Regression models, R solution day7

Transcription

1 Regression models, R solution day7 Exercise 1 In this exercise, we shall look at the differences in vitamin D status for women in 4 European countries Read and prepare the data: vit <- read.table(" header=t) vit$country <- factor(vit$country, levels=c(1,2,4,6), labels=c("denmark","finland","ireland","poland")) vit$sunexp <- factor(vit$sunexp, levels=c(1,2,3), labels=c("avoid sun","sometimes in sun","prefer sun")) vit$logvitd <- log(vit$vitd) vit$logintake <- log(vit$vitdintake) 1. Compare the vitamin D level (on an appropriate scale) for the four countries. m1 <- glm(vitd~country, data=vit) par(mfrow=c(1,2)) plot(m1, which=1:2) Residuals vs Fitted Normal Q Q Residuals Std. deviance resid Predicted values Theoretical Quantiles From the residual plot it looks like the residual variance differ between the countries. m1log <- glm(logvitd~country, data=vit) par(mfrow=c(1,2)) plot(m1log, which=1:2) 1

2 Residuals vs Fitted Normal Q Q Residuals Std. deviance resid Predicted values Theoretical Quantiles With the log-transformed vitd, the residual variances are similar and we prefer this though the Q-Q plot looks slightly worse due to a few data points. However, if is not very clear which model to choose, and if you think it makes more sense to present the estimates as absolute differences instead of relative, you could choose the untransformed vitd. We test whether the vitamin D level is the same in all four countries: fit <- aov(logvitd~country, data=vit) summary(fit) Df Sum Sq Mean Sq F value Pr(>F) country e-05 *** Residuals Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 We use the Tukey post-hoc test to get all pairwise comparisons: testt <- TukeyHSD(fit, 'country') exp(testt$country) diff lwr upr p adj Finland-Denmark Ireland-Denmark Poland-Denmark Ireland-Finland Poland-Finland Poland-Ireland There is a significant difference in vitamin D levels between Poland and both Denmark, Finland, and Ireland. The residual variation can be extracted from an lm-summary: summary(lm(logvitd~country, data=vit))$sigma 2

3 [1] Draw a model diagram for the situation, including now the covariate BMI, and include the effect of BMI in the analysis. Country may affect both BMI and Vitamin D level. BMI may affect Vitamin D level. m2 <- glm(logvitd~country+bmi, data=vit) summary(m2) Call: glm(formula = logvitd ~ country + bmi, data = vit) Deviance Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) < 2e-16 *** countryfinland countryireland countrypoland ** bmi ** --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for gaussian family taken to be ) Null deviance: on 212 degrees of freedom Residual deviance: on 208 degrees of freedom AIC: Number of Fisher Scoring iterations: 2 The effect of BMI on vitamin D is significant (p=0.001) when adjusting for country. The residual variation can also be extracted as the square root of the dispersion parameter, and is now slightly lower than for the model with only country: sqrt(summary(m2)$dispersion) [1] The vitamin D level in Poland is 76.9% of the vitamin D level in Demark, which is slightly higher than without adjustment for BMI but still a significant difference (if we a priori have specified that we want to compare only Denmark and Poland, we need not to adjust the confidence intervals for multiple testing): c( Ratio = exp(coef(summary(m2))[4,1]), CI = exp(confint(m2)[4,]) ) Ratio CI.2.5 % CI.97.5 %

4 3. Could any of the covariates (sun exposure habits and vitamin D intake) be confounders for the comparison of countries? Depends on the scientific question Do they appear to be unequally distributed among the countries? The distribution of sun exposure habits in the four countries (addmargins() does as the name indicates: add the sums/totals to the table, with margin specifying whether it should be row sums, column sums, or both): t <- with(vit, table(country,sunexp)) addmargins(t) sunexp country avoid sun sometimes in sun prefer sun Sum Denmark Finland Ireland Poland Sum tprop <- with(vit, round(prop.table(table(country,sunexp),1),3) ) addmargins(tprop, margin=2) sunexp country avoid sun sometimes in sun prefer sun Sum Denmark Finland Ireland Poland chisq.test(table(vit$country,vit$sunexp)) Pearson's Chi-squared test data: table(vit$country, vit$sunexp) X-squared = , df = 6, p-value = The chi-square test shows that sun exposure habits differ significantly between the countries. The distribution of log vitamin D intake in the four countries: summary(vit$logintake[vit$country=="denmark"]) Min. 1st Qu. Median Mean 3rd Qu. Max summary(vit$logintake[vit$country=="finland"]) Min. 1st Qu. Median Mean 3rd Qu. Max summary(vit$logintake[vit$country=="ireland"]) 4

5 Min. 1st Qu. Median Mean 3rd Qu. Max summary(vit$logintake[vit$country=="poland"]) Min. 1st Qu. Median Mean 3rd Qu. Max m4 <- glm(logintake~country, data=vit) anova(m4, test="f") Analysis of Deviance Table Model: gaussian, link: identity Response: logintake Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev F Pr(>F) NULL country e-05 *** --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 The anova shows that vitamin D intake cannot be assumed to be the same in the four countries. 5. Why would age not be expected to be an important confounder for the country comparison? The estimated differences between countries can be afffected by adjustment for age only if the age distribution differs between the countries, which it does not: tapply(vit$age,vit$country,mean) Denmark Finland Ireland Poland Perform an analysis including sunexp and intake, and comment on your findings m6 <- glm(logvitd~country+bmi+logintake+sunexp, data=vit) library(car) Anova(m6, type="iii", test.statistic="f") Analysis of Deviance Table (Type III tests) Response: logvitd Error estimate based on Pearson residuals SS Df F Pr(>F) country ** bmi ** 5

6 logintake e-11 *** sunexp Residuals Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 logintake is significant, but sunexp is not. c( Ratio = exp(coef(summary(m6))[4,1]), CI = exp(confint(m6)[4,]) ) Ratio CI.2.5 % CI.97.5 % The vitamin D level in Poland is 80.2% of the level in Denmark (slightly higher than when only adjusted for bmi). The residual variation is even lower now with more variables added to the model: sqrt( summary(m6)$dispersion ) [1] Discuss possible interactions between the covariates Interaction between country and sun exposure habits (if we think the attitude towards sun bathing -and the strength of the sun- differ between the countries): m71 <- glm(logvitd~ country*sunexp + bmi + logintake, data=vit) anova(m6,m71, test="f") Analysis of Deviance Table Model 1: logvitd ~ country + bmi + logintake + sunexp Model 2: logvitd ~ country * sunexp + bmi + logintake Resid. Df Resid. Dev Df Deviance F Pr(>F) * --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 summary(m71) Call: glm(formula = logvitd ~ country * sunexp + bmi + logintake, data = vit) Deviance Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value (Intercept) countryfinland countryireland

7 countrypoland sunexpsometimes in sun sunexpprefer sun bmi logintake countryfinland:sunexpsometimes in sun countryireland:sunexpsometimes in sun countrypoland:sunexpsometimes in sun countryfinland:sunexpprefer sun countryireland:sunexpprefer sun countrypoland:sunexpprefer sun Pr(> t ) (Intercept) < 2e-16 *** countryfinland countryireland * countrypoland sunexpsometimes in sun sunexpprefer sun bmi ** logintake 2.45e-12 *** countryfinland:sunexpsometimes in sun countryireland:sunexpsometimes in sun countrypoland:sunexpsometimes in sun countryfinland:sunexpprefer sun countryireland:sunexpprefer sun * countrypoland:sunexpprefer sun Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for gaussian family taken to be ) Null deviance: on 212 degrees of freedom Residual deviance: on 199 degrees of freedom AIC: Number of Fisher Scoring iterations: 2 Interaction between country and vitamin D intake (if we think the diet/vitamin D sources differs between the countries): m72 <- glm(logvitd~ country*logintake + bmi + sunexp, data=vit) anova(m6,m72, test="f") Analysis of Deviance Table Model 1: logvitd ~ country + bmi + logintake + sunexp Model 2: logvitd ~ country * logintake + bmi + sunexp Resid. Df Resid. Dev Df Deviance F Pr(>F) ** --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 summary(m72) 7

8 Call: glm(formula = logvitd ~ country * logintake + bmi + sunexp, data = vit) Deviance Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) < 2e-16 *** countryfinland countryireland *** countrypoland logintake e-11 *** bmi ** sunexpsometimes in sun sunexpprefer sun countryfinland:logintake * countryireland:logintake *** countrypoland:logintake * --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for gaussian family taken to be ) Null deviance: on 212 degrees of freedom Residual deviance: on 202 degrees of freedom AIC: Number of Fisher Scoring iterations: 2 8. Make appropriate residual plots and diagnostic plots to make sure that the analyses are reasonable m8 <- glm(logvitd~ country + bmi + sunexp + logintake + country:logintake + country:sunexp, data=vit) Anova(m8, type="iii", test.statistic="f") Analysis of Deviance Table (Type III tests) Response: logvitd Error estimate based on Pearson residuals SS Df F Pr(>F) country *** bmi ** sunexp logintake e-11 *** country:logintake ** country:sunexp * Residuals Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 8

9 plot(m8, which=1:2) Residuals vs Fitted Residuals Predicted values glm(logvitd ~ country + bmi + sunexp + logintake + country:logintake + coun... Normal Q Q 35 Std. deviance resid Theoretical Quantiles glm(logvitd ~ country + bmi + sunexp + logintake + country:logintake + coun Consider a Danish woman of median height (e.g., 165 cm), usually trying to avoid the sun and with a log(intake) value of, e.g How much different is the predicted vitamin D level for similar women who either: a) weigh 10 kg less?, b) have a 25% higher vitamin D intake?, c) sunbathe?, d) live in Ireland? 9

10 We keep reference level Denmark and avoid sun and creates two new variables: bmi_10kg165cm (1 unit increase corresponds to a 10 kg difference) and logintakeper25pct_1_1 (equal to 0 when log(intake=1.1) and 1 unit increase corresponds to 25% increase): vit$bmi_10kg165cm <- vit$bmi/(10/(1.65^2)) vit$logintakeper25pct_1_1 <- (vit$logintake-1.1)/log(1.25) m9 <- lm(logvitd~ country + bmi_10kg165cm + sunexp + logintakeper25pct_1_1 + country:logintakeper25pct_1_1 + country:sunexp, data=vit) summary(m9) Call: lm(formula = logvitd ~ country + bmi_10kg165cm + sunexp + logintakeper25pct_1_1 + country:logintakeper25pct_1_1 + country:sunexp, data = vit) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value (Intercept) countryfinland countryireland countrypoland bmi_10kg165cm sunexpsometimes in sun sunexpprefer sun logintakeper25pct_1_ countryfinland:logintakeper25pct_1_ countryireland:logintakeper25pct_1_ countrypoland:logintakeper25pct_1_ countryfinland:sunexpsometimes in sun countryireland:sunexpsometimes in sun countrypoland:sunexpsometimes in sun countryfinland:sunexpprefer sun countryireland:sunexpprefer sun countrypoland:sunexpprefer sun Pr(> t ) (Intercept) < 2e-16 *** countryfinland countryireland ** countrypoland bmi_10kg165cm ** sunexpsometimes in sun sunexpprefer sun logintakeper25pct_1_1 2.53e-11 *** countryfinland:logintakeper25pct_1_ ** countryireland:logintakeper25pct_1_ ** countrypoland:logintakeper25pct_1_ * countryfinland:sunexpsometimes in sun countryireland:sunexpsometimes in sun countrypoland:sunexpsometimes in sun countryfinland:sunexpprefer sun countryireland:sunexpprefer sun * 10

11 countrypoland:sunexpprefer sun Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: on 196 degrees of freedom Multiple R-squared: 0.421, Adjusted R-squared: F-statistic: on 16 and 196 DF, p-value: 3.247e-16 a) corresponds to a change in logvitd of , b) corresponds to a change in logvitd of , c) corresponds to a change in logvitd of 0.151, and d) corresponds to a change in logvitd of Hence, the Irish woman has the highest predicted serum vitamin D compared to a Danish woman. 9. alternative Another way of solving this exercise is to use the glht()-function multcomp-package where the linfct should be a matrix where the order (only one row!) follows the order of coefficients in the output from the model (hence, 1 intercept 2:4 country, 5 bmi, 6:7 sunexp, 8 logintake, 9:11 country:logintake, 12:17) country:sunexp). library(multcomp) a) weigh 10 kg less? A weight reduction of 10 kg: 10/(1.65ˆ2)=3.67 K1 <- matrix(c(0,0,0,0,-3.67,0,0,0,0,0,0,0,0,0,0,0,0),1) t1 <- glht(m8, linfct = K1) summary(t1) Simultaneous Tests for General Linear Hypotheses Fit: glm(formula = logvitd ~ country + bmi + sunexp + logintake + country:logintake + country:sunexp, data = vit) Linear Hypotheses: Estimate Std. Error z value Pr(> z ) 1 == ** --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Adjusted p values reported -- single-step method) b) have a 25% higher vitamin D intake? An increase in intake of 25%: log(1.25)=0.223 K2 <- matrix(c(0,0,0,0,0,0,0,0.223,0,0,0,0,0,0,0,0,0), 1) t2 <- glht(m8, linfct = K2) summary(t2) Simultaneous Tests for General Linear Hypotheses Fit: glm(formula = logvitd ~ country + bmi + sunexp + logintake + country:logintake + country:sunexp, data = vit) Linear Hypotheses: Estimate Std. Error z value Pr(> z ) 1 == e-12 *** --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 11

12 (Adjusted p values reported -- single-step method) c) sunbathe? K3 <- matrix(c(0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0), 1) t3 <- glht(m8, linfct = K3) summary(t3) Simultaneous Tests for General Linear Hypotheses Fit: glm(formula = logvitd ~ country + bmi + sunexp + logintake + country:logintake + country:sunexp, data = vit) Linear Hypotheses: Estimate Std. Error z value Pr(> z ) 1 == (Adjusted p values reported -- single-step method) d) live in Ireland? K4 <- matrix(c(0,0,1,0,0,0,0,0,0,1.1,0,0,0,0,0,0,0), 1) t4 <- glht(m8, linfct = K4) summary(t4) Simultaneous Tests for General Linear Hypotheses Fit: glm(formula = logvitd ~ country + bmi + sunexp + logintake + country:logintake + country:sunexp, data = vit) Linear Hypotheses: Estimate Std. Error z value Pr(> z ) 1 == ** --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Adjusted p values reported -- single-step method) 12