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ANOVA Homework Solutions

EPP 245/298

Statistical Analysis of

Laboratory Data

November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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Exercise 6.1

> library(ISwR)Loading required package: survival Loading required package: splines > data(zelazo)> zelazo$active[1] 9.00 9.50 9.75 10.00 13.00 9.50

$passive[1] 11.00 10.00 10.00 11.75 10.50 15.00

$none[1] 11.50 12.00 9.00 11.50 13.25 13.00

$ctr.8w[1] 13.25 11.50 12.00 13.50 11.50

November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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> age.walk <- c(zelazo$active,zelazo$passive,zelazo$none,zelazo$ctr.8w)> group <- rep(c("active","passive","none","ctr.8w"),c(6,6,6,5))> group <- as.factor(group)> group [1] active active active active active active passive passive passive[10] passive passive passive none none none none none none [19] ctr.8w ctr.8w ctr.8w ctr.8w ctr.8w Levels: active ctr.8w none passive> anova(lm(age.walk ~ group))Analysis of Variance Table

Response: age.walk Df Sum Sq Mean Sq F value Pr(>F)group 3 14.778 4.926 2.1422 0.1285Residuals 19 43.690 2.299> plot(age.walk ~ group)

November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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> mgroup <- rep(c("active","passive","none"),c(6,6,11))> mgroup <- as.factor(mgroup) > anova(lm(age.walk ~ mgroup))Analysis of Variance Table

Response: age.walk Df Sum Sq Mean Sq F value Pr(>F) mgroup 2 13.655 6.827 3.0471 0.06996 .Residuals 20 44.812 2.241 ---Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1> t.test(zelazo$active,c(zelazo$none,zelazo$ctr.8w))

Welch Two Sample t-test

data: zelazo$active and c(zelazo$none, zelazo$ctr.8w) t = -2.6574, df = 9.327, p-value = 0.02539alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -3.4626053 -0.2873947 sample estimates:mean of x mean of y 10.125 12.000

November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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Exercise 6.2

• Lung data set has columns– volume = measured lung volume– method = method of measurement– subject = subject

• Compare the methods. Are they different? Which ones differ?

November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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> attach(lung)> lm(volume ~ method + subject)

Call:lm(formula = volume ~ method + subject)

Coefficients:(Intercept) methodB methodC subject2 subject3 subject4 3.17222 0.28333 0.60000 -0.83333 0.10000 -0.06667 subject5 subject6 -0.03333 -0.60000

> lung.lm <- lm(volume ~ method + subject)> anova(lung.lm)Analysis of Variance Table

Response: volume Df Sum Sq Mean Sq F value Pr(>F) method 2 1.08111 0.54056 6.4953 0.01557 *subject 5 2.18278 0.43656 5.2457 0.01271 *Residuals 10 0.83222 0.08322 ---Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1

November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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> tapply(volume,method,mean) A B C 2.933333 3.216667 3.533333 > tapply(volume,subject,mean) 1 2 3 4 5 6 3.466667 2.633333 3.566667 3.400000 3.433333 2.866667 > diff(sort(tapply(volume,method,mean))) B C 0.2833333 0.3166667

> plot(volume ~ method)> plot(lung.lm)Hit <Return> to see next plot: Hit <Return> to see next plot: Hit <Return> to see next plot: Hit <Return> to see next plot: > help(plot.lm)> plot(lung.lm$resid ~ method)

November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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November 10, 2004 EPP 245 Statistical Analysis of Laboratory Data

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LSD (.975, ) MSE(2 / )

(.975,10) (0.0832)(2 / 6)

2.228 (0.0832)(2 / )

0.3710

t df n

t

n

> anova(lung.lm)Analysis of Variance Table

Response: volume Df Sum Sq Mean Sq F value Pr(>F) method 2 1.08111 0.54056 6.4953 0.01557 *subject 5 2.18278 0.43656 5.2457 0.01271 *Residuals 10 0.83222 0.08322 > diff(sort(tapply(volume,method,mean))) B C 0.2833333 0.3166667