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1 One-Factor Analysis of ANOVA Reading Assignment: O/L Sections 8.1–8.4 Model: Yij = i + eij, i = 1, ..., g = 4, j = 1, ..., n = 10; eij iid NORM(0, ). Note: Simulated data here: 1 = 110, 2 = 110, 3 = 120, 4 = 100, and = 10. Data: Entered into c1 - c4 of MTB worksheet ("Unstacked" format). Data Display Row 1 2 3 4 5 6 7 8 9 10 D1 103 112 113 103 118 123 114 105 124 103 D2 111 126 117 101 99 131 108 114 103 128 D3 107 113 137 127 134 140 103 128 125 99 Pl 82 112 113 89 92 107 116 106 105 91 One-way ANOVA: Response versus Group Copyright © 2007 by Bruce E. Trumbo. All rights reserved. Department of Statistics and Biostatistics. California State University East Bay (Hayward). This is a draft. Corrections/Permissions: [email protected] 2 Source Group Error Total Level 1 2 3 4 DF 3 36 39 N 10 10 10 10 SS 2043 5001 7044 Mean 111.80 113.80 121.30 101.30 MS 681 139 F 4.90 StDev 8.15 11.54 14.74 11.80 P 0.006 Individual 95% CIs For Mean Based on Pooled StDev ------+---------+---------+---------+--(-------*------) (-------*------) (------*-------) (------*-------) ------+---------+---------+---------+--100 110 120 130 Pooled StDev = 11.79 Copyright © 2007 by Bruce E. Trumbo. All rights reserved. Department of Statistics and Biostatistics. California State University East Bay (Hayward). This is a draft. Corrections/Permissions: [email protected] 3 Problems: 1) Compute the ANOVA table for the original simulation by hand from the data table and compare your results with those from Minitab. The possible differences are due to rounding differently. 2) Make a column of the 40 residuals using Minitab. Verify several values. Residuals can be calculated and stored when running ANOVA in Minitab. See the handout for One-way Anova for detail. And check a few of the residuals by the definition of residual: ˆij yij y j . 3) Do several simulations with the same parameters as shown, except = 15. Comment on the results. Please refer to p. 185 of the textbook O/L for how to generate random data. Here is the brief version: Minitab: Calc>> random data>> normal and type the required number of observations as the number of rows and the required mean and standard deviation (15). Repeat this for each of the four groups. Then run Anova and see if we can detect the difference among groups. 4) How big would n (for each group) have to be in order for the F-test to have 95% power? Copyright © 2007 by Bruce E. Trumbo. All rights reserved. Department of Statistics and Biostatistics. California State University East Bay (Hayward). This is a draft. Corrections/Permissions: [email protected] 4 Minitab: Stat>>Power and sample size>> one-way anova and type in the required number of levels (groups) and other information and leave sample size blank. Power and Sample Size One-way ANOVA Alpha = 0.05 SS Means 200 Assumed standard deviation = 15 Number of Levels = 4 Sample Target Maximum Size Power Actual Power Difference 21 0.95 0.956730 20 When the maximum difference in group means is 20 and = 15, then sample size should be 21 for each group. 5) Do a simulation with the population means and variance of the original example. Also use n1 = n2 = n3 = 10, but use n4 = 15. Explain the entries in the DF column. Explain why the CIs for the group means have different lengths. The df of total changes (45-1=44) since the total sample size is increased to be 45. Therefore, the df of error term changes too: df of total – df of group = 44 - 3=41. The CIs for the D3 groups have the same width while the CI for the Placebo group is much narrower since the standard deviation of sample mean is /sqrt(n). The larger the sample size is, the narrower the CI is. Copyright © 2007 by Bruce E. Trumbo. All rights reserved. Department of Statistics and Biostatistics. California State University East Bay (Hayward). This is a draft. Corrections/Permissions: [email protected]