Dear 

 i analyzed ur data  in spss 22  and  i found R .690 and R Square.448

so  i suggest u to use spss better  than excel 

with warm regards

Thanks a lot for your replies. I have set the intercept equal to zero on the basis of the required model. This had given as R2 value of 0.8986 both in excel as well as in SPSS. But the graphical plot shows  a spread in the data and hence it a bit surprising to have such a high R2 value.

In my opinion, the two coefficients of determination are not comparable. Without the intercept, the R2 denominator is $ \sum Yi2 $. With the intercept, it is $ \sum (Yi - mean(Y))2 $. 

You have a set of points very distant from zero and want to build a linear model that passes through zero. My suggestion is: work the model with the intercept and if this parameter has a significant and important value, maybe the complete model is not linear. You only have a linear approximation in the sampled region.

With your data, you have a non significant positive intercept. In my opinion it is probable that for the lower values of x the model without intercept will underestimate y

Mithun - 

I just moved and don't have access to my computer yet, just my phone, so I can't read your excel file. Could you put a scatterplot of your data on this thread?  It sounds interesting. Such a scatterplot may be of general interest to others, as well. Also, it would be helpful to know what is your subject matter/applicatIon.  

From what you have said, it sounds like you have a theoretical reason for y to be zero when x is zero.  You mention "spread," so are you saying that you see heteroscedasticity? (That is generally present, even when not visually noticeable on a scatterplot.)  

Do you have a theoretical reason for expecting linearity? Jorge has a good point that I understand can occur, say in a lab equipment calibration test, where only for a certain range of x does this seem linear, but it may be nonlinear overall. Thus there would be an apparent intercept if you used a linear regression. If you obtain such an estimated intercept that is substantially larger than its estimated standard error, when y should be 0 if x is 0, then this could be an indication that the overall nonlinearity is substantial. There may be substantial nonlinearity anyway, if there is curvature one way and then the other.  

If, however, you expect a zero intercept, and overall linearity, you might be interested in the link attached, which accounts for a degree of heteroscedasticity that is generally robust.  If you expect linearity and a zero intercept, you should use regression through the origin, especially if an estimated intercept is not much larger than its estimated standard error. 

Finally, Willett and Singer had articles, I think in The American Statistician saying how R2 can be adjusted to compare between models, but I think this is volatile, and subject to software quirks, as I also seem to have found an odd such result using SAS once, as I recall. 

Cheers - Jim 

PS - I assume these are continuous data?  Is that correct?

Data CRE Prediction 'Bounds' and Graphs Example for Section 4 of ...

Mithun - 

I was able to download your scatterplot, as shown here, and these data do not appear to be linear through the origin. 

Perhaps some information on the nature of the subject matter would help further discussion. 

Jim 

Dear Mithun,

I never deal data with Excel, so I could not answer if there is a bug with Excel. However, I do not suggest to drop the constant term in regression at all times, unless there is a really strong theoretical reason that supports us to do so, even that could increase the value of R2. I agree with James, we might need more information about your data for further suggestion.

Sincerely, Kuan-Wei

Hi Mithun,

I’m agree with Jason about a problem with Excel, if you see the graphic of residuals in both cases you can note that the dispersion is greatest when you forced the regression line to a intercept=0.0, then your R2 value must be lower than R2 of the “normal” regression line, the output of regression that gives R2=0.899 is wrong, and believe it or not the R2 that is calculated directly of the scatter plot is wright (R2=0.3769).

See the residuals graphics in the attached file.

Best regards

Thank you all for your suggestions and I apologize for this late reply. I am working in the field of traffic engineering and was trying to compare some field data with my modeled data. So if the two data matches, they should lie on a 45 degree line when they are plotted against each other and hence, the regression line should pass through zero. So I am a bit reluctant to include any constants.

Mithun -

That does sound like the origin should be included, and if linear, as with the continuous data shown in the example excel sheet found through the link I attached above, you should be able to put confidence around predicted y-values. But perhaps there may be some reason that this relationship might be nonlinear - a smooth curve here rather than a straight line?  -  Can you think of a possible reason for that in your application?

Jim 

Hi Jim,

Thank you for your reply. As I had mentioned earlier, I am checking if my model (hoping that it is correct) closely resembles field conditions or not. So, if my model is correct, the output from it should be same as that obtained from the field. And if I plot the two against each other, the points should fall on 45 degree line. I used regression to see how close my model predicted values are to the field value. That's when I came across this issue. Hope I was able to explain this problem in a better way.

Regards,

Mithun

Hi.  -  What I meant by possible nonlinearity is "Could your model have parameters in it that would make it perform differently over the range of x?"  

I did a great deal of work with applications of a type where a linear regression through the origin was very much appropriate, but it is always a concern that the relationship between x and y not change appreciably over x, or you may need to consider nonlinearity, or at least model in segments over ranges of x, which could lead in your case to a spline regression, which would be another form of nonlinearity, though a smoother one might be better. 

Even when doing the work I did where linear regression through the origin worked very well, you still should consider heteroscedasticity. That is considered in the excel program I linked to a previous response to this question. 

But you have to consider how your model works. 

I suggest you start by using that excel program to see whether or not your data make sense within reasonable confidence intervals, to assume a linear regression.  At a glance, however, I suspect nonlinearity in your data, meaning that your model performs differently as x increases. 

Note also that even if your situation did call for a linear regression, your model could always give you a constant percent difference, and not yield a 45 degree relationship as you described. You would only get what you described if your model exactly "predicts" reality within an interval based on the variance of the prediction error. Models are biased because they can never be exactly right. 

So if linear, and instead of b =  1, if you estimated b = 1.1, then y is always 10% more than x (give or take residuals).  Otherwise, you are assuming that except for random error, your model exactly reflects reality. 

Not knowing the model, I was saying you have to consider your application. 

Cheers. 

Hi Jim,

Thank you once again for your valuable suggestion. Let me just look at my model from a different perspective as you have suggested,

Thanks once again.

Regards,

Mithun

Mithun -

Best wishes.  It sounds like you have an interesting project. 

Note that George Box is famously quoted as saying (something like) "All models are wrong, but some are useful."  I have found models to be extremely useful in work that I did, but it depends.  In your case, I suspect modeling may be quite useful.

But remember that you are examining estimated residuals.  Your regression is an approximation to the real, unknowable model, and thus the residual is represented by e, rather than epsilon.

Jim 

PS - Ah! I guess the word "famously," regarding George Box, stuck in my head because of something Bruce Weaver said in answer to a related question. The link for that question is attached below.

Cheers!  

https://www.researchgate.net/post/Is_linear_regression_valid_when_the_outcome_dependant_variable_not_normally_distributed

hey mithun sorry for so much delay in reply but as you said i saw your file and i found that value is changing only after consider constant zero in regression and intercept zero otherwise they are showing same value. i also try this with my data also but i am not acing such kind of problem after set intercept to zero.

The same problem happens in SAS 9.3 as well when in regression without intercept (intercept = 0). They may share same module somewhere. Too bad. 

The following resources will answer the concerns posted here. R2 is defined differently for linear regression models with and without a constant; the two should not be directly compared.

https://online.stat.psu.edu/~ajw13/stat501/SpecialTopics/Reg_thru_origin.pdf

http://stats.stackexchange.com/questions/26176/removal-of-statistically-significant-intercept-term-increases-r2-in-linear-mo

Mithun -

I know this is an old question now, and hopefully all was resolved satisfactorily, but I just realized that I had missed something here. 

You noted above that you were "... trying to compare some field data with [your] modeled data."  Later you said "So, if my model is correct, the output from it should be same as that obtained from the field."  Then, say y_i, say "y sub i" is the ith observed y-value in the field.  The predicted value of y from your model, if OLS regression is used, is generally written y with a "hat" over it, but the more general weighted least squares (WLS) regression, for which OLS is a special case, using notation by G.S. Maddala, may use y* as the predicted value of y. 

So, if a "field"/observed value is y, and the modeled value is y*, then y - y* is the estimated residual.  You might, therefore, be interested in "graphical residual analysis."  If you aren't familiar with that, you might start with the following:

 https://onlinecourses.science.psu.edu/stat501/node/279/

Generally, when evaluating the performance of a regression model, besides graphical residual analyses, it would be good to have saved some test data to avoid overfitting.  In your case, I am not certain of the situation.  Were all field data actually new observations that could act as test data for an already established model?   Might you want to use some of these "field" data as test data, but most of it as new data for model selection ... or perhaps to re-estimate regression coefficients? 

If comparing the performances of two or more models, the following may contain something that might be of some use to you:

 https://www.researchgate.net/publication/313925533_Comparing_Model_Performances_Graphically

Cheers - Jim