Dear Ka Ting

You can attempt the quadratic regression treating one dependent variable. If you have theoretical basis for the curvilinear or no relationship pattern, then also u can go for curvilinear regression. It will be better than correlation coefficient, as CC measures the linear relationship between two variables. 

Dear Rajiv, 

Thank you very much for your advice. 

I am not sure how to confirm does the curvilinear relationship really present on my graph? 

I judge it on the scatter plot based on my bare eyes. I have run the quadratic on one dependent variable as you said this morning and it seems not much different. 

I just wonder would there be any way that i could sure whether is there any curvilinear relatioonship on my scatter plot? 

Many thanks for your time. 

Ka

Replace X and Y with your own variable names in the following SPSS syntax:

* Fit quadratic function using REGRESSION procedure.

COMPUTE Xsq = X**2.

REGRESSION

  /DEPENDENT Y

  /METHOD=ENTER X Xsq.

* Fit quadratic function using UNIANOVA procedure.

UNIANOVA Y WITH X /DESIGN=X X*X.

* Rsq value is show below the ANOVA summary table.

HTH.

Pls find the graph here.

Many thanks.

Looking at the graph, it's difficult to know what kind of model is best.

You _could_ make it a linear model, or a quadratic model.

But I suspect these models don't fit particularly well.  One problem is that, using Artery as the x-axis, the variance around x = 0.25 is tiny, and the variance around x = 1.25 is huge.

What to do?  One of the following:

1) It looks to me like like switching the x- and y- axes would make the data more applicable for a general linear model.

2) Here's what the data say to me:  When artery is low, vein is low.  When artery is high, vein is anywhere from slightly higher to very high.  Because of spread of the data, you might try a Cate-Nelson approach.  See:  http://rcompanion.org/rcompanion/h_02.html  . This just divides bi-variate into groups of low-low and high-high, but is really useful when the data spread out in an L shape.

4) Switch the x- and y- axes and use a linear- plateau model.  See the section called "Examples of linear plateau and quadratic plateau models" in this book: http://rcompanion.org/documents/RHandbookProgramEvaluation.pdf

3) Use a nonparametric correlation, and leave it at that.  Kendall or Spearman correlation.

I have tried to use curvilinear quadratic regression, but it seems that it didnt have much improvement to the formula. The R2 changes is very subtle/minimal after using the quadratic approach. 

Hence, i questioned myself whether will that scatter plot is actually a linear relationship instead of curvilinear relationship. 

What do you think? 

Many thanks 

Dear Ka Ting Ng,

In case of curve linear relationship is suspected, take the logarithm transformation of the y data. If correlation improves  after the transformation means there exist a curve linear relationship. If the change in correlation is not much, then you can proceed  for further analysis and interpretation assuming  that there exist a linear relationship.   

Dear Ramnath Takiar, 

Thank you so much for your suggestion. 

That is a very sensible way to try. 

Have a nice day. 

Yours sincerely,

Ka

Dear Ramnath, 

Should I use logarithm transformation for both x- and y-axis data? 

I have run a normality test on both variables. X-variable is non-normal distributed, Y-variable is normal distributed. 

I have transformed Y-data as you suggested. Parametric value (increased from 0.67 to 0.80); whilst non parametric no changes (remained as 0.68).

1. Just out my curiosity, which correlation result should I take? Presumably is non-parametric one as my variables are one non-normal and one normal distributed.?

2. Should I also transformed both variables?

Many thanks for your time. 

Yours sincerely,

Ka

I think by parametric value, you mean correlation. Right. OK. As per your description, your aim is to predict y given x, so it can be only done by parametric approach. Find out the regression coefficient, intercept, thus the regression equation for prediction. By doing x transformation, if correlation improves, you can do so. I am eager to know what is the sample size in your project? If r = 0.8, it is quite good for a model and prediction purposes.  It is generally argued that a parametric test is often more powerful than a non-parametric test. So, if normality assumption is violated to a less extent then one should not bother. 

I suggest you to find out the prediction values of y and see how much they differ from their original values. If it is within 5 to 10% fine or In case the average value of the differences is below 10%, it is fine. Please, do this and inform me then I can suggest you further. 

1. My sample size is 23.

Will do it later after my work