Lecture 11 Analysis of variance

Transcription

1 Lecture 11 Analysis of variance Dr. Wim P. Krijnen Lecturer Statistics University of Groningen Faculty of Mathematics and Natural Sciences Johann Bernoulli Institute for Mathematics and Computer Science October 18, 2010

2 Lecture overview one-way, two-way analysis of variance fixed effects; skip random effects parametric as well as nonparametric inference paired group comparisons correction for multiple testing purpose: finding evidence for valid statistical inferences distinguish: assumptions, hypotheses, conclusions order of approach: first concentrate on methods, than on data data preparation in the book is useful, but initially confusing Why not repeated t-testing? 1. ANOVA has more power! 2. Avoid multiple testing 2

3 Example: consumed gasoline by pickup trucks gas consumption in gallons of 5 pickup trucks driving 500 miles from manufactures Chevrolet, Dogde, and Ford Chevy Dodge Ford Question: Are the means significantly different? H 0 : µ = µ 1 = µ 2 = µ 3 versus H A : not all mu s equal Population sample description y ij Y ij observation j in group i µ Y total mean µ i = µ + α i Y i mean for group i (identification i α i = 0)) H 0 : µ = µ 1 = µ 2 = µ 3 H 0 : α 1 = α 2 = α 3 = 0 equals means equivalent to no effects 3

4 Computing means and variances organize the data in a matrix and use apply Chevy <- c(15.2,15.4,14.8,14.4,14.7) Dodge <- c(14.8,14.4,14.3,14.1,14.4) Ford <- c(15.1,14.3,14.6,13.9,14.6) truck.ma <- cbind(chevy,dodge,ford) > apply(truck.ma,2,mean) Chevy Dodge Ford > apply(truck.ma,2,var) Chevy Dodge Ford

5 Decomposition of sums of squares assume the model: y ij = µ + α i + ε ij, where ε ij iid N(0, σ 2 ) normally distributed for all ij; equal group variances Y ij observation j in group i n i number of observations in group i k number of groups Y ij Y = k n i (Y ij Y ) 2 i=1 j=1 } {{ } Total SS Y ij Y i }{{} within group deviation = k n i (Y ij Y i ) 2 i=1 j=1 } {{ } Within SS + Y i Y }{{} between group deviation + k n i (Y i Y ) 2 i=1 Total SS = Within SS + Between SS j=1 } {{ } Between SS 5

6 F-test of ANOVA WithinMS = 1 n k k n i (Y ij Y i ) 2 BetweenMS = 1 k 1 i=1 j=1 Between MS F = Within MS H 0 : µ 1 = µ 2 = = µ k ; equal group means in population k n i (Y i Y ) 2 i=1 j=1 reject H 0 if p-value = P(F k 1,n k > F) = 1-pf(F,k-1,n-k) < α large population difference in means generates differences in sample means with high probability large sample differences in means large Between MS large F-value small p-value general idea: make inferences about means by analysis of variance (in the F!) 6

7 Elementary computation of sums of squares Chevy <- c(15.2,15.4,14.8,14.4,14.7) Dodge <- c(14.8,14.4,14.3,14.1,14.4) Ford <- c(15.1,14.3,14.6,13.9,14.6) k <- 3; ni <- 5; n <- 15 truck.ma <- cbind(chevy,dodge,ford) Total.mean <- sum(truck.ma)/n Within.SS <- 0; Between.SS <- 0 for (i in 1:k) Within.SS <- Within.SS + sum((truck.ma[,i] - mean(truck.ma[,i]))ˆ2) for (i in 1:k) Between.SS <- Between.SS + ni * (mean(truck.ma[,i]) - Total.mean)ˆ2 > (Within.MS <- Within.SS/(n-k)) [1] 0.14 > (Between.MS <- Between.SS/(k-1)) [1] 0.35 > (F <- Between.MS/Within.MS)# gives 2.5 > 1-pf(F,k-1,n-k) [1] #not reject H0

8 Example: F-test for gasoline consumption data > Chevy <- c(15.2,15.4,14.8,14.4,14.7) > Dodge <- c(14.8,14.4,14.3,14.1,14.4) > Ford <- c(15.1,14.3,14.6,13.9,14.6) > truck.dat <- c(chevy,dodge,ford) > truck.fac <- gl(3,5) > summary(aov(truck.dat truck.fac)) # or > anova(lm(truck.dat truck.fac)) Analysis of Variance Table Response: truck.dat Df Sum Sq Mean Sq F value Pr(>F) truck.fac Residuals k = 3, n = 15, Between SS = 0.70, Between MS = 0.70/2 = 0.35, Within SS = 1.68, Within MS = 1.68/12 =.14, F =.35/0.14 = 2.5, p-value =.1237 > α not reject H 0 8

9 Extract predictor matrix by model.matrix illustrate by an example: extract the predictor matrix > (X <- model.matrix(model)) (Intercept) truck.fac2 truck.fac

10 Matrix computation of predicted values perform regression analysis wrt X and compute ŷ i by two methods X <- model.matrix(model) y <- c(chevy,dodge,ford) betahat <- (solve(t(x) %*% X)) %*% t(x) %*% y yhat <- X %*% betahat > sqrt(sum((fitted(model)-yhat)ˆ2)) [1] e-14 euclidian distance between ŷ i from lm and regression analysis 0 10

11 Summary of model estimation > summary(lm(truck.dat truck.fac)) Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) <2e-16 *** truck.fac truck.fac Residual standard error: on 12 degrees of freedo Multiple R-squared: , Adjusted R-squared: F-statistic: 2.5 on 2 and 12 DF, p-value: estimated effects α 1 =.9 (not printed), α 2 =.5, α 3 =.4 standard error follows from least squares t-value from estimate divided by standard error p-value from R function pt Conclusion: do not reject H 0 : µ 1 = µ 2 = µ 3 ; α 1 = α 2 = α 3 = 0 11

12 Testing validity of assumptions model implies error y ij µ + α i = ε ij (iid) N(0, σ 2 ) homoscedasticity assumption: equal group variances Bartlett s procedure tests H 0 : σ 2 = σ 2 1 = σ2 2 = σ2 3 versus H A : not all group variances equal Shapiro-Wilk procedure tests for normality of ε ij > model <- lm(truck.dat truck.fac) > shapiro.test(residuals(model)) Shapiro-Wilk normality test data: residuals(model) W = , p-value = > bartlett.test(truck.dat truck.fac) Bartlett test of homogeneity of variances data: truck.dat by truck.fac Bartlett s K-squared = , df = 2, p-value = Conclusion: Do not reject normality, homoscedaticity 12

13 Paired Comparisons test the difference between experimental effects H 0 : α i = α j versus H A : α i α j ( ( ) under H 0 Y i Y j has density φ 0, σ ) 1ni 2 + 1nj substitute estimator S 2 for σ 2 in the test testistic T ij = Y i Y j SE(Y i Y j ) = Y i Y j S 2 ( 1ni + 1nj ) reject H 0 if p-value = P(t n k > T ij ) = 1 pt(abs(tij),n-k) < α/2 13

14 Example: blood coagulation times Example: coagulation blood coagulation times from 24 animals receiving 4 different diets Box, Hunter& Hunter(1978) > library(faraway) > data(coagulation) > anova(lm(coag diet, data=coagulation)) Analysis of Variance Table Response: coag Df Sum Sq Mean Sq F value Pr(>F) diet e-05 *** Residuals Conclusion: There are difference in means Question: Which? 14

15 Answer: Detect Least Significant Differences library(agricolae); library(faraway) model <- aov(coag diet, data=coagulation) df <- df.residual(model) MS.error <- deviance(model)/df > LSD.test(coag,diet,df,MS.error,group=FALSE) Difference pvalue sig LCL UCL B - A ** C - A *** A - D C - B B - D *** C - D *** Conclusion: starred differences are significant (equality rejected) 15

16 Correction for Multiple testing we just made ( 4 2 ) = 6 paired comparisons probability of false positive is binomially distributed with π = 0.05, n = 6 P(X 1) = sum(dbinom(1:6, 6, 0.05)) = > 0.05 Solution 1: adjust alpha α = α ( 4 2 ) Solution 2: multiply raw p-values by ( 4 2 ); call them adjusted > LSD.test(coag,diet,df,MS.error,group=FALSE, p.adj="bonferroni") Difference pvalue sig LCL UCL B - A * C - A ** A - D C - B B - D ** C - D *** > * 6 [1]

17 Tuckey s honest significant differences (HSD) Example: > TukeyHSD(aov(coag diet, data=coagulation)) Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = coag diet, data = coagulation) $diet diff lwr upr p adj B-A e C-A e D-A e C-B e D-B e D-C e

18 Conclusion: 95% Confidence Intervals not containing zero indicate significant differences (H 0 rejected) 18

19 Kruskal-Wallis test compute the ranks of all observations and R i the sum of the ranks per treatment if all observations are different (no ties), then ( ) χ 12 k Ri 2 = 3(n 1) n(n + 1) n i i=1 reject H0: mean ranks are equal among groups if p-value = 1-pf(chi,k-1) < α good power properties (Lehmann, 1998, Nonparametrics ) > kruskal.test(truck.dat truck.fac) Kruskal-Wallis rank sum test data: truck.dat by truck.fac Kruskal-Wallis chi-squared = , df = 2, p-value =

20 Air Quality in New York Daily air quality measurements in New York, May to September Ozone= Mean ozone in parts per billion from 1300 to 1500 hours at Roosevelt Island > model <- lm(ozone Month, data = airquality) > shapiro.test(residuals(model)) Shapiro-Wilk normality test data: residuals(model) W = , p-value = 1.022e-07 # reject normality > bartlett.test(ozone Month, data = airquality) Bartlett test of homogeneity of variances data: Ozone by Month Bartlett s K-squared = , df = 4, p-value = # reject homoscedasticity > kruskal.test(ozone Month, data = airquality) Kruskal-Wallis rank sum test data: Ozone by Month Kruskal-Wallis chi-squared = , df = 4, p-value = 6.901e-06 #reject equal distribution

21 Two-way analysis of variance, fixed effects two factors α and β y ijk = µ + α i + β j + γ ij + ε ijk y ijk measurement of case k, level i factor α, level j factor β µ overall population mean α i effect of level i of factor α ( i α i = 0) β j effect of level j of factor β ( j β j = 0) γ ij interaction effect between level i of α, level j of β ( i j γ ij = 0) ε ijk error (iid error normal mean 0, var σ 2 ) model formula in R y ij = µ + α i + ε ij y a y ijk = µ + α i + β j + ε ijk y a + b y ijk = µ + α i + β j + γ ij + ε ijk y a + b + a : b y ijk = µ + α i + β j + γ ij + ε ijk y a * b H 0 : α 1 = α 2 = α 3 = 0; H 0 : β 1 = β 2 = β 3 = 0; H 0 : γ 1 = γ 2 = γ 3 = 0 21

22 Weight gain due to diet weight gain (gr.) in rats due to different types of diets two factors investigated (Hand, et al., 1993): source of protein: beef, cereal type of diet : low, high amount of protein Beef Cereal Low High Low High

23 > library(hsaur) > anova(lm(weightgain type * source, data=weightgain)) Analysis of Variance Table Response: weightgain Df Sum Sq Mean Sq F value Pr(>F) type * source type:source Residuals pmrt: population mean, main effects type, main effect source, interactioneffect Residuals SS = Within SS = , Within MS = /36 = , Between type SS = , df = nr levers -1 = 1, Between MS = /1, F = / = , p-value = < α Conclusion: Reject type of diet H 0 ; not reject source H 0 23