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This analysis applies ANOVA and t-tests to evaluate the impact of different groups on physical activity and lung volume measures from laboratory data. The dataset includes active, passive, and control groups analyzed for their walking age outcomes. Further comparison of measurement methods for lung volume is performed, identifying significant differences in methods and subjects. The findings utilize R's statistical functions for linear models and ANOVA, providing insight into the effects of group classifications and measurement techniques.
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ANOVA Homework Solutions EPP 245/298 Statistical Analysis of Laboratory Data
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 $ctr.8w [1] 13.25 11.50 12.00 13.50 11.50 EPP 245 Statistical Analysis of Laboratory Data
> age.walk <- c(zelazo$active,zelazo$passive,zelazo$none,zelazo$ctr.8w) > group <- rep(c("active","passive","none","ctr.8w"),c(6,6,6,5)) > group <- as.factor(group) > group [1] active active active active active active passive passive passive [10] passive passive passive none none none none none none [19] ctr.8w ctr.8w ctr.8w ctr.8w ctr.8w Levels: active ctr.8w none passive > anova(lm(age.walk ~ group)) Analysis of Variance Table Response: age.walk Df Sum Sq Mean Sq F value Pr(>F) group 3 14.778 4.926 2.1422 0.1285 Residuals 19 43.690 2.299> plot(age.walk ~ group) EPP 245 Statistical Analysis of Laboratory Data
> mgroup <- rep(c("active","passive","none"),c(6,6,11)) > mgroup <- as.factor(mgroup) > anova(lm(age.walk ~ mgroup)) Analysis of Variance Table Response: age.walk Df Sum Sq Mean Sq F value Pr(>F) mgroup 2 13.655 6.827 3.0471 0.06996 . Residuals 20 44.812 2.241 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 > t.test(zelazo$active,c(zelazo$none,zelazo$ctr.8w)) Welch Two Sample t-test data: zelazo$active and c(zelazo$none, zelazo$ctr.8w) t = -2.6574, df = 9.327, p-value = 0.02539 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -3.4626053 -0.2873947 sample estimates: mean of x mean of y 10.125 12.000 EPP 245 Statistical Analysis of Laboratory Data
Exercise 6.2 • Lung data set has columns • volume = measured lung volume • method = method of measurement • subject = subject • Compare the methods. Are they different? Which ones differ? EPP 245 Statistical Analysis of Laboratory Data
> attach(lung) > lm(volume ~ method + subject) Call: lm(formula = volume ~ method + subject) Coefficients: (Intercept) methodB methodC subject2 subject3 subject4 3.17222 0.28333 0.60000 -0.83333 0.10000 -0.06667 subject5 subject6 -0.03333 -0.60000 > lung.lm <- lm(volume ~ method + subject) > anova(lung.lm) Analysis of Variance Table Response: volume Df Sum Sq Mean Sq F value Pr(>F) method 2 1.08111 0.54056 6.4953 0.01557 * subject 5 2.18278 0.43656 5.2457 0.01271 * Residuals 10 0.83222 0.08322 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 EPP 245 Statistical Analysis of Laboratory Data
> tapply(volume,method,mean) A B C 2.933333 3.216667 3.533333 > tapply(volume,subject,mean) 1 2 3 4 5 6 3.466667 2.633333 3.566667 3.400000 3.433333 2.866667 > diff(sort(tapply(volume,method,mean))) B C 0.2833333 0.3166667 > plot(volume ~ method) > plot(lung.lm) Hit <Return> to see next plot: Hit <Return> to see next plot: Hit <Return> to see next plot: Hit <Return> to see next plot: > help(plot.lm) > plot(lung.lm$resid ~ method) EPP 245 Statistical Analysis of Laboratory Data
> 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 EPP 245 Statistical Analysis of Laboratory Data