Lecture 5:SLR Diagnostics (Continued)CorrelationIntroduction to Multiple Linear Regression

BMTRY 701Biostatistical Methods II

From last lecture

What were the problems we diagnosed?

We shouldn’t just give up!

Some possible approaches for improvement• remove the outliers: does the model change?

• transform LOS: do we better adhere to model assumptions?

Outlier Quandry

To remove or not to remove outliers

Are they real data?

If they are truly reflective of the data, then what does removing them imply?

Use caution!• better to be true to the data• having a perfect model should not be at the expense

of using ‘real’ data!

Removing the outliers: How to?

I am always reluctant.

my approach in this example:• remove each separately• remove both together• compare each model with the model that includes

outliers

How to decide: compare slope estimates.

SENIC Data

> par(mfrow=c(1,2))> hist(data$LOS)> plot(data$BEDS, data$LOS)

Histogram of data$LOS

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How to fit regression removing outlier(s)?

> keep.remove.both <- ifelse(data$LOS<16,1,0)> keep.remove.20 <- ifelse(data$LOS<19,1,0)> keep.remove.18 <- ifelse(data$LOS<16 | data$BEDS<600,1,0)> > table(keep.remove.both)keep.remove.both 0 1 2 111 > table(keep.remove.20)keep.remove.20 0 1 1 112 > table(keep.remove.18)keep.remove.18 0 1 1 112

Regression Fitting

reg <- lm(LOS ~ BEDS, data=data)

reg.remove.both <- lm(LOS ~ BEDS, data=data[keep.remove.both==1,])

reg.remove.20 <- lm(LOS ~ BEDS, data=data[keep.remove.20==1,])

reg.remove.18 <- lm(LOS ~ BEDS, data=data[keep.remove.18==1,])

How much do our inferences change?

reg remove

both

remove 20

remove 18

β1 estimate

0.00406 0.00299 0.00393 0.00314

se(β1) 0.00086 0.00070 0.00073 0.00085

% change 0

(ref)

26% 3% 23%

Why is “18” a bigger outlier than “20”?

Leverage and Influence

Leverage is a function of the explanatory variable(s) alone and measures the potential for a data point to affect the model parameter estimates.

Influence is a measure of how much a data point actually does affect the estimated model.

Leverage and influence both may be defined in terms of matrices

More later in MLR (MPV ch. 6)

Graphically

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R code

par(mfrow=c(1,1))plot(data$BEDS, data$LOS, pch=16)# old plain old regression modelabline(reg, lwd=2)# plot “20” to show which point we are removing, then# add regression linepoints(data$BEDS[keep.remove.20==0], data$LOS[keep.remove.20==0],

col=2, cex=1.5, pch=16)abline(reg.remove.20, col=2, lwd=2)# plot “18” and then add regressionlinepoints(data$BEDS[keep.remove.18==0], data$LOS[keep.remove.18==0],

col=4, cex=1.5, pch=16)abline(reg.remove.18, col=4, lwd=2)# add regression line where we removed both outliersabline(reg.remove.both, col=5, lwd=2)# add a legend to the plotlegend(1,19, c("reg","w/out 18","w/out 20","w/out both"),

lwd=rep(2,4), lty=rep(1,4), col=c(1,2,4,5))

What to do?

Let’s try something else What was our other problem?

• heteroskedasticity (great word…try that at scrabble)

• non-normality of outliers

Common way to solve: transform the outcome

Determining the Transformation

Box-Cox transformation approach

Finds the “best” power transformation to achieve closest distribution to normality

Can apply to• a variable• to a linear regression model

When applied to a regression model, result tells you what is the ‘best’ power transform of Y to achieve normal residuals

Review of power transformation

Assume we want to transform Y Box-Cox considers Ya for all values of a Solution is the a that provides the “most normal”

looking Ya

Practical powers• a = 1: identity• a = ½ : square-root• a = 0: log• a = -1: 1/Y. usually we also take negative so that

order is maintained (see example) Often not practical interpretation: Y-0.136

Box-Cox for linear regression

library(MASS)

bc <- boxcox(reg)

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ty <- -1/data$LOS

plot(data$LOS, ty)

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New regression: transform is -1/LOS

plot(data$BEDS, ty, pch=16)reg.ty <- lm(ty ~ data$BEDS)abline(reg.ty, lwd=2)

More interpretable?

LOS is often analyzed in the literature Common transform is log

• it is well-known that LOS is skewed in most applications

• most people take the log• people are used to seeing and interpreting it on the

log scale

How good is our model if we just take the log?

Regression with log(LOS)

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Let’s Compare: |Residuals|

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Let’s Compare: QQ-plot

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R code logy <- log(data$LOS)par(mfrow=c(1,2))plot(data$LOS, logy)plot(data$BEDS, logy, pch=16)reg.logy <- lm(logy ~ data$BEDS)abline(reg.logy, lwd=2)

par(mfrow=c(1,2))plot(data$BEDS, reg.ty$residuals, pch=16)abline(h=0, lwd=2)plot(data$BEDS, reg.logy$residuals, pch=16)abline(h=0, lwd=2)

boxplot(reg.ty$residuals)title("Residuals where Y = -1/LOS")boxplot(reg.logy$residuals)title("Residuals where Y = log(LOS)")

qqnorm(reg.ty$residuals, main="TY")qqline(reg.ty$residuals)qqnorm(reg.logy$residuals, main="LogY")qqline(reg.logy$residuals)

Regression results> summary(reg.ty)

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -1.169e-01 2.522e-03 -46.371 < 2e-16 ***data$BEDS 3.953e-05 7.957e-06 4.968 2.47e-06 ***---

> summary(reg.logy)

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.1512591 0.0251328 85.596 < 2e-16 ***data$BEDS 0.0003921 0.0000793 4.944 2.74e-06 ***---

Let’s compare: results ‘untransformed’

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R code

par(mfrow=c(1,2))plot(data$BEDS, data$LOS, pch=16)abline(reg, lwd=2)lines(sort(data$BEDS), -1/sort(reg.ty$fitted.values),lwd=2, lty=2)lines(sort(data$BEDS), exp(sort(reg.logy$fitted.values)), lwd=2, lty=3)

plot(data$BEDS, data$LOS, pch=16, ylim=c(7,12))abline(reg, lwd=2)lines(sort(data$BEDS), -1/sort(reg.ty$fitted.values),lwd=2, lty=2)lines(sort(data$BEDS), exp(sort(reg.logy$fitted.values)), lwd=2, lty=3)

So, what to do?

What are the pros and cons of each transform?

Should we transform at all?!

Switching Gears: Correlation

“Pearson” Correlation Measures linear association between two

variables A natural by-product of linear regression Notation: r or ρ (rho)

Correlation versus slope?

Measure different aspects of the association between X and Y

Slope: measures if there is a linear trend Correlation: provides measure of how close the

datapoints fall to the line

Statistical significance is IDENTICAL• p-value for testing that correlation is 0 is the SAME as

the p-value for testing that the slope is 0.

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Example: Same slope, different correlation

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Example: Same correlation, different slope

r = 0.46, b1=4 r = 0.46, b1=2

Correlation

Scaled version of Covariance between X and Y Recall Covariance:

Estimating the Covariance:

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Correlation tells how closely two variables “track” one another

Provides information about ability to predict Y from X

Regression output:• look for R2

• for SLR, sqrt(R2) = correlation Can have low correlation yet significant

association With correlation, 95% confidence interval is

helpful

LOS ~ BEDS

> summary(lm(data$LOS ~ data$BEDS))

Call:lm(formula = data$LOS ~ data$BEDS)

Residuals: Min 1Q Median 3Q Max -2.8291 -1.0028 -0.1302 0.6782 9.6933

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 8.6253643 0.2720589 31.704 < 2e-16 ***data$BEDS 0.0040566 0.0008584 4.726 6.77e-06 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.752 on 111 degrees of freedomMultiple R-squared: 0.1675, Adjusted R-squared: 0.16 F-statistic: 22.33 on 1 and 111 DF, p-value: 6.765e-06

409.01675.0

95% Confidence Interval for Correlation

The computation of a confidence interval on the population value of Pearson's correlation (ρ) is complicated by the fact that the sampling distribution of r is not normally distributed. The solution lies with Fisher's z' transformation described in the section on the sampling distribution of Pearson's r. The steps in computing a confidence interval for ρ are:• Convert r to z' • Compute a confidence interval in terms of z' • Convert the confidence interval back to r.

freeware!• http://www.danielsoper.com/statcalc/calc28.aspx• http://glass.ed.asu.edu/stats/analysis/rci.html• http://faculty.vassar.edu/lowry/rho.html

log(LOS) ~ BEDS

> summary(lm(log(data$LOS) ~ data$BEDS))

Call:lm(formula = log(data$LOS) ~ data$BEDS)

Residuals: Min 1Q Median 3Q Max -0.296328 -0.106103 -0.005296 0.084177 0.702262

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.1512591 0.0251328 85.596 < 2e-16 ***data$BEDS 0.0003921 0.0000793 4.944 2.74e-06 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1618 on 111 degrees of freedomMultiple R-squared: 0.1805, Adjusted R-squared: 0.1731 F-statistic: 24.44 on 1 and 111 DF, p-value: 2.737e-06

425.01805.0

Multiple Linear Regression

Most regression applications include more than one covariate

Allows us to make inferences about the relationship between two variables (X and Y) adjusting for other variables

Used to account for confounding. Especially important in observational studies

• smoking and lung cancer• we know people who smoke tend to expose themselves to other

risks and harms• if we didn’t adjust, we would overestimate the effect of smoking

on the risk of lung cancer.

Importance of including ‘important’ covariates

If you leave out relevant covariates, your estimate of β1 will be biased

How biased?

Assume: • true model:

• fitted model:

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Implications

The bias is a function of the correlation between the two covariates, X1 and X2

If the correlation is high, the bias will be high If the correlation is low, the bias may be quite

small. If there is no correlation between X1 and X2, then

omitting X2 does not bias inferences

However, it is not a good model for prediction if X2 is related to Y

Example: LOS ~ BEDS analysis.

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> cor(cbind(data$BEDS, data$NURSE, data$LOS)) [,1] [,2] [,3][1,] 1.0000000 0.9155042 0.4092652[2,] 0.9155042 1.0000000 0.3403671[3,] 0.4092652 0.3403671 1.0000000

R code

reg.beds <- lm(log(data$LOS) ~ data$BEDS)reg.nurse <- lm(log(data$LOS) ~ data$NURSE)reg.beds.nurse <- lm(log(data$LOS) ~ data$BEDS + data$NURSE)summary(reg.beds)summary(reg.nurse)summary(reg.beds.nurse)

SLRs

Call:lm(formula = log(data$LOS) ~ data$BEDS)

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.1512591 0.0251328 85.596 < 2e-16 ***data$BEDS 0.0003921 0.0000793 4.944 2.74e-06 ***---

Call:lm(formula = log(data$LOS) ~ data$NURSE)

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.1682138 0.0250054 86.710 < 2e-16 ***data$NURSE 0.0004728 0.0001127 4.195 5.51e-05 ***---

BEDS + NURSE> summary(reg.beds.nurse)

Call:lm(formula = log(data$LOS) ~ data$BEDS + data$NURSE)

Residuals: Min 1Q Median 3Q Max -0.291537 -0.108447 -0.006711 0.087594 0.696747

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.1522361 0.0252758 85.150 <2e-16 ***data$BEDS 0.0004910 0.0001977 2.483 0.0145 * data$NURSE -0.0001497 0.0002738 -0.547 0.5857 ---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1624 on 110 degrees of freedomMultiple R-squared: 0.1827, Adjusted R-squared: 0.1678 F-statistic: 12.29 on 2 and 110 DF, p-value: 1.519e-05