Re: odds ratio to chi square conversion

Posted by Rich Ulrich on Jul 16, 2012; 7:27pm
URL: http://spssx-discussion.165.s1.nabble.com/Re-odds-ratio-to-chi-square-conversion-tp5714198p5714261.html

Two samples separated by 0.5 SD will have their
overlap of histograms at z +/-  0.25.

My convenient normal table tells me that a z-score
of 0.25 (half of 0.5) contains about 10% of the area.

So the implied sample separation is 60/40 vs. 40/60 --
  OR = 3600/1600 = 9/4 = 2.25.
 
If you don't have the formulas handy to manipulate
(or maybe, even if you do), the easy way to find
what the equation would be is to dummy up some
100 lines of data to be approximately Normal(0, 1);
adjust to mean=0 exactly; multiply by 10 to create
variance= 100; then add 45 to half, 50 to the other
half, for Group 1 and 2.  Then run logistic on it.

--
Rich Ulrich

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Date: Mon, 16 Jul 2012 14:18:14 -0400
From: [hidden email]
Subject: Re: odds ratio to chi square conversion
To: [hidden email]

Thanks, Rich. My question is how. For instance, suppose x is a true-enough dichotomy and y is a continuous variable.

Let

n(1) = 50; mean(1) = 45; SD(1) = 10

n(2) = 50; mean(2) = 50; SD(2) = 10

Analyzed as a t-test, the standard ‘d’ effect size is 0.5.

 

But, what would the computation to get, most importantly, the corresponding OR and, secondarily, the intercept?

 

Thanks, Gene

 

 

From: Rich Ulrich [mailto:[hidden email]]
Sent: Monday, July 16, 2012 2:01 PM
To: Maguin, Eugene; SPSS list
Subject: RE: odds ratio to chi square conversion

 

Diana K. emphasized the problem of assumptions.
But, to turn assertion around the other way -- If you make
assumptions about normality, you surely can get estimates
of the coefficients of the simple logistic regression.

And I say "about normality" because assuming that they
are "normal" is not the only choice.  You could assume
some degree of skewness (say), and use Monte Carlo
randomizations to estimate what the LR results would be
under various assumptions.

However, it does seem to me that the "effect sizes" in
terms of mean differences, etc., is what you will need
for any power analysis, rather than the LR results.