Two-Way Analysis of Variance: Comparing Means in Two-Factor Experiments - Prof. James W. D, Study notes of Data Analysis & Statistical Methods

Download Two-Way Analysis of Variance: Comparing Means in Two-Factor Experiments - Prof. James W. D and more Study notes Data Analysis & Statistical Methods in PDF only on Docsity! Lecture 10, Chapter 13 Two-Way Analysis of Variance: Two-way ANOVA compares the means of populations that are classified two ways or the mean responses in two-factor experiments. Examples: 1. The strength of concrete depends upon the formula used to prepare it. An experiment compares six different mixtures. Nine specimens of concrete are poured from each mixture. Three of these specimens are subjected to 0 cycles of freezing and thawing, three are subjected to 100 cycles, and three specimens are subjected to 500 cycles. The strength of each specimen is then measured. 2. Four methods for teaching sign language are to be compared. Sixteen students in special education and sixteen students majoring in other areas are the subjects for the study. Within each group they are randomly assigned to the methods. Scores on a final exam are compared. Why is it better to do a Two-Way ANOVA than to just do 2 separate One-Way ANOVAs:  It is more efficient to study two factors simultaneously rather than separately. Your sample size does not have to be as large so experiments with several factors are an efficient use of resources.  We can reduce the residual variation in a model by including a second factor thought to influence the response variable (lurking variable). We are reducing σ and increasing the power of the test.  We can investigate the interactions between factors. Assumptions for Two-Way ANOVA: 1. We have two factors. We have I factor levels for the first factor (call it factor A) and J factor levels for the second factor (call it factor B). We have I x J combinations of individual factor levels. 2. We have independent SRSs of size ijn from each of I x J populations. 3. Each of the I x J populations are normally distributed. 4. Each of the I x J populations have the same standard deviation σ. Lecture 10, Chapter 13 Page 1 Model for the Two-Way ANOVA: Let ijkx represent the kth observation from the population having factor A at level I and factor B at level j. The statistical model is ijk ij ijk x    for i = 1,……,I and j = 1,……..,J and k = 1,………, ijn , where the deviations ijk  are from an N(0,σ) distribution. Examples: For the two examples above, identify the response variable and both factors, and state the number of levels for each factor (I and J) and total number of observations (N). 1. The strength of concrete depends upon the formula used to prepare it. 2. Four methods for teaching sign language are to be compared. Sixteen students in special education and sixteen students majoring in other areas are the subjects for the study. Lecture 10, Chapter 13 Page 2 a. Make a table giving the sample size, mean, and standard deviation for each of the material-by-time combinations. Is it reasonable to pool the variances? (Because the sample sizes in this experiment are very small, we expect a large amount of variability in the sample standard deviations. Although they vary more than we would prefer, we will proceed with the ANOVA.) SPSS: Data/Split file, select compare groups, move one categorical variable (material) into groups box. Analyze/compare means/Means, move the other categorical variable (weeks) into independent list, move gpi into dependent list. Descriptive Statistics Dependent Variable: gpi material weeks Mean Std. Deviation N ECM1 2 weeks 70.00 5.000 3 4 weeks 65.00 8.660 3 8 weeks 63.33 2.887 3 Total 66.11 6.009 9 ECM2 2 weeks 65.00 5.000 3 4 weeks 63.33 2.887 3 8 weeks 63.33 5.774 3 Total 63.89 4.167 9 ECM3 2 weeks 71.67 10.408 3 4 weeks 73.33 2.887 3 8 weeks 73.33 5.774 3 Total 72.78 6.180 9 MAT1 2 weeks 48.33 2.887 3 4 weeks 23.33 2.887 3 8 weeks 21.67 5.774 3 Total 31.11 13.411 9 MAT2 2 weeks 10.00 5.000 3 4 weeks 6.67 2.887 3 8 weeks 6.67 2.887 3 Total 7.78 3.632 9 MAT3 2 weeks 26.67 2.887 3 4 weeks 11.67 2.887 3 8 weeks 10.00 5.000 3 Total 16.11 8.580 9 Total 2 weeks 48.61 24.363 18 4 weeks 40.56 28.330 18 8 weeks 39.72 28.619 18 Total 42.96 26.961 54 Lecture 10, Chapter 13 Page 5 b. Make a table giving the sample size, mean, and standard deviation for each of the material types. Give a short summary of the % Gpi depends on material. SPSS: Data/Split file, remove the compare groups. Analyze/compare means/Means, move material into independent list, move gpi into dependent list. Report gpi 66.11 9 6.009 63.89 9 4.167 72.78 9 6.180 31.11 9 13.411 7.78 9 3.632 16.11 9 8.580 42.96 54 26.961 material ECM1 ECM2 ECM3 MAT1 MAT2 MAT3 Total Mean N Std. Deviation c. Make a table of the sample size, mean, and standard deviation for the weeks after the repair. Give a short summary of the % Gpi depends on the weeks after repair. SPSS: same only move weeks into independent box. Report gpi 48.61 18 24.363 40.56 18 28.330 39.72 18 28.619 42.96 54 26.961 weeks 2 weeks 4 weeks 8 weeks Total Mean N Std. Deviation Lecture 10, Chapter 13 Page 6 d. Run the analysis of variance. Report the F statistics with degrees of freedom and P-values for the main effects and the interaction. What are the hypotheses you are testing? What do you conclude? SPSS Instructions: Analyze>General Linear>Univariate. Move the appropriate variables to the dependent and fixed factor boxes. Select Model and do not remove the “checkmark” beside the “Include intercept in model”. Tests of Between-Subjects Effects Dependent Variable: gpi Source Type III Sum of Squares df Mean Square F Sig. Corrected Model 37609.259(a) 17 2212.309 86.883 .000 Intercept 99674.074 1 99674.074 3914.473 .000 material 35659.259 5 7131.852 280.087 .000 weeks 867.593 2 433.796 17.036 .000 material * weeks 1082.407 10 108.241 4.251 .001 Error 916.667 36 25.463 Total 138200.000 54 Corrected Total 38525.926 53 a R Squared = .976 (Adjusted R Squared = .965) Lecture 10, Chapter 13 Page 7