Lecture 6:Multiple Linear RegressionAdjusted Variable Plots

BMTRY 701Biostatistical Methods II

Graphical Displays in MLR

No more one simple scatterplot: need to look at multiple pairs of variables

“pairs” in R. but, we can’t look at all covariates in regards to

the way they enter the model solution: adjusted variable plot aka: partial regression plot

Adjusted Variable Plots

Adjusted variable plots are useful in linear regression for outlier detection and for qualitative evaluation of the fit of a model.

With two covariates: Shows the association between X and Y adjusted for another variable, Z.

With more than two covariates: Shows the association between X and Y adjusted for many other covariates

In our example, association between logLOS and number of nurses, adjusted for number of beds

Approach

Assume we want to look at the association of Y and X, adjusted for Z

Step 1: Regress Y on X and save residuals (res.xy)

Step 2: Regress Z on X and save residuals (res.xz)

Step 3: plot res.xy versus res.xz Optional step 4:

• perform regression of res.xy on res.xz• compare slope to that of MLR of Y on X and Z

MPV: section 4.2.4

SENIC

INFRISK

0 200 400 600 800

23

45

67

8

020

040

060

080

0

BEDS

2 3 4 5 6 7 8 2.0 2.2 2.4 2.6 2.8 3.0

2.0

2.2

2.4

2.6

2.8

3.0

logLOS

R

pairs(~INFRISK+BEDS+logLOS, data=data, pch=16)

# adjusted variable plot approach# look at the association between INFRISK and logLOS, # adjusting for BEDS

reg.xy <- lm(logLOS ~ BEDS, data=data)res.xy <- reg.xy$residuals

reg.xz <- lm(INFRISK ~ BEDS, data=data)res.xz <- reg.xz$residuals

plot(res.xz, res.xy, pch=16)reg.res <- lm(res.xy ~ res.xz)abline(reg.res, lwd=2)reg.infrisk.beds <- lm(logLOS ~ BEDS + INFRISK, data=data)

-2 -1 0 1 2 3

-0.2

0.0

0.2

0.4

0.6

res.xz

res.

xy

Why is this important or interesting?

It shows us the ‘adjusted’ relationship it can help us determine if

• it is an important variable (at all)• if another form of X is more appropriate• if the correlation is high vs. low after adjustment• we need to/want to adjust for this variable

It also informs us about why a variable ‘loses’ significance

Most important: check for non-linearity Example: logLOS ~ NURSE

What about BEDS and NURSE?

# why NURSE is not associated, after adjustment for BEDS?

reg.nurse <- lm(logLOS ~ NURSE, data=data)reg.nurse.beds <- lm(logLOS ~ NURSE + BEDS, data=data)

reg.xy <- lm(logLOS ~ BEDS, data=data)res.xy <- reg.xy$residuals

reg.xz <- lm(NURSE ~ BEDS, data=data)res.xz <- reg.xz$residuals

plot(res.xz, res.xy, pch=16)reg.res <- lm(res.xy ~ res.xz)abline(reg.res, lwd=2)

-200 -100 0 100 200

-0.2

0.0

0.2

0.4

0.6

res.xz

res.

xy

What about the other way around?

######################## what about the other way? what about why BEDS is # assoc after adjustment for NURSE?

reg.xy <- lm(logLOS ~ NURSE, data=data)res.xy <- reg.xy$residuals

reg.xz <- lm(BEDS ~ NURSE, data=data)res.xz <- reg.xz$residuals

plot(res.xz, res.xy, pch=16)reg.res <- lm(res.xy ~ res.xz)abline(reg.res, lwd=2)reg.nurse.beds <- lm(logLOS ~ NURSE + BEDS, data=data)

-200 -100 0 100 200 300

-0.2

0.0

0.2

0.4

0.6

res.xz

res.

xy

Interpretation in MLR

“Adjusted for” “Controlled for “ “Holding all else constant”

In MLR, you need to include one of these phrases (or something like one of them) when interpreting a regression coefficient

LOS ~ INFRISK + BEDS

> reg.infrisk.beds <- lm(LOS ~ BEDS + INFRISK, data=data)> summary(reg.infrisk.beds)

Call:lm(formula = LOS ~ BEDS + INFRISK, data = data)

Residuals: Min 1Q Median 3Q Max -2.8624 -0.9904 -0.1996 0.6671 8.4219

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 6.2703521 0.5038751 12.444 < 2e-16 ***BEDS 0.0024747 0.0008236 3.005 0.00329 ** INFRISK 0.6323812 0.1184476 5.339 5.08e-07 ***---

Hard to interpret with so many decimal places!

> data$beds100 <- data$BEDS/100> reg.infrisk.beds <- lm(LOS ~ beds100 + INFRISK, data=data)> summary(reg.infrisk.beds)

Call:lm(formula = LOS ~ beds100 + INFRISK, data = data)

Residuals: Min 1Q Median 3Q Max -2.8624 -0.9904 -0.1996 0.6671 8.4219

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 6.27035 0.50388 12.444 < 2e-16 ***beds100 0.24747 0.08236 3.005 0.00329 ** INFRISK 0.63238 0.11845 5.339 5.08e-07 ***---

logLOS ~ INFRISK + BEDS

> reg.infrisk.beds <- lm(logLOS ~ BEDS + INFRISK, data=data)> summary(reg.infrisk.beds)

Call:lm(formula = logLOS ~ BEDS + INFRISK, data = data)

Residuals: Min 1Q Median 3Q Max -0.314377 -0.079979 -0.008026 0.072108 0.580675

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.926e+00 4.611e-02 41.767 < 2e-16 ***BEDS 2.407e-04 7.538e-05 3.194 0.00183 ** INFRISK 6.048e-02 1.084e-02 5.579 1.75e-07 ***---

Hard to interpret with so many decimal places!

> data$beds100 <- data$BEDS/100> reg.infrisk.beds100 <- lm(logLOS ~ beds100 + INFRISK, data=data)> summary(reg.infrisk.beds100)

Call:lm(formula = logLOS ~ beds100 + INFRISK, data = data)

Residuals: Min 1Q Median 3Q Max -0.314377 -0.079979 -0.008026 0.072108 0.580675

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.926040 0.046114 41.767 < 2e-16 ***beds100 0.024075 0.007538 3.194 0.00183 ** INFRISK 0.060477 0.010840 5.579 1.75e-07 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1435 on 110 degrees of freedomMultiple R-squared: 0.3612, Adjusted R-squared: 0.3496 F-statistic: 31.1 on 2 and 110 DF, p-value: 1.971e-11

How to interpret?

Pick two values of BEDS• e.g. 100 to 200• e.g. 400 to 500

Estimate the difference in logLOS for each value

What do we plug in for INFRISK?

INFRISKbeds

INFRISKbedsSLO

*060.0100*024.093.1

ˆ100ˆˆˆlog 210

How to interpret? Remember that our inferences are “holding all else

constant” To compare two hospitals with the same INFRISK, it

doesn’t matter what you put in (as long as it is the same)

024.0

1*024.02*024.0

)*060.01*024.093.1(

)*060.02*024.093.1(]1100|ˆ[log]2100|ˆ[log

*060.02*024.093.1

ˆ100ˆˆ]2100|ˆ[log

*060.01*024.093.1

ˆ100ˆˆ]1100|ˆ[log

210

210

INFRISK

INFRISKbedsSLObedsSLO

INFRISK

INFRISKbedsbedsSLO

INFRISK

INFRISKbedsbedsSLO

How to interpret?

02.1)024.0exp(

)1(ˆ)2(ˆ

)1(ˆ)2(ˆ

logexp))1(ˆlog)2(ˆexp(log

SLO

SLO

SLO

SLOSLOSLO

Comparing two hospitals whose number of beds differ by 100 andassuming the same infection risk in the two hospitals is the same, theratio of average LOS in the two hospitals is 1.02 with the hospital with more beds having the longer stay.

difference of 400 beds?

10.1)024.0*4exp(

)1(ˆ)5(ˆ

)1(ˆ)5(ˆ

logexp))1(ˆlog)5(ˆexp(log

SLO

SLO

SLO

SLOSLOSLO

When outcome is log transformed

interpretation of coefficients must be made as RATIOS instead of DIFFERENCES

Need to exponentiate the coefficient. its interpretation is the ratio for a one-unit

difference in the predictor.

Why differences do not work

Consider comparing two hospitals with 400 and 300 beds:

)*06.093.1exp(*0263.0

)075.1101.1)(*06.093.1exp(

))3*024.0exp()4*024.0)(exp(*06.093.1exp(

)*06.0exp(*)3*024.0exp(*)93.1exp(

)*06.0exp(*)4*024.0exp(*)93.1exp(]3100|ˆ[]4100|ˆ[

)*06.0exp(*)3*024.0exp(*)93.1exp(

)*06.03*024.093.1exp(]3100|ˆ[

*060.03*024.093.1

ˆ100ˆˆ]3100|ˆ[log

)*06.0exp(*)4*024.0exp(*)93.1exp(

)*06.04*024.093.1exp(]4100|ˆ[

*060.04*024.093.1

ˆ100ˆˆ]4100|ˆ[log

210

210

INFRISK

INFRISK

INFRISK

INFRISK

INFRISKbedsSLObedsSLO

INFRISK

INFRISKbedsSOL

INFRISK

INFRISKbedsbedsSLO

INFRISK

INFRISKbedsSOL

INFRISK

INFRISKbedsbedsSLO

Why differences do not work

Consider comparing two hospitals with 800 and 700 beds:

)*06.093.1exp(*0291.0

)183.1212.1)(*06.093.1exp(

))7*024.0exp()8*024.0)(exp(*06.093.1exp(

)*06.0exp(*)7*024.0exp(*)93.1exp(

)*06.0exp(*)8*024.0exp(*)93.1exp(]7100|ˆ[]8100|ˆ[

)*06.0exp(*)7*024.0exp(*)93.1exp(

)*06.07*024.093.1exp(]7100|ˆ[

*060.07*024.093.1

ˆ100ˆˆ]7100|ˆ[log

)*06.0exp(*)8*024.0exp(*)93.1exp(

)*06.08*024.093.1exp(]8100|ˆ[

*060.08*024.093.1

ˆ100ˆˆ]8100|ˆ[log

210

210

INFRISK

INFRISK

INFRISK

INFRISK

INFRISKbedsSLObedsSLO

INFRISK

INFRISKbedsSOL

INFRISK

INFRISKbedsbedsSLO

INFRISK

INFRISKbedsSOL

INFRISK

INFRISKbedsbedsSLO

Results in the log scale

0 2 4 6 8

2.0

2.2

2.4

2.6

2.8

3.0

data$beds100

da

ta$

log

LO

S

INFRISK=2INFRISK=5

Results on the “linear” scale: not huge differences

0 2 4 6 8

81

01

21

41

61

82

0

data$beds100

da

ta$

LO

S

INFRISK=2INFRISK=5

Differences can be seen on a larger scale plot

0 10 20 30 40 50

10

15

20

25

30

data$beds100

da

ta$

LO

S

INFRISK=2INFRISK=5