R help - Jun 2010 - Calculation of r squared from a linear regression

Hi,

I'm trying to verify the calculation of coefficient of determination (r
squared) for linear regression. I've done the calculation manually with a
simple test case and using the definition of r squared outlined in summary(lm)
help. There seems to be a discrepancy between the what R produced and the manual
calculation. Does anyone know why this is so? What does the multiple r squared
reported in summary(lm) represent?

# The test case:
x <- c(1,2,3,4)
y <- c(1.6,4.4,5.5,8.3)
dummy <- data.frame(x, y)
fm1 <- lm(y ~ x-1, data = dummy)
summary(fm1)
betax <- fm1$coeff[x] * sd(x) / sd(y) 
# cd is coefficient of determination
cd <- betax * cor(y, x)

Thanks.

On 2010-06-11 2:16, Sandra Hawthorne wrote:
> Hi,
>
> I'm trying to verify the calculation of coefficient of determination (r
squared) for linear regression. I've done the calculation manually with a
simple test case and using the definition of r squared outlined in summary(lm)
help. There seems to be a discrepancy between the what R produced and the manual
calculation. Does anyone know why this is so? What does the multiple r squared
reported in summary(lm) represent?
>
> # The test case:
> x<- c(1,2,3,4)
> y<- c(1.6,4.4,5.5,8.3)
> dummy<- data.frame(x, y)
> fm1<- lm(y ~ x-1, data = dummy)
> summary(fm1)
> betax<- fm1$coeff[x] * sd(x) / sd(y)
> # cd is coefficient of determination
> cd<- betax * cor(y, x)
The discrepancy is due to incorrect manual calculation.
You're using (incorrectly, at that) formulas for simple
regression _with an intercept term_ whereas you model
has _no_ intercept term.

What summary.lm reports is clearly described on the
help page. See r.squared in the Value section.

   -Peter Ehlers

On Fri, 2010-06-11 at 01:16 -0700, Sandra Hawthorne
wrote:
> Hi,
> 
> I'm trying to verify the calculation of coefficient of determination (r
squared) for linear regression. I've done the calculation manually with a
simple test case and using the definition of r squared outlined in summary(lm)
help. There seems to be a discrepancy between the what R produced and the manual
calculation. Does anyone know why this is so? What does the multiple r squared
reported in summary(lm) represent?
> 
> # The test case:
> x <- c(1,2,3,4)
> y <- c(1.6,4.4,5.5,8.3)
> dummy <- data.frame(x, y)
> fm1 <- lm(y ~ x-1, data = dummy)
> summary(fm1)
> betax <- fm1$coeff[x] * sd(x) / sd(y) 
> # cd is coefficient of determination
> cd <- betax * cor(y, x)
> 
> Thanks.

Sorry Sandra,

But the problem in yours script. Look this
x <- c(1,2,3,4)
y <- c(1.6,4.4,5.5,8.3)
dummy <- data.frame(x, y)
fm1 <- lm(y ~ x, data = dummy)
summary(fm1)

Call:
lm(formula = y ~ x, data = dummy)

Residuals:
    1     2     3     4 
-0.17  0.51 -0.51  0.17 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  -0.3500     0.6584  -0.532   0.6481  
x             2.1200     0.2404   8.818   0.0126 *
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 

Residual standard error: 0.5376 on 2 degrees of freedom
Multiple R-squared: 0.9749,	Adjusted R-squared: 0.9624 
F-statistic: 77.76 on 1 and 2 DF,  p-value: 0.01262 

betax <- fm1$coeff[2] * sd(x) / sd(y) 
# cd is coefficient of determination
cd <- betax * cor(y, x)
cd
       x 
0.974924 

The formula "fm1$coeff[2] * sd(x) / sd(y)" is valid only the model
have
a intercept...

-- 
Bernardo Rangel Tura, M.D,MPH,Ph.D
National Institute of Cardiology
Brazil

Hi, as pointed out previously, the problem is in using the canned routine
(lm) without including an intercept term. Here is a working, generic example
with commented code.

#Simulate data
x=rnorm(100)
e=rnorm(100)
y=x+e

#Create X matrix with intercept
X=cbind(1,x)

#Projection matrix
P=X%*%solve(t(X)%*%X)%*%t(X)

#Fitted values
fv=P%*%y

#Run canned regression
reg=lm(y~x)

#Canned and hand computed fitted values
cbind(fitted(reg),fv)

#Are they all equal?
all.equal(as.vector(as.numeric(fitted(reg))),as.vector(as.numeric(fv)))
#This already implies that the R-squared is equal

#Compute R-squared by hand
R.sq=1-sum((y-fv)^2)/sum((y-mean(y))^2)

#Is this equal to the R-squared from the canned routine?
summary(reg)$r.squared==R.sq

Daniel

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