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Hi Micheal, Thank you for your reply. I tried the solution by using the same method that you have suggested. I assumed a value of n so that Z = X^n. Now it forms an ordinary linear equation Y = A + B*Z. I tried by assuming various values of n. I plotted a graph with n values on the x-axis and R square values on the y-axis. Finally I chose the value of n that gives maximum R square. With that value of n, I was able to estimate A and B. The value of n I chose is 0.196 I was able to do this as I had an idea about the range of n. But if I don't have an inkling about the range of n, such assumptions may make my predictions erroneous. To make sure that my predictions are correct, can you tell me the parameters that should be investigated (apart from R square) ? Some say that one should not go blindly by the value of R square. The difference in predicted and actual values should be considered.i.e, Standard Error should be considered. Can you suggest some tests to make sure that predicted n value is correct ? Thanks, Sankar. >From: "Graziano, Michael" <[log in to unmask]> >Reply-To: Concerned with the initial learning and teaching of statistics > <[log in to unmask]> >To: [log in to unmask] >Subject: Re: Polynomial Regression >Date: Mon, 19 Mar 2001 13:27:11 -0500 > >Hi Sankara, You can transform the data as follows. For each x, define z = x^n. Then, if you work with the z's you have Y = A + B*Z which is solved via OLE (Ordinary Linear Estimation). The A and B estimated in this >fashion are also the A and B required in the original problem. >Hope that helps. > >Rgds > >mjg ([log in to unmask] ; 203-353-8100 x277) > >======================================================================= > >I have a set of values for Y and X. Y is the response variable and X is the >independent variable. > >The trend is as follows: > >Y = A + B*X^n > >I want to get the values of A,B and n. Can you help me in obtaining the >regression solution ? _________________________________________________________________________ Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com.