variable . i.e: First Independent Variable: dosage Level 1: 0mg Level 2: 50 mg Level 3 100 mg Second Independent Variable: school attended Level 1: High school students Level 2: University students

59. Cont.. We will be comparing 6 groups ( 3 levels of dosage x 2 levels of the stage of education ). The procedure by which we analyze the sums of squares among the 6 groups based on 2 independent variables ( stage of education and dosage ) is called Factorial ANOVA . Thus we will organize the data for easy analysis

60. dosage High school college O mg 3 1 4 2 5 1 3 1 4 1 3 2 50 mg 5 5 4 4 3 3 5 5 5 5 5 4 100 mg 7 9 8 8 7 9 8 8 7 7 7 9

61. State the null hypothesis There is no significant difference between the level of anxiety caused by drug and the level education There is no significant difference between the level of anxiety and drug dosage There is no significant interaction between the level of education and drug dosage to cause a change in anxiety

62. next we begin with calculating stage of education Sums of Squares We organize the data set with stage of education in the headers,

63. High school college students 3 1 4 2 5 1 3 1 4 1 3 2 5 5 4 4 3 3 5 5 5 5 5 4 7 9 8 8 7 9 8 8 7 7 7 9 5.17 4.67 4.92 4.92 0.0625 0.0625 1.125 1.125 2.25 Mean (m) Grand mean (gm) Deviation ² (m-gm)² Weighted sq deviation (m-gm)²× 18 Sum of squares 2.25

64. Next we calculate the dosage Sums of Squares We reorder the data so that we can calculate sums of squares for the dosage

65. we will compute the between group sums of squares for dosage Group 0mg 50mg 100mg 3 5 7 4 4 8 5 3 7 3 5 8 4 5 7 3 5 7 1 5 9 2 4 8 1 3 9 1 5 8 1 5 7 2 4 9 2.5 4.42 7.83 4.92 4.92 4.92 5.86 0.25 8.47 70.32 3 101.64 MEAN Grand mean Dv² Weighted sq deviation 174.96

66. Here is how we reorder the data to calculate the error ( within groups sums of squares) (mark that the tables are broken into preceding slides due to the amount of data being calculated)

67. dosage Stage of education Group mean deviation Deviation ² O mg High schl 3 3.6 -0.6 0.36 O mg High schl 4 3.6 0.4 0.16 O mg High schl 5 3.6 1.4 1.96 O mg High schl 3 3.6 -0.6 0.36 O mg High schl 4 3.6 0.4 0.16 O mg High schl 3 3.6 - 0.6 0.36 O mg College st 1 1.3 - 0.3 0.09 O mg College st 2 1.3 0.7 0.49 O mg College st 1 1.3 - 0.3 0.09 O mg College st 1 1.3 - 0.3 0.09

68. Dosage stage of education Group mean deviation deviation² O mg College st 1 1.3 - 0.3 0.09 O mg College st 2 1.3 0.7 0.49 50 mg High schl 5 4.5 0.5 0.25 50 mg High schl 4 4.5 - 0.5 0.25 50 mg High schl 3 4.5 - 1.5 2.25 50mg High schl 5 4.5 0.5 0.25 50 mg High schl 5 4.5 0.5 0.25 50 mg High schl 5 4.5 0.5 0.25 50 mg College st 5 4.3 0.7 0.49 50 mg College st 4 4.3 - 0.3 0.09 50 mg College st 3 4.3 - 1.3 1.69

69. Dosage Level of education Group mean deviation deviation² 50 mg College st 5 4.3 0.7 0.49 50 mg College st 5 4.3 0.7 0.49 50 mg College st 4 4.3 - 0.3 0.09 100 mg High schl 7 7.3 -0.3 0.09 100 mg High schl 8 7.3 0.7 0.49 100 mg High schl 7 7.3 -0.3 0.09 100 mg High schl 8 7.3 0.7 0.49 100 mg High schl 7 7.3 - 0.3 0.09 100 mg High schl 7 7.3 - 0.3 0.09 100 mg College st 9 8.3 0.7 0.49 100 mg College st 8 8.3 - 0.3 0.09

70. Dosage Stage of education Group mean deviation Deviation² 100 mg College st 9 8.3 0.7 0.49 100 mg College st 8 8.3 - 0.3 0.09 100 mg College st 7 8.3 - 1.3 1.69 100 mg College st 9 8.3 0.7 0.49 TOTAL SUM OF SQUARES= 16.22

71. Here is a simple way we go about calculating sums of squares for the interaction between dosage and stage of education Dependent variable : anxiety source sum of sq df ms f Sig level Level of education 2.25 1 dosage 174.96 2 Level of education× dosage error 16.22 30 Total SS

72. Calculate the mean sum of squares for interaction effect Here is a simple way we go about calculating sums of squares for the interaction between dosage and stage of education We simply sum up the total sums of squares for the data being analyzed and then subtract it from the other sums of squares we calculated earlier i.e Total sum of squares ( SS of dosage SS for age SS for error) SS for interaction effect

73. So we will calculate total sum of squares to help us get the interaction effect Here is our data again

74. Mean deviation Deviation ² 3 4.92 -1.92 3.69 4 4.92 -0.92 0.85 5 4.92 0.08 0.0064 3 4.92 -1.92 3.69 4 4.92 -0.92 0.85 3 4.92 -1.92 3.69 5 4.92 0.08 0.0064 4 4.92 -0.92 0.85 3 4.92 -1.92 3.69 5 4.92 0.08 0.0064 5 4.92 0.08 0.0064 5 4.92 0.08 0.0064 7 4.92 2.08 4.33 8 4.92 3.08 9.49 7 4.92 2.08 4.33 8 4.92 3.08 9.49 7 4.92 2.08 4.33 7 4.92 2.08 4.33

75. Mean deviation deviation² 1 4.92 -3.92 15.37 2 4.92 -2.92 8.53 1 4.92 -3.92 15.37 1 4.92 -3.92 15.37 1 4.92 -3.92 15.37 2 4.92 -2.92 8.53 5 4.92 0.08 0.0064 4 4.92 -0.92 0.85 3 4.92 -1.92 3.69 5 4.92 0.08 0.0064 5 4.92 0.08 0.0064 4 4.92 -0.92 0.85 9 4.92 4.08 16.65 8 4.92 3.08 9.49 9 4.92 4.08 16.65 8 4.92 3.08 9.49 7 4.92 2.08 0.0064 9 4.92 4.08 16.65 Total 177 Sum of sq = 206.53

76. to get interaction effect (level of education × dosage) = total SS – (error + dosage + level of education) source sum of sq df ms f Sig level Level of education 2.25 1 dosage 174.96 2 Level of education× dosage 13.1 2 error 16.22 30 Total SS 206.53 Total dosage Level of education Error L.E ×dosage 206.53 – 174.96 – 2.25 – 16.22 = 13.1

77. to determine the degrees of freedom for error . we take the number of subjects (36) and subtract that number by the number of subgroups (6): = 30

78. source sum of sq df Level of education 2.25 1 dosage 174.96 2 Level of education× dosage 13.1 2 error 16.22 3o Total SS 206.53 MS 2.25 87.48 6.55 0.54 F 4.17 162 12.13 sig 0.05

79. If the F ratio is greater than the F critical, we would reject the null hypothesis and determine that the result is statistically significant. If the F ratio is smaller than the F critical then we would fail to reject the null hypothesis.

80. In this case we reject the three stated null hypotheses Main Effect for dosage : There is no significant difference between the level of anxiety caused by drug and the level education Main Effect for level of education : There is no significant difference between the level of anxiety and drug dosage Interaction Effect Between dosage and level of education : There is no significant interaction between the level of education and drug dosage to cause a change in anxiety

81. . Having rejected the three null hypothesis the researcher goes ahead to carry out post hoc test using either tukey hsd, bonferroni or Scheffe's procedure In this case you can use tukey honest significant difference by using the formula and procedure discussed in the previous slides..

82. LETS HAVE A LOOK AT HOW FACTORIAL ANALYSIS IS DONE ON SPSS USING OUR EXAMPLE HERE IS OUR QUESTION AGAIN. Researchers wanted to test a new anti-anxiety medication. They measured the anxiety of 36 participants on 3 different dosages of the medication: 0mg, 50mg and 100mg. Participants are also divided based on what school they are attending, which researchers hypothesized will also affect anxiety levels. Anxiety is rated on a scale of 1-10, with 10 being ‘high anxiety’ and 1 being ‘low anxiety’. Use a p value of 0.05 to conduct your analysis

83. 0 mg 50mg 100mg High school students 3 4 5 3 4 3 5 4 3 5 5 5 7 8 7 8 7 7 College students 1 2 1 1 1 2 5 4 3 5 5 4 9 8 9 8 7 9 We have 2 independent variable: school and dosage and 1 dependent variable: anxiety level.

84. State the null hypothesis There is no significant difference between the level of anxiety caused by drug and the level education There is no significant difference between the level of anxiety and drug dosage There is no significant interaction between the level of education and drug dosage to cause a change in anxiety

85. * Remember to properly code the variables

86. The data view will look like this;

87. Moving on… Analyze- general linear model

88. select Univariate

89. After clicking ‘ok’ the output looks like this…

90. cont Our factorial Anova tested for 3 things: school, dosage and the interaction between school and dosage School had an f value of 4.17 and a significance level of 0.04988 we can therefore reject the null hypothesis and say there is a difference between high school and college in terms of anxiety

91. Cont.. For dosage the significance level is zero so we reject the null hypothesis: there is a significant difference between anxiety levels of students administered 0mg, 50mg and 100mg. For the interaction between school and dosage we shall reject the null hypothesis: there is a significant interaction between dosage and school.

92. Select Analyze → General Linear Model → Univariate → Add dependent and fixed factor(s) → Click on PLOTS → Transfer independent variables from factors into horizontal and separate lines → Add → Continue

93. 8) Click the POST HOC button to get Univariate: Post Hoc multiple comparisons for observed means 9) Transfer the Independent variable (IV) from factors to Post Hoc tests, then select Tukey Hsd, bonferroni, scheffes 10) Click CONTINUE button to return to Univariate dialogue box. 11) Click OPTIONS button

94. 12) Transfer IV from factor(s) and factor interactions box into the display means for box → TICK Descriptive Stat Option 13) Click CONTINUE 14) Click OK *...End..*

95. references J. Neter , W. Wasserman and M.H. Kutner (1985). Applied Linear Statistical Models, Second Edition, Irwin, Inc. Y. Hochberg and A.C. Tamhane (1987). Multiple Comparison Procedures. John Wiley & Sons, New York. L.S. Nelson (1974). “Factors for the Analysis of Means,” Journal of Quality Technology, 6, pp.175–181 J.C. Hsu (1996). Multiple Comparisons, Theory and methods, Chapman & Hall, New York. R.A. Olshen (1973). “The conditional level of the F-test,” Journal of the American Statistical Association, 68, pp.692–698