Ch. 12 The Analysis of Variance. can be modelled as normal random variables. come from populations having the same variance.

Transcription

1 Ch. 12 The Analysis of Variance Residual Analysis and Model Assessment Model Assumptions: The measurements X i,j can be modelled as normal random variables. are statistically independent. come from populations having the same variance. These assumptions can be checked by examining plots of residuals.

2 For the completely randomized design, the residuals are e i,j = x i,j x i for i = 1, 2,..., I, j = 1, 2,..., J. Compare: ε i,j = x i,j µ i e i,j = x i,j x i e i,j contains information about variability left unexplained by effects of the differences in the treatment groups.

3 If the model assumptions are reasonable, then the residuals should appear to be normally distributed should be close to statistically independent have a constant variance; the residual variance should not differ for different treatment groups

4 Residual Plots: Normal QQ Plot: Residuals should be scattered about a straight line. Run Chart: There should be no systematic pattern. Plot of Residuals Versus Fitted Values ( x i ): There should be no systematic pattern.

5 Remedial Measures: If the residuals are not close to normal, a data transformation might help A pattern on the run chart may be indicating that the measurements are not independent. Run order may be an important factor it should be included in the experimental design

6 If variability in the residuals is increasing with the treatment means, two common approaches are: 1. work on square root scale if treatment variances are approximately proportional to treatment means 2. work on log scale if treatment standard deviations are approximately proportional to treatment means

7 Example. An experiment was conducted to see if temperature had an effect on the number of surface flaws resulting after painting car exteriors. 24 cars were assigned randomly to each of 3 temperature groups (10 C, 20 C, 30 C) A boxplot of the number of surface flaws against temperature suggests that there might be a difference in mean number of flaws at the different temperatures.

8 Boxplot of Car Surface Defects Number of Surface Flaws temperature

9 The treatment means are x 1 = 4.875, x 2 = 1.375, and x 3 = Residuals: e 1,1 = = 1.875, etc. Can the analysis of variance be used? Check Residual Plots: 1. Are the residuals normally distributed? No they are discrete... but

10 QQ Plot of Car Finish Residuals Surface Flaw Residuals Theoretical Quantiles

11 This normal QQ-plot suggests no serious departure from normality 2. Are the residuals independent?

12 Residuals Residuals Time Time Residuals Time

13 Systematic patterns on the run charts are not evident 3. Are the residual variances constant?

14 Residuals vs Fitted Residuals Fitted values aov(formula = SurfaceFlaws ~ factor(temperature))

15 the treatment variance appears to increase with treatment mean Remedy: For count data, a square root transformation often works. Strategy: Take square root of all 24 defect counts. Compute treatment means on the square root scale: x 1 = 2.16, x 2 = 0.985, x 3 = 2.41

16 QQ Plot of Car Finish Residuals Surface Flaw Residuals Theoretical Quantiles Residuals vs Fitted Residuals

17 Normal probability plot of residuals indicates a departure from normality; ANOVA results will be somewhat conservative Plot of residuals versus treatment means indicates a constant variance. Proceed to ANOVA using measurements on a square root scale.

18 ANOVA Results: SST = 16.69, d.f. = 23 SSTr = 9.26, d.f. = 2 Ultimately, we find F = which is very large relative to the tabulated F values: e.g. f.001,2,21 = Thus, we conclude that the mean number of surface flaws is not always the same for the different temperatures.

19 Multiple Comparisons ANOVA leads to 1 of two possible conclusions: 1. Treatment group means are the same. 2. Treatment group means are different. If treatment means are the same, STOP. If treatment means are different, you may conduct a Post-Hoc study: 1. Which means are different? 2. Find a confidence interval for a linear combination of the true treatment means. Tukey s Procedure: gives simultaneous confidence intervals for all possible pairs of treatment means.

20 We can have (1 α)100% confidence that the true difference µ i µ k lies in the interval for each i, k = 1, 2,..., I. x i x k ± Q α,i,i(j 1) MSE/J The Q factor is tabulated in Appendix A.10.

21 Trusses Example. Find simultaneous 95% confidence intervals for all pairs of true treatment group means. First, recall that we demonstrated that mean axial stiffness index is different for trusses constructed with plates of different lengths, it makes sense to find out which treatment means are different.

22 Estimated mean axial stiffness for each of the plate lengths: (4 in.) 368 (6 in.) (8 in.) (10 in.) (12 in.) I = 5, J = 7, α =.05 Q.05,5,30 = 4.1 (m = 5, ν = 30) M SE: w = Q MSE/J = /7 = 50.2

23 The 95% confidence intervals for µ 1 µ i for i = 2, 3, 4, 5 are ± 50.2 = 34.8 ± ± 50.2 = 41.9 ± ± 50.2 = 74.2 ± ± 50.2 = 104 ± 50.2 For µ 2 µ i, i = 3, 4, 5, we have 7.1 ± ± ± 50.2

24 For µ 3 µ i, i = 4, 5, we have 32.3 ± ± 50.2 and µ 4 µ 5 is in 29.8 ± 50.2 Lots of intervals. In some cases, the signal is stronger than the noise. These are the significant differences at level α=.05 A quick way of seeing which treatment means are significantly different is to list the means in ascending order and identify all means that are within the margin of error (50.2) of each other.

25 Summary: to compute simultaneous confidence intervals for true treatment means, Tukey s method is as follows: 1. Obtain Q α,i,i(j 1) from Tukey s studentized range table 2. Compute w = Q MSE/J 3. List the treatment group means in ascending order and underline those which do not differ by more than w.

26 Confidence intervals for linear combinations of treatment group means e.g. I i=1 a i µ i µ 1 + µ 2 + µ 3 3 µ 4 + µ 5 2 Point estimator: p.e. = x 1 + x 2 + x 3 3 x 4 + x 5 2 Standard error: square root of σ 2 = MSE so 3σ 2 9J + 2σ2 4J

27 Estimated Standard error is s.e. = MSE 3 J A (1 α)100% confidence interval for this linear combination is then p.e. ± t α/2,i(j 1) s.e. Trusses Example. Compute p.e. and s.e. and a 95% confidence interval for the above linear combination.

28 10.3 Unbalanced Designs Unequal sample sizes for different treatment groups SST: numerator of the sample variance of all measurements SSE: sum of treatment group variance numerators SSTr = SST - SSE

29 Total d.f. = number of measurements - 1 Treatment d.f. = I - 1 Error d.f. = number of measurements - I ANOVA table and F -test: as in the balanced case slight change to multiple comparisons formula (see page 425)

30 Example The yield of tomatoes (kg/plot) for four different salinity levels (measured in terms of electrical conductivity EC) are given. EC levels = 1.6, 3.8, 6.0 and 10.2 nmhos/cm 1.6: : : : The scatterplot of the tomato yields for the various salinities suggests that there is a difference. It also indicates that the within treatment variance is constant.

31 Scatterplot of Tomato Data tomato yield (kg/plot) salinity (EC) level

32 Use the F test at level α =.05 to test for any differences in true average yield due to the different salinity levels.

33 SST = 581 The sample variances at each EC level are 13.0, 7.1, 5.9, and 8.4, giving us SSE = After filling out the ANOVA table, we find that the observed F value is f.05,3,14 = 3.34 so we conclude that there is a difference among the mean yields. Salinity level makes a difference.

34 Compute a 95% confidence interval for µ (µ µ µ 6 ).