

Hi,
I have been using lm in R to do a linear regression and find the slope
coefficients and value for Rsquared. The Rsquared value reported by R
(R^2 = 0.9558) is very different than the Rsquared value when I use the
same equation in Exce (R^2 = 0.328). I manually computed Rsquared and the
Excel value is correct. I show my code for the determination of R^2 in R.
When I do not set 0 as the intercept, the R^2 value is the same in R and
Excel. In both cases the slope coefficient from R and from Excel are
identical.
k is a data frame with two columns.
M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different
R^2 values in R and Excel
M2 = lm(k[,1]~k[,2])
sumM1 = summary(M1)
sumM2 = summary(M2) ## get same value as Excel when intercept is not
set to 0
Below is what R returns for sumM1:
lm(formula = k[, 1] ~ k[, 2] + 0)
Residuals:
Min 1Q Median 3Q Max
0.057199 0.015857 0.003793 0.013737 0.056178
Coefficients:
Estimate Std. Error t value Pr(>t)
k[, 2] 1.05022 0.04266 24.62 <2e16 ***

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
Residual standard error: 0.02411 on 28 degrees of freedom
Multiple Rsquared: 0.9558, Adjusted Rsquared: 0.9543
Fstatistic: 606.2 on 1 and 28 DF, pvalue: < 2.2e16
Way manual determination was performed. The value returned coincides with
the value from Excel:
#### trying to figure out why the R^2 for R and Excel are so different.
sqerr = (k[,1]  predict(M1))^2
sqtot = (k[,1]  mean(k[,1]) ^2
R2 = 1  sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as
excel value
I am very puzzled by this. How does R compute the value for R^2 in this
case? Did i write the lm incorrectly?
Thanks
Pam
PS In case you are interested, the data I am using for hte two columns is
below.
k[, 1]
1]
[1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441
0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484
0.06358752 0.07945473
[15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827
0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638
0.08419016 0.11995240
[29] 0.14425678
> k[,2]
[1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10
0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12

Pam KroneDavis
Project Research Assistant and Grant Manager
PO Box 22122
Carmel, CA 93922
(831)5823684 (o)
(831)3240391 (h)
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


What does Excel give for the following data, where the byhand formula
you gave is obviously wrong?
> x < c(1, 2, 3)
> y < c(13.1, 11.9, 11.0)
> M1 < lm(y~x+0)
> sqerr < (y predict(M1)) ^ 2
> sqtot < (y  mean(y)) ^ 2
> 1  sum(sqerr)/sum(sqtot)
[1] 37.38707
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> Original Message
> From: [hidden email] [mailto: [hidden email]] On
> Behalf Of Pamela KroneDavis
> Sent: Friday, July 13, 2012 9:01 AM
> To: [hidden email]
> Subject: [R] Rsquared with Intercept set to 0 (zero) for linear regression in R is
> incorrect
>
> Hi,
>
> I have been using lm in R to do a linear regression and find the slope
> coefficients and value for Rsquared. The Rsquared value reported by R
> (R^2 = 0.9558) is very different than the Rsquared value when I use the
> same equation in Exce (R^2 = 0.328). I manually computed Rsquared and the
> Excel value is correct. I show my code for the determination of R^2 in R.
> When I do not set 0 as the intercept, the R^2 value is the same in R and
> Excel. In both cases the slope coefficient from R and from Excel are
> identical.
>
> k is a data frame with two columns.
>
> M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different
> R^2 values in R and Excel
> M2 = lm(k[,1]~k[,2])
> sumM1 = summary(M1)
> sumM2 = summary(M2) ## get same value as Excel when intercept is not
> set to 0
>
> Below is what R returns for sumM1:
>
> lm(formula = k[, 1] ~ k[, 2] + 0)
>
> Residuals:
> Min 1Q Median 3Q Max
> 0.057199 0.015857 0.003793 0.013737 0.056178
>
> Coefficients:
> Estimate Std. Error t value Pr(>t)
> k[, 2] 1.05022 0.04266 24.62 <2e16 ***
> 
> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>
> Residual standard error: 0.02411 on 28 degrees of freedom
> Multiple Rsquared: 0.9558, Adjusted Rsquared: 0.9543
> Fstatistic: 606.2 on 1 and 28 DF, pvalue: < 2.2e16
>
> Way manual determination was performed. The value returned coincides with
> the value from Excel:
>
> #### trying to figure out why the R^2 for R and Excel are so different.
> sqerr = (k[,1]  predict(M1))^2
> sqtot = (k[,1]  mean(k[,1]) ^2
>
> R2 = 1  sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as
> excel value
>
> I am very puzzled by this. How does R compute the value for R^2 in this
> case? Did i write the lm incorrectly?
>
> Thanks
> Pam
>
> PS In case you are interested, the data I am using for hte two columns is
> below.
>
> k[, 1]
> 1]
> [1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441
> 0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484
> 0.06358752 0.07945473
> [15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827
> 0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638
> 0.08419016 0.11995240
> [29] 0.14425678
>
> > k[,2]
> [1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10
> 0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12
>
>
> 
> Pam KroneDavis
> Project Research Assistant and Grant Manager
> PO Box 22122
> Carmel, CA 93922
> (831)5823684 (o)
> (831)3240391 (h)
>
> [[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


You might want to look at
http://support.microsoft.com/kb/214230entitled
Incorrect output is returned when you use the Linear Regression (LINEST) function in Excel
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> Original Message
> From: [hidden email] [mailto: [hidden email]] On
> Behalf Of William Dunlap
> Sent: Friday, July 13, 2012 10:04 AM
> To: Pamela KroneDavis; [hidden email]
> Subject: Re: [R] Rsquared with Intercept set to 0 (zero) for linear regression in R is
> incorrect
>
> What does Excel give for the following data, where the byhand formula
> you gave is obviously wrong?
> > x < c(1, 2, 3)
> > y < c(13.1, 11.9, 11.0)
> > M1 < lm(y~x+0)
> > sqerr < (y predict(M1)) ^ 2
> > sqtot < (y  mean(y)) ^ 2
> > 1  sum(sqerr)/sum(sqtot)
> [1] 37.38707
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>
> > Original Message
> > From: [hidden email] [mailto: [hidden email]] On
> > Behalf Of Pamela KroneDavis
> > Sent: Friday, July 13, 2012 9:01 AM
> > To: [hidden email]
> > Subject: [R] Rsquared with Intercept set to 0 (zero) for linear regression in R is
> > incorrect
> >
> > Hi,
> >
> > I have been using lm in R to do a linear regression and find the slope
> > coefficients and value for Rsquared. The Rsquared value reported by R
> > (R^2 = 0.9558) is very different than the Rsquared value when I use the
> > same equation in Exce (R^2 = 0.328). I manually computed Rsquared and the
> > Excel value is correct. I show my code for the determination of R^2 in R.
> > When I do not set 0 as the intercept, the R^2 value is the same in R and
> > Excel. In both cases the slope coefficient from R and from Excel are
> > identical.
> >
> > k is a data frame with two columns.
> >
> > M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different
> > R^2 values in R and Excel
> > M2 = lm(k[,1]~k[,2])
> > sumM1 = summary(M1)
> > sumM2 = summary(M2) ## get same value as Excel when intercept is not
> > set to 0
> >
> > Below is what R returns for sumM1:
> >
> > lm(formula = k[, 1] ~ k[, 2] + 0)
> >
> > Residuals:
> > Min 1Q Median 3Q Max
> > 0.057199 0.015857 0.003793 0.013737 0.056178
> >
> > Coefficients:
> > Estimate Std. Error t value Pr(>t)
> > k[, 2] 1.05022 0.04266 24.62 <2e16 ***
> > 
> > Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
> >
> > Residual standard error: 0.02411 on 28 degrees of freedom
> > Multiple Rsquared: 0.9558, Adjusted Rsquared: 0.9543
> > Fstatistic: 606.2 on 1 and 28 DF, pvalue: < 2.2e16
> >
> > Way manual determination was performed. The value returned coincides with
> > the value from Excel:
> >
> > #### trying to figure out why the R^2 for R and Excel are so different.
> > sqerr = (k[,1]  predict(M1))^2
> > sqtot = (k[,1]  mean(k[,1]) ^2
> >
> > R2 = 1  sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as
> > excel value
> >
> > I am very puzzled by this. How does R compute the value for R^2 in this
> > case? Did i write the lm incorrectly?
> >
> > Thanks
> > Pam
> >
> > PS In case you are interested, the data I am using for hte two columns is
> > below.
> >
> > k[, 1]
> > 1]
> > [1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441
> > 0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484
> > 0.06358752 0.07945473
> > [15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827
> > 0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638
> > 0.08419016 0.11995240
> > [29] 0.14425678
> >
> > > k[,2]
> > [1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10
> > 0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12
> >
> >
> > 
> > Pam KroneDavis
> > Project Research Assistant and Grant Manager
> > PO Box 22122
> > Carmel, CA 93922
> > (831)5823684 (o)
> > (831)3240391 (h)
> >
> > [[alternative HTML version deleted]]
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained, reproducible code.
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


Pamela
R squared with a nonzero, and with a zero intercept can be very different as the regression line that you get with and without a zero intercept can be very different. Have you plotted your data plot(k[,2],k[,1]) to see if a zero intercept is reasonable for your data? Have you drawn the regression lines that you get from your models and compared the lines to the plots of your data?
John
>>> Pamela KroneDavis < [hidden email]> 7/13/2012 12:00:36 PM >>>
Hi,
I have been using lm in R to do a linear regression and find the slope
coefficients and value for Rsquared. The Rsquared value reported by R
(R^2 = 0.9558) is very different than the Rsquared value when I use the
same equation in Exce (R^2 = 0.328). I manually computed Rsquared and the
Excel value is correct. I show my code for the determination of R^2 in R.
When I do not set 0 as the intercept, the R^2 value is the same in R and
Excel. In both cases the slope coefficient from R and from Excel are
identical.
k is a data frame with two columns.
M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different
R^2 values in R and Excel
M2 = lm(k[,1]~k[,2])
sumM1 = summary(M1)
sumM2 = summary(M2) ## get same value as Excel when intercept is not
set to 0
Below is what R returns for sumM1:
lm(formula = k[, 1] ~ k[, 2] + 0)
Residuals:
Min 1Q Median 3Q Max
0.057199 0.015857 0.003793 0.013737 0.056178
Coefficients:
Estimate Std. Error t value Pr(>t)
k[, 2] 1.05022 0.04266 24.62 <2e16 ***

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
Residual standard error: 0.02411 on 28 degrees of freedom
Multiple Rsquared: 0.9558, Adjusted Rsquared: 0.9543
Fstatistic: 606.2 on 1 and 28 DF, pvalue: < 2.2e16
Way manual determination was performed. The value returned coincides with
the value from Excel:
#### trying to figure out why the R^2 for R and Excel are so different.
sqerr = (k[,1]  predict(M1))^2
sqtot = (k[,1]  mean(k[,1]) ^2
R2 = 1  sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as
excel value
I am very puzzled by this. How does R compute the value for R^2 in this
case? Did i write the lm incorrectly?
Thanks
Pam
PS In case you are interested, the data I am using for hte two columns is
below.
k[, 1]
1]
[1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441
0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484
0.06358752 0.07945473
[15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827
0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638
0.08419016 0.11995240
[29] 0.14425678
> k[,2]
[1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10
0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12

Pam KroneDavis
Project Research Assistant and Grant Manager
PO Box 22122
Carmel, CA 93922
(831)5823684 (o)
(831)3240391 (h)
[[alternative HTML version deleted]]
Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


While excluding the intercept may make sense, your formula for r^2 assumes
that there was an intercept (that is why mean(y) is in your expression for
sqtot).
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
From: Pamela KroneDavis [mailto: [hidden email]]
Sent: Friday, July 13, 2012 10:32 AM
To: William Dunlap
Subject: Re: [R] Rsquared with Intercept set to 0 (zero) for linear regression in R is incorrect
Hi William,
Thanks for getting back to me. When I use the values you provided, it would not make sense to set an intercept of 0. For my data, 0 does make sense as an intercept. When I do not set a 0 intercept using your data points, I get the same value for Rsquared in R and in Excel and manually.
Thanks
Pam
On Fri, Jul 13, 2012 at 10:03 AM, William Dunlap < [hidden email]<mailto: [hidden email]>> wrote:
What does Excel give for the following data, where the byhand formula
you gave is obviously wrong?
> x < c(1, 2, 3)
> y < c(13.1, 11.9, 11.0)
> M1 < lm(y~x+0)
> sqerr < (y predict(M1)) ^ 2
> sqtot < (y  mean(y)) ^ 2
> 1  sum(sqerr)/sum(sqtot)
[1] 37.38707
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com< http://tibco.com>
> Original Message
> From: [hidden email]<mailto: [hidden email]> [mailto: [hidden email]<mailto: [hidden email]>] On
> Behalf Of Pamela KroneDavis
> Sent: Friday, July 13, 2012 9:01 AM
> To: [hidden email]<mailto: [hidden email]>
> Subject: [R] Rsquared with Intercept set to 0 (zero) for linear regression in R is
> incorrect
>
> Hi,
>
> I have been using lm in R to do a linear regression and find the slope
> coefficients and value for Rsquared. The Rsquared value reported by R
> (R^2 = 0.9558) is very different than the Rsquared value when I use the
> same equation in Exce (R^2 = 0.328). I manually computed Rsquared and the
> Excel value is correct. I show my code for the determination of R^2 in R.
> When I do not set 0 as the intercept, the R^2 value is the same in R and
> Excel. In both cases the slope coefficient from R and from Excel are
> identical.
>
> k is a data frame with two columns.
>
> M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different
> R^2 values in R and Excel
> M2 = lm(k[,1]~k[,2])
> sumM1 = summary(M1)
> sumM2 = summary(M2) ## get same value as Excel when intercept is not
> set to 0
>
> Below is what R returns for sumM1:
>
> lm(formula = k[, 1] ~ k[, 2] + 0)
>
> Residuals:
> Min 1Q Median 3Q Max
> 0.057199 0.015857 0.003793 0.013737 0.056178
>
> Coefficients:
> Estimate Std. Error t value Pr(>t)
> k[, 2] 1.05022 0.04266 24.62 <2e16 ***
> 
> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>
> Residual standard error: 0.02411 on 28 degrees of freedom
> Multiple Rsquared: 0.9558, Adjusted Rsquared: 0.9543
> Fstatistic: 606.2 on 1 and 28 DF, pvalue: < 2.2e16
>
> Way manual determination was performed. The value returned coincides with
> the value from Excel:
>
> #### trying to figure out why the R^2 for R and Excel are so different.
> sqerr = (k[,1]  predict(M1))^2
> sqtot = (k[,1]  mean(k[,1]) ^2
>
> R2 = 1  sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as
> excel value
>
> I am very puzzled by this. How does R compute the value for R^2 in this
> case? Did i write the lm incorrectly?
>
> Thanks
> Pam
>
> PS In case you are interested, the data I am using for hte two columns is
> below.
>
> k[, 1]
> 1]
> [1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441
> 0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484
> 0.06358752 0.07945473
> [15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827
> 0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638
> 0.08419016 0.11995240
> [29] 0.14425678
>
> > k[,2]
> [1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10
> 0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12
>
>
> 
> Pam KroneDavis
> Project Research Assistant and Grant Manager
> PO Box 22122
> Carmel, CA 93922
> (831)5823684 (o)
> (831)3240391 (h)
>
> [[alternative HTML version deleted]]

Pam KroneDavis
Project Research Assistant and Grant Manager
PO Box 22122
Carmel, CA 93922
(831)5823684 (o)
(831)3240391 (h)
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


S+ and, I assume, R compute r^2 when there is no intercept as
sum(fitted(M1)^2) / sum(y^2)
where M1 is the fitted model and y the response.
See, for example,
http://web.ist.utl.pt/~ist11038/compute/errtheory/,regression/regrthroughorigin.pdffor the derivation of this formula.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
From: Pamela KroneDavis [mailto: [hidden email]]
Sent: Friday, July 13, 2012 11:05 AM
To: William Dunlap
Subject: Re: [R] Rsquared with Intercept set to 0 (zero) for linear regression in R is incorrect
Thanks William,
I have actually tried a couple of different formulas for determining R^2. The second formula does return a different value and assumes no intercept. The second formula is attached in the image.
However, when I use the manual formula for R^2 that is shown, I get an R^2 value that matches the second formula when I don't set the intercept. ISo I think the formula I showed works for both cases.
M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0
M2 = lm(k[,1]~k[,2])
sqerrM2 = (k[,1]  predict(M2))^2
sqtotM2 = (k[,1]  mean(k[,1])) ^2
sqerrM1 = (k[,1]  predict(M1))^2
sqtotM1 = (k[,1]  mean(k[,1])) ^2
R2M1 = 1  sum(sqerrM1)/sum(sqtotM1) ## get 0.328 same as excel value
R2M2 = 1  sum(sqerrM2)/sum(sqtotM2) ## same as Excel 0.408 and as R in this case
> R2M1
[1] 0.3284381
> R2M2
[1] 0.4083052
How does R compute the Rsquared value?
Thanks
Pam
On Fri, Jul 13, 2012 at 10:38 AM, William Dunlap < [hidden email]<mailto: [hidden email]>> wrote:
While excluding the intercept may make sense, your formula for r^2 assumes
that there was an intercept (that is why mean(y) is in your expression for
sqtot).
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com< http://tibco.com>
From: Pamela KroneDavis [mailto: [hidden email]<mailto: [hidden email]>]
Sent: Friday, July 13, 2012 10:32 AM
To: William Dunlap
Subject: Re: [R] Rsquared with Intercept set to 0 (zero) for linear regression in R is incorrect
Hi William,
Thanks for getting back to me. When I use the values you provided, it would not make sense to set an intercept of 0. For my data, 0 does make sense as an intercept. When I do not set a 0 intercept using your data points, I get the same value for Rsquared in R and in Excel and manually.
Thanks
Pam
On Fri, Jul 13, 2012 at 10:03 AM, William Dunlap < [hidden email]<mailto: [hidden email]>> wrote:
What does Excel give for the following data, where the byhand formula
you gave is obviously wrong?
> x < c(1, 2, 3)
> y < c(13.1, 11.9, 11.0)
> M1 < lm(y~x+0)
> sqerr < (y predict(M1)) ^ 2
> sqtot < (y  mean(y)) ^ 2
> 1  sum(sqerr)/sum(sqtot)
[1] 37.38707
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com< http://tibco.com>
> Original Message
> From: [hidden email]<mailto: [hidden email]> [mailto: [hidden email]<mailto: [hidden email]>] On
> Behalf Of Pamela KroneDavis
> Sent: Friday, July 13, 2012 9:01 AM
> To: [hidden email]<mailto: [hidden email]>
> Subject: [R] Rsquared with Intercept set to 0 (zero) for linear regression in R is
> incorrect
>
> Hi,
>
> I have been using lm in R to do a linear regression and find the slope
> coefficients and value for Rsquared. The Rsquared value reported by R
> (R^2 = 0.9558) is very different than the Rsquared value when I use the
> same equation in Exce (R^2 = 0.328). I manually computed Rsquared and the
> Excel value is correct. I show my code for the determination of R^2 in R.
> When I do not set 0 as the intercept, the R^2 value is the same in R and
> Excel. In both cases the slope coefficient from R and from Excel are
> identical.
>
> k is a data frame with two columns.
>
> M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different
> R^2 values in R and Excel
> M2 = lm(k[,1]~k[,2])
> sumM1 = summary(M1)
> sumM2 = summary(M2) ## get same value as Excel when intercept is not
> set to 0
>
> Below is what R returns for sumM1:
>
> lm(formula = k[, 1] ~ k[, 2] + 0)
>
> Residuals:
> Min 1Q Median 3Q Max
> 0.057199 0.015857 0.003793 0.013737 0.056178
>
> Coefficients:
> Estimate Std. Error t value Pr(>t)
> k[, 2] 1.05022 0.04266 24.62 <2e16 ***
> 
> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>
> Residual standard error: 0.02411 on 28 degrees of freedom
> Multiple Rsquared: 0.9558, Adjusted Rsquared: 0.9543
> Fstatistic: 606.2 on 1 and 28 DF, pvalue: < 2.2e16
>
> Way manual determination was performed. The value returned coincides with
> the value from Excel:
>
> #### trying to figure out why the R^2 for R and Excel are so different.
> sqerr = (k[,1]  predict(M1))^2
> sqtot = (k[,1]  mean(k[,1]) ^2
>
> R2 = 1  sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as
> excel value
>
> I am very puzzled by this. How does R compute the value for R^2 in this
> case? Did i write the lm incorrectly?
>
> Thanks
> Pam
>
> PS In case you are interested, the data I am using for hte two columns is
> below.
>
> k[, 1]
> 1]
> [1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441
> 0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484
> 0.06358752 0.07945473
> [15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827
> 0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638
> 0.08419016 0.11995240
> [29] 0.14425678
>
> > k[,2]
> [1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10
> 0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12
>
>
> 
> Pam KroneDavis
> Project Research Assistant and Grant Manager
> PO Box 22122
> Carmel, CA 93922
> (831)5823684<tel:%28831%295823684> (o)
> (831)3240391<tel:%28831%293240391> (h)
>
> [[alternative HTML version deleted]]

Pam KroneDavis
Project Research Assistant and Grant Manager
PO Box 22122
Carmel, CA 93922
(831)5823684<tel:%28831%295823684> (o)
(831)3240391<tel:%28831%293240391> (h)

Pam KroneDavis
Project Research Assistant and Grant Manager
PO Box 22122
Carmel, CA 93922
(831)5823684 (o)
(831)3240391 (h)
[[alternative HTML version deleted]]
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