## Calculating linear fit with respect from origin (0,0)

### Sharah (view profile)

on 28 Sep 2016
Latest activity Edited by Massimo Zanetti

on 28 Sep 2016

### Massimo Zanetti (view profile)

I have a data of DataX and Data Y, which is plotted in the blue line as in picture.I am trying to analyse how much the blue line is deviated from the black dotted line x and y = [0:1:7]. The thing is, if i want to compare the linear fit, the linear fit of my data does not start from 0 (yellow line). Can anyone suggest what kind of metric that I can use here? If possible, I would like a way to measure say for example if the fit of my data falls below the dotted line, the value will be <0, and if it is above the >0.

Massimo Zanetti

### Massimo Zanetti (view profile)

on 28 Sep 2016
There is no data above the dotted line. Can you explain better? Do you want to measure the distance of the Data with respect to the Linear Fit Data?
Sharah

### Sharah (view profile)

on 28 Sep 2016
Say, if I have data line the blue line, but it falls above the dotted line. What I want to do is to calculate how much the blue data is different from the dotted line.
The following is one of the method that I used, but not sure if this is accurate
function [rsq] = linearity(xData, yData)
xlin = 0:max(xData)/(length(xData)-1):max(xData);
ylin = 0:max(xData)/(length(xData)-1):max(xData);
p = polyfit(xlin, ylin,1);
yfit = p(1) * xlin +p(2); %similar to polyval(p, x)
yresid = yData - yfit';
SSresid = sum(yresid.^2);
SStotal = (length(yData)-1)*var(yData);
rsq= 1-SSresid/SStotal;
In which I calculated the Rsquare of my data relative to the linear fit of blue line? Will that be correct?

### Tags ### Massimo Zanetti (view profile)

on 28 Sep 2016
Edited by Massimo Zanetti

### Massimo Zanetti (view profile)

on 28 Sep 2016

Ok, you just have to use the right statistics. Look here:
%random points
x=1:10;
y=rand(1,10);
%fit data
p=polyfit(x,y,1);
yfit=p(2)+x*p(1);
%residuals
yres = y-yfit;
%deviation from mean
ymean = mean(y(:));
ydev = y-ymean;
%compute R2
SSR = sum(yres.^2);
SST = sum(ydev.^2);
R2 = 1-SSR/SST;
%plot fitting
plot(x,y,'b',x,yfit,'r');
The R2 number is Rsquared statistics explaining how the fit line well represents the points.

Sharah

### Sharah (view profile)

on 28 Sep 2016
The naswer that you are suggesting over here would calculate how well the data explain to the polyfit of your data. what I am trying to do is to see how well the data is explaining to the linear line of for example xlin=ylin =1:7 (which will be the same size with the data)
function [rsq] = linearity(xData, yData)
xlin = 0:max(xData)/(length(xData)-1):max(xData);
ylin = 0:max(xData)/(length(xData)-1):max(xData);
p = polyfit(xlin, ylin,1);
yfit = p(1) * xlin +p(2); %similar to polyval(p, x)
yresid = yData - yfit';
SSresid = sum(yresid.^2);
SStotal = (length(yData)-1)*var(yData);
rsq= 1-SSresid/SStotal;
will this one works? instead of taking the Rsquared to the polyfit of orginial data, I take 'rsquared' to the linear line that I want (the reference line)
Massimo Zanetti

### Massimo Zanetti (view profile)

on 28 Sep 2016
Ok, so it is much simple. Just replace the polyfit line with reference one.
%random points
x=1:10;
y=rand(1,10);
yref=1:10;
%residuals
yres_ref = y-yref;
%deviation from mean
ymean = mean(y(:));
ydev = y-ymean;
%compute R2
SSR = sum(yres_ref.^2);
SST = sum(ydev.^2);
R2 = 1-SSR/SST;
%plot fitting
plot(x,y,'b',x,yref,'r');