# R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect

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## R-squared 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 R-squared.  The R-squared value reported by R (R^2 = 0.9558) is very different than the R-squared value when I use the same equation in Exce (R^2 = 0.328).  I manually computed R-squared 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   <2e-16 *** --- Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1 Residual standard error: 0.02411 on 28 degrees of freedom Multiple R-squared: 0.9558,     Adjusted R-squared: 0.9543 F-statistic: 606.2 on 1 and 28 DF,  p-value: < 2.2e-16 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 Krone-Davis Project Research Assistant and Grant Manager PO Box 22122 Carmel, CA 93922 (831)582-3684 (o) (831)324-0391 (h)         [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect

 What does Excel give for the following data, where the by-hand 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 Krone-Davis > Sent: Friday, July 13, 2012 9:01 AM > To: [hidden email] > Subject: [R] R-squared 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 R-squared.  The R-squared value reported by R > (R^2 = 0.9558) is very different than the R-squared value when I use the > same equation in Exce (R^2 = 0.328).  I manually computed R-squared 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   <2e-16 *** > --- > Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > > Residual standard error: 0.02411 on 28 degrees of freedom > Multiple R-squared: 0.9558,     Adjusted R-squared: 0.9543 > F-statistic: 606.2 on 1 and 28 DF,  p-value: < 2.2e-16 > > 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 Krone-Davis > Project Research Assistant and Grant Manager > PO Box 22122 > Carmel, CA 93922 > (831)582-3684 (o) > (831)324-0391 (h) > > [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect

 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 Krone-Davis; [hidden email] > Subject: Re: [R] R-squared with Intercept set to 0 (zero) for linear regression in R is > incorrect > > What does Excel give for the following data, where the by-hand 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 Krone-Davis > > Sent: Friday, July 13, 2012 9:01 AM > > To: [hidden email] > > Subject: [R] R-squared 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 R-squared.  The R-squared value reported by R > > (R^2 = 0.9558) is very different than the R-squared value when I use the > > same equation in Exce (R^2 = 0.328).  I manually computed R-squared 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   <2e-16 *** > > --- > > Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > > > > Residual standard error: 0.02411 on 28 degrees of freedom > > Multiple R-squared: 0.9558,     Adjusted R-squared: 0.9543 > > F-statistic: 606.2 on 1 and 28 DF,  p-value: < 2.2e-16 > > > > 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 Krone-Davis > > Project Research Assistant and Grant Manager > > PO Box 22122 > > Carmel, CA 93922 > > (831)582-3684 (o) > > (831)324-0391 (h) > > > > [[alternative HTML version deleted]] > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect

 In reply to this post by Pamela Krone-Davis Pamela R squared with a non-zero, 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 Krone-Davis <[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 R-squared.  The R-squared value reported by R (R^2 = 0.9558) is very different than the R-squared value when I use the same equation in Exce (R^2 = 0.328).  I manually computed R-squared 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   <2e-16 *** --- Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1 Residual standard error: 0.02411 on 28 degrees of freedom Multiple R-squared: 0.9558,     Adjusted R-squared: 0.9543 F-statistic: 606.2 on 1 and 28 DF,  p-value: < 2.2e-16 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 Krone-Davis Project Research Assistant and Grant Manager PO Box 22122 Carmel, CA 93922 (831)582-3684 (o) (831)324-0391 (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/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect

 In reply to this post by William Dunlap 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 Krone-Davis [mailto:[hidden email]] Sent: Friday, July 13, 2012 10:32 AM To: William Dunlap Subject: Re: [R] R-squared 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 R-squared in R and in Excel and manually. Thanks Pam On Fri, Jul 13, 2012 at 10:03 AM, William Dunlap <[hidden email]> wrote: What does Excel give for the following data, where the by-hand 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 Krone-Davis > Sent: Friday, July 13, 2012 9:01 AM > To: [hidden email] > Subject: [R] R-squared 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 R-squared.  The R-squared value reported by R > (R^2 = 0.9558) is very different than the R-squared value when I use the > same equation in Exce (R^2 = 0.328).  I manually computed R-squared 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   <2e-16 *** > --- > Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > > Residual standard error: 0.02411 on 28 degrees of freedom > Multiple R-squared: 0.9558,     Adjusted R-squared: 0.9543 > F-statistic: 606.2 on 1 and 28 DF,  p-value: < 2.2e-16 > > 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 Krone-Davis > Project Research Assistant and Grant Manager > PO Box 22122 > Carmel, CA 93922 > (831)582-3684 (o) > (831)324-0391 (h) > >       [[alternative HTML version deleted]] -- Pam Krone-Davis Project Research Assistant and Grant Manager PO Box 22122 Carmel, CA 93922 (831)582-3684 (o) (831)324-0391 (h)         [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.