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Date:         Wed, 18 Oct 2006 20:10:38 -0400
Reply-To:     Peter Flom <flom@NDRI.ORG>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Peter Flom <flom@NDRI.ORG>
Subject:      Re: How can I do this type of analysis
Comments: To: sasncsu@GMAIL.COM
Content-Type: text/plain; charset=US-ASCII

>>> Xu Yuan <sasncsu@GMAIL.COM> 10/18/06 6:13 PM >>> wrote <<< I am a rookie in statistics, so please don't laught at me if my question is silly. >>>

People who laugh at beginners are, I think, displaying their own insecurity or something.

<<< I have two columns of data (X and Y), which come from the same group of experiment units but using two different methods. My objective is to see if the two methods are (statistically) equivalent. The ideal situation is Y=X, but this world is never ideal. I tried to fit linear regression in MS Excel and the default model is Y=a*X+b with a R2. My question is what technique should I use to achieve my goal (compare the two methods). Should I force the intercept to zero? Or should I specify the slope to 1? (if so, please tell me how). Please advise me. I can work in either MS Excel or SAS or R. I am considering using paired t-test, but I am not if this is the right procedure. >>>

Never use Excel for statistics. Don't use a hammer to cut things (especially if the hammer is defective). R is an excellent package, but has, I think, a steeper learning curve than SAS. Besides, you wrote to a SAS list.....There is also an R list, but it's shark infested waters, whereas SAS-L is like a kiddie pool :-)

What are X and Y, and what are the methods? Where did the data come from? What is sample size? I'm going to take a shot and assume they are both continuous variables.

What do you mean by 'equivalent'? If you want to see how strong the connection is between the two variables, and are willing to assume there is a linear relationship, then you can correlate them (PROC CORR).

But I really need to know more to give a good answer. If you write back (to SAS L, not just me) with some more detail, I am sure a better answer will follow

Peter


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