Week 8 Practice Problems

Author

Affiliation

Alex Kaizer

University of Colorado-Anschutz Medical Campus

This page includes optional practice problems, many of which are structured to assist you on the homework with Solutions provided on a separate page. Data sets, if needed, are provided on the BIOS 6618 Canvas page for students registered for the course.

This week’s extra practice exercises are focusing on examining the diagnostic plots for various regression models simulated to violate various regression assumptions. The interpretation of models with transformations are also explored for further practice.

Exercise 1: Diagnostic Examination

For each of the following sets of plots, identify which, if any, assumptions may be violated for the simple linear regression model.

1a. Scenario 1

1b. Scenario 2

1c. Scenario 3

1d. Scenario 4

1e. Scenario 5

Exercise 2: Data Transformation Interpretations

We will use the blood storage dataset from our course site:

Code

dat <- read.csv('../../.data/Blood_Storage.csv')

The dataset includes 316 men who had undergone radical prostatectomy and received transfusion during or within 30 days of the surgical procedure and had available PSA follow-up data .

For each of the following models, what are the estimates of the intercept and slope and how would you interpret them? If possible, for transformations of the dependent variable, $Y$, also transform the estimate of your slope and its 95% confidence interval back to its non-transformed scale and interpret.

2a. $Y$, $X$

Code

summary(lm(PVol ~ Age, data=dat))


Call:
lm(formula = PVol ~ Age, data = dat)

Residuals:
    Min      1Q  Median      3Q     Max 
-39.527 -15.817  -4.367   6.451 205.105 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) -13.3823    14.1790  -0.944    0.346    
Age           1.1443     0.2308   4.959 1.18e-06 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 29.1 on 305 degrees of freedom
  (9 observations deleted due to missingness)
Multiple R-squared:  0.07461,   Adjusted R-squared:  0.07158 
F-statistic: 24.59 on 1 and 305 DF,  p-value: 1.178e-06

2b. $log(Y)$, $X$

Code

summary(lm(log(PVol) ~ Age, data=dat))


Call:
lm(formula = log(PVol) ~ Age, data = dat)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.85316 -0.25833 -0.02639  0.18992  1.56272 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  2.83345    0.18124   15.63  < 2e-16 ***
Age          0.01820    0.00295    6.17 2.17e-09 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.3719 on 305 degrees of freedom
  (9 observations deleted due to missingness)
Multiple R-squared:  0.111, Adjusted R-squared:  0.108 
F-statistic: 38.07 on 1 and 305 DF,  p-value: 2.169e-09

2c. $\sqrt{Y}$, $X$

Code

summary(lm(sqrt(PVol) ~ Age, data=dat))


Call:
lm(formula = sqrt(PVol) ~ Age, data = dat)

Residuals:
    Min      1Q  Median      3Q     Max 
-2.6025 -0.9659 -0.1915  0.5813  8.4549 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  3.05356    0.75625   4.038 6.83e-05 ***
Age          0.07016    0.01231   5.700 2.82e-08 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.552 on 305 degrees of freedom
  (9 observations deleted due to missingness)
Multiple R-squared:  0.09628,   Adjusted R-squared:  0.09332 
F-statistic:  32.5 on 1 and 305 DF,  p-value: 2.819e-08

2d. $Y$, $log(X)$

Code

summary(lm(PVol ~ log(Age), data=dat))


Call:
lm(formula = PVol ~ log(Age), data = dat)

Residuals:
    Min      1Q  Median      3Q     Max 
-37.901 -15.970  -4.164   6.732 206.188 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  -215.75      55.63  -3.878 0.000129 ***
log(Age)       66.33      13.55   4.895 1.59e-06 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 29.13 on 305 degrees of freedom
  (9 observations deleted due to missingness)
Multiple R-squared:  0.07285,   Adjusted R-squared:  0.06981 
F-statistic: 23.97 on 1 and 305 DF,  p-value: 1.593e-06

2e. $log(Y)$, $log(X)$

Code

summary(lm(log(PVol) ~ log(Age), data=dat))


Call:
lm(formula = log(PVol) ~ log(Age), data = dat)

Residuals:
    Min      1Q  Median      3Q     Max 
-0.8618 -0.2634 -0.0178  0.1913  1.5541 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  -0.4363     0.7104  -0.614     0.54    
log(Age)      1.0673     0.1730   6.169 2.18e-09 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.3719 on 305 degrees of freedom
  (9 observations deleted due to missingness)
Multiple R-squared:  0.1109,    Adjusted R-squared:  0.108 
F-statistic: 38.06 on 1 and 305 DF,  p-value: 2.181e-09

--- title: "Week 8 Practice Problems" author: name: Alex Kaizer roles: "Instructor" affiliation: University of Colorado-Anschutz Medical Campus toc: true toc_float: true toc-location: left format: html: code-fold: show code-overflow: wrap code-tools: true --- ```{r, echo=F, message=F, warning=F} library(kableExtra) library(dplyr) ``` This page includes optional practice problems, many of which are structured to assist you on the homework with [Solutions provided on a separate page](/labs/prac8s/index.qmd). Data sets, if needed, are provided on the BIOS 6618 Canvas page for students registered for the course. This week's extra practice exercises are focusing on examining the diagnostic plots for various regression models simulated to violate various regression assumptions. The interpretation of models with transformations are also explored for further practice. # Exercise 1: Diagnostic Examination For each of the following sets of plots, identify which, if any, assumptions may be violated for the simple linear regression model. ## 1a. Scenario 1 ```{r, echo=F} # Code to generate figures for violation of linearity set.seed(303) x <- rnorm(n=100, mean=10, sd=3) # simulate a single continuous predictor with mean=10 and SD=3 reg <- 15 + 5 * x - 1 * x^2 - 0.25 * x^3 # set the regression equation so the intercept=10 and the slope=3 y <- rnorm(n=100, mean=reg, sd=8) # simulate the outcome based on the conditional mean mod1 <- glm(y ~ x) par(mfrow=c(2,2), mar=c(4.1,4.1,3.1,2.1)) plot(x=x, y=y, xlab='X', ylab='Y', main='Scatterplot', cex=0.7); abline( mod1 ) plot(x=x, y=rstudent(mod1), xlab='X', ylab='Jackknife Residual', main='Residual Plot', cex=0.7); abline(h=0, lty=2, col='gray65') hist(rstudent(mod1), xlab='Jackknife Residual', main='Histogram of Residuals', freq=F, breaks=seq(-6,1,0.25)); curve( dnorm(x,mean=0,sd=1), lwd=2, col='blue', add=T) plot( ppoints(length(rstudent(mod1))), sort(pnorm(rstudent(mod1))), xlab='Observed Cumulative Probability', ylab='Expected Cumulative Probability', main='Normal Probability Plot', cex=0.7); abline(a=0,b=1, col='gray65', lwd=1) ``` ## 1b. Scenario 2 ```{r, echo=F} # Code to generate figures for violation of normality set.seed(515) x <- rnorm(n=100, mean=10, sd=3) # simulate a single continuous predictor with mean=10 and SD=3 reg <- 15 + 5 * x # set the regression equation so the intercept=10 and the slope=3 y <- rexp(n=100, rate=1/reg) # simulate the outcome based on the conditional mean mod1 <- glm(y ~ x) par(mfrow=c(2,2), mar=c(4.1,4.1,3.1,2.1)) plot(x=x, y=y, xlab='X', ylab='Y', main='Scatterplot', cex=0.7); abline( mod1 ) plot(x=x, y=rstudent(mod1), xlab='X', ylab='Jackknife Residual', main='Residual Plot', cex=0.7); abline(h=0, lty=2, col='gray65') hist(rstudent(mod1), xlab='Jackknife Residual', main='Histogram of Residuals', freq=F, breaks=seq(-4,7,0.25)); curve( dnorm(x,mean=0,sd=1), lwd=2, col='blue', add=T) plot( ppoints(length(rstudent(mod1))), sort(pnorm(rstudent(mod1))), xlab='Observed Cumulative Probability', ylab='Expected Cumulative Probability', main='Normal Probability Plot', cex=0.7); abline(a=0,b=1, col='gray65', lwd=1) ``` ## 1c. Scenario 3 ```{r, echo=F} # Code to generate figures for no violations set.seed(720) x <- rnorm(n=100, mean=10, sd=3) # simulate a single continuous predictor with mean=10 and SD=3 reg <- 15 + 5 * x # set the regression equation so the intercept=10 and the slope=3 y <- rnorm(n=100, mean=reg, sd=8) # simulate the outcome based on the conditional mean mod1 <- glm(y ~ x) par(mfrow=c(2,2), mar=c(4.1,4.1,3.1,2.1)) plot(x=x, y=y, xlab='X', ylab='Y', main='Scatterplot', cex=0.7); abline( mod1 ) plot(x=x, y=rstudent(mod1), xlab='X', ylab='Jackknife Residual', main='Residual Plot', cex=0.7); abline(h=0, lty=2, col='gray65') hist(rstudent(mod1), xlab='Jackknife Residual', main='Histogram of Residuals', freq=F, breaks=seq(-4,4,0.25)); curve( dnorm(x,mean=0,sd=1), lwd=2, col='blue', add=T) plot( ppoints(length(rstudent(mod1))), sort(pnorm(rstudent(mod1))), xlab='Observed Cumulative Probability', ylab='Expected Cumulative Probability', main='Normal Probability Plot', cex=0.7); abline(a=0,b=1, col='gray65', lwd=1) ``` ## 1d. Scenario 4 ```{r, echo=F} # Code to generate figures for violation of homoscedasticity set.seed(660) x <- rnorm(n=100, mean=10, sd=3) # simulate a single continuous predictor with mean=10 and SD=3 reg <- 15 + 5 * x # set the regression equation so the intercept=10 and the slope=3 y <- rnorm(n=100, mean=reg, sd=x^2) # simulate the outcome based on the conditional mean mod1 <- glm(y ~ x) par(mfrow=c(2,2), mar=c(4.1,4.1,3.1,2.1)) plot(x=x, y=y, xlab='X', ylab='Y', main='Scatterplot', cex=0.7); abline( mod1 ) plot(x=x, y=rstudent(mod1), xlab='X', ylab='Jackknife Residual', main='Residual Plot', cex=0.7); abline(h=0, lty=2, col='gray65') hist(rstudent(mod1), xlab='Jackknife Residual', main='Histogram of Residuals', freq=F, breaks=seq(-6,4,0.25)); curve( dnorm(x,mean=0,sd=1), lwd=2, col='blue', add=T) plot( ppoints(length(rstudent(mod1))), sort(pnorm(rstudent(mod1))), xlab='Observed Cumulative Probability', ylab='Expected Cumulative Probability', main='Normal Probability Plot', cex=0.7); abline(a=0,b=1, col='gray65', lwd=1) ``` ## 1e. Scenario 5 ```{r, echo=F} # Code to generate figures for violation of homoscedasticity, linearity, and normality set.seed(660) x <- rnorm(n=100, mean=10, sd=3) # simulate a single continuous predictor with mean=10 and SD=3 reg <- 15 + 5 * x + 0.5 * x^2 # set the regression equation so the intercept=10 and the slope=3 y <- rgamma(n=100, shape=reg, scale=x^2) # simulate the outcome based on the conditional mean mod1 <- glm(y ~ x) par(mfrow=c(2,2), mar=c(4.1,4.1,3.1,2.1)) plot(x=x, y=y, xlab='X', ylab='Y', main='Scatterplot', cex=0.7); abline( mod1 ) plot(x=x, y=rstudent(mod1), xlab='X', ylab='Jackknife Residual', main='Residual Plot', cex=0.7); abline(h=0, lty=2, col='gray65') hist(rstudent(mod1), xlab='Jackknife Residual', main='Histogram of Residuals', freq=F, breaks=seq(-2,7,0.25)); curve( dnorm(x,mean=0,sd=1), lwd=2, col='blue', add=T) plot( ppoints(length(rstudent(mod1))), sort(pnorm(rstudent(mod1))), xlab='Observed Cumulative Probability', ylab='Expected Cumulative Probability', main='Normal Probability Plot', cex=0.7); abline(a=0,b=1, col='gray65', lwd=1) ``` # Exercise 2: Data Transformation Interpretations We will use the blood storage dataset from our course site: ```{r} dat <- read.csv('../../.data/Blood_Storage.csv') ``` The dataset includes 316 men who had undergone radical prostatectomy and received transfusion during or within 30 days of the surgical procedure and had available PSA follow-up data . For each of the following models, what are the estimates of the intercept and slope and how would you interpret them? If possible, for transformations of the dependent variable, $Y$, also transform the estimate of your slope and its 95% confidence interval back to its non-transformed scale and interpret. ## 2a. $Y$, $X$ ```{r} summary(lm(PVol ~ Age, data=dat)) ``` ## 2b. $log(Y)$, $X$ ```{r} summary(lm(log(PVol) ~ Age, data=dat)) ``` ## 2c. $\sqrt{Y}$, $X$ ```{r} summary(lm(sqrt(PVol) ~ Age, data=dat)) ``` ## 2d. $Y$, $log(X)$ ```{r} summary(lm(PVol ~ log(Age), data=dat)) ``` ## 2e. $log(Y)$, $log(X)$ ```{r} summary(lm(log(PVol) ~ log(Age), data=dat)) ```

Week 8 Practice Problems

Exercise 1: Diagnostic Examination

1a. Scenario 1

1b. Scenario 2

1c. Scenario 3

1d. Scenario 4

1e. Scenario 5

Exercise 2: Data Transformation Interpretations

2a. \(Y\), \(X\)

2b. \(log(Y)\), \(X\)

2c. \(\sqrt{Y}\), \(X\)

2d. \(Y\), \(log(X)\)

2e. \(log(Y)\), \(log(X)\)