odds ratio to chi square conversion
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odds ratio to chi square conversion
 
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Is there a way to convert an odds ratio to a chi square value? Specifically, given a 2x2 crosstabulation of x against y, the association between x and y can be expressed as an odds ratio (OR: bc/ad), as exp(B) from a logistic regression or a chi square. Knowing only either OR or B, can a chi-square be computed? One of the places this problem comes up is in meta-analyses where authors don't report sufficient data in primary reports.
Thanks, Gene Maguin ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD |
Re: odds ratio to chi square conversion
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No. You need both sets of marginal totals to convert
an OR to a chi squared. You can show this by taking a fixed total-N and constructing various tables with the same OR and different tests. For the meta-analysis: after making sure that the information really is not there, contact the authors. They probably will oblige. A worthwhile meta-analysis is seldom easy. It wants expertise both in statistics and in the content, so, good luck. -- Rich Ulrich > Date: Fri, 29 Jun 2012 09:45:35 -0400 > From: [hidden email] > Subject: odds ratio to chi square conversion > To: [hidden email] > > Is there a way to convert an odds ratio to a chi square value? Specifically, given a 2x2 crosstabulation of x against y, the association between x and y can be expressed as an odds ratio (OR: bc/ad), as exp(B) from a logistic regression or a chi square. Knowing only either OR or B, can a chi-square be computed? One of the places this problem comes up is in meta-analyses where authors don't report sufficient data in primary reports. > > Thanks, Gene Maguin > ... |
Re: odds ratio to chi square conversion
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The researchers are not even giving you enough information to tell whether the OR is significantly different from 1. Here are 2 hypothetical tables with identical OR and N. chi-square and significance are materially different 6 12 12 6 Chi-squ = 4.00, p = .045 2 4 20 10 Chi-squ = 2.34, p = .127 Those marginals MATTER How they can imagine that reasers would not want to know the raw odds in each group I cannot imagine! Best Diana On 29/06/2012 15:58, "Rich Ulrich" <rich-ulrich@...> wrote: No. You need both sets of marginal totals to convert Emeritus Professor Diana Kornbrot email: d.e.kornbrot@... web: http://dianakornbrot.wordpress.com/ Work School of Psychology University of Hertfordshire College Lane, Hatfield, Hertfordshire AL10 9AB, UK voice: +44 (0) 170 728 4626 fax: +44 (0) 170 728 5073 Home 19 Elmhurst Avenue London N2 0LT, UK voice: +44 (0) 208 444 2081 mobile: +44 (0) 740 318 1612 |
Re: odds ratio to chi square conversion
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FYI: ~~~~~~~~~~~ Scott R Millis, PhD, ABPP, CStat, PStat® Board Certified in Clinical Neuropsychology, Clinical Psychology, & Rehabilitation Psychology Professor Wayne State University School of Medicine Email: [hidden email] Email: [hidden email] Tel: 313-993-8085 |
Re: odds ratio to chi square conversion
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That's an interesting note. I'm not sure why it makes good sense, to divide the ln(OR) by 1.81. But the writer does make good sense in suggesting that a good measure of effect size is log(OR), and what you need for further information is not the N, but the standard error. (I think he is suggesting that.) The original question was about converting an OR to a chi-squared test, in the context of meta-analysis. I doubt why anyone should want to do that, except as an intermediate step to finding the error term -- the chi-squared statistic itself is a *test* statistic, and is poorly suited for "effect size". Where it is appropriate, the OR is a fine measure of effect, and log(OR) is the version that serves as an interval-scaled measure. -- Rich Ulrich Date: Sat, 30 Jun 2012 07:44:43 -0700 From: [hidden email] Subject: Re: odds ratio to chi square conversion To: [hidden email] FYI: ~~~~~~~~~~~ Scott R Millis, PhD, ABPP, CStat, PStat® ... |
Automatic reply: odds ratio to chi square conversion
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I am currently out of the office and will return your message on Monday, July 9th.
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In reply to this post by Rich Ulrich
Rich, I haven't read the article, but my guess is that they suggested dividing by 1.81 since that value approximates the standard deviation of the standard logistic distribution, which is pi/3.
Ryan On Sat, Jun 30, 2012 at 3:46 PM, Rich Ulrich <[hidden email]> wrote:
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*Correction: sd = pi / sqrt(3) ~ 1.81 Ryan
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Re: odds ratio to chi square conversion
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In reply to this post by Rich Ulrich
Thanks, Rich. I agree with you on all points. Much appreciated. Scott ~~~~~~~~~~~ Scott R Millis, PhD, ABPP, CStat, PStat® Board Certified in Clinical Neuropsychology, Clinical Psychology, & Rehabilitation Psychology Professor Wayne State University School of Medicine Email: [hidden email] Email: [hidden email] Tel: 313-993-8085 From: Rich Ulrich <[hidden email]> To: [hidden email]; SPSS list <[hidden email]> Sent: Saturday, June 30, 2012 3:46 PM Subject: RE: odds ratio to chi square conversion That's an interesting note. I'm not sure why it makes good sense, to divide the ln(OR) by 1.81. But the writer does make good sense in suggesting that a good measure of effect size is log(OR), and what you need for further information is not the N, but the standard error. (I think he is suggesting that.) The original question was about converting an OR to a chi-squared test, in the context of meta-analysis. I doubt why anyone should want to do that, except as an intermediate step to finding the error term -- the chi-squared statistic itself is a *test* statistic, and is poorly suited for "effect size". Where it is appropriate, the OR is a fine measure of effect, and log(OR) is the version that serves as an interval-scaled measure. -- Rich Ulrich Date: Sat, 30 Jun 2012 07:44:43 -0700 From: [hidden email] Subject: Re: odds ratio to chi square conversion To: [hidden email] FYI: http://www.ncbi.nlm.nih.gov/pubmed/11113947 ~~~~~~~~~~~ Scott R Millis, PhD, ABPP, CStat, PStat® ... |
Re: odds ratio to chi square conversionfor meta-anlaysis
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In reply to this post by Ryan
Meta-analysis for 2by2 tables requires ALL 4 cells, for all studies. Do not shoot the messenger Why? Because a mean measure of effect size across all studies requires a mean effect size, WEIGHTED by some estimate of standard error of effect size measure. Total N is simply not sufficient to estimate standard errors of effect sizes. My previous example showing different Pearson chi-square for same effect size and total N is an illustration. For Ln (OR) se^2 = 1/a+1/b+1/c+1/d. No amount of jiggery-pokery can get round this. Of course this problem applies just as much to Cohen’s d comparing means for independent groups. Total N does not allow one to give greater weight to equal n/group studies than to studies with 10* as many observations in 1 group than another. Of course such inappropriate weighting, based on the dubious assumption of all groups having same variance, is extremely prevalent – but that does not mean its a good idea. Unequal n studies are prevalent in health where a disease is rare and so disease group is consderably smaller than control group Best Diana On 30/06/2012 21:38, "R B" <ryan.andrew.black@...> wrote: Rich, Emeritus Professor Diana Kornbrot email: d.e.kornbrot@... web: http://dianakornbrot.wordpress.com/ Work School of Psychology University of Hertfordshire College Lane, Hatfield, Hertfordshire AL10 9AB, UK voice: +44 (0) 170 728 4626 fax: +44 (0) 170 728 5073 Home 19 Elmhurst Avenue London N2 0LT, UK voice: +44 (0) 208 444 2081 mobile: +44 (0) 740 318 1612 |
Re: odds ratio to chi square conversion
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In reply to this post by Ryan
Ryan is correct. From the article (http://www.aliquote.org/pub/odds_meta.pdf):
"The standard logistic distribution [7] has variance pi^2/3, so a difference in ln(odds) can be converted to an approximate difference in NED [i.e., Normal equivalent deviate] by dividing by pi / SQRT(3), which is 1.81 to 2 decimal places." I suspect that the R function I mentioned in the other thread uses this method.
--
Bruce Weaver bweaver@lakeheadu.ca http://sites.google.com/a/lakeheadu.ca/bweaver/ "When all else fails, RTFM." PLEASE NOTE THE FOLLOWING: 1. My Hotmail account is not monitored regularly. To send me an e-mail, please use the address shown above. 2. The SPSSX Discussion forum on Nabble is no longer linked to the SPSSX-L listserv administered by UGA (https://listserv.uga.edu/). |
Re: odds ratio to chi square conversion
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I wanted to thank all that have replied and especially Ryan, via Bruce, for responding with the Chinn article. In the process i found out something that I didn't know because I ran a simulation yesterday. Chinn said that the normal equivalent deviate, the 'NED', which I think is the same as the d statistic [(m1-m1)/sd], given that sd=1.0, is proportional to the logit, [Ln(odds)], in the range -2 = NED = 2. The proportionality constant being pi/sqrt(3), i.e., NED = logit/(pi/sqrt(3)) or NED = logit/1.8138. I am interested in the situation with a continuous predictor, 'x'. My little simulation showed that d = sd(x)*Ln(odds) to within a few tenths of a percent. So, different from Chinn.
Gene Maguin -----Original Message----- From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of Bruce Weaver Sent: Monday, July 16, 2012 3:19 PM To: [hidden email] Subject: Re: odds ratio to chi square conversion Ryan is correct. From the article (http://www.aliquote.org/pub/odds_meta.pdf): "The standard logistic distribution [7] has variance pi^ 2/3, so a difference in ln(odds) can be converted to an approximate difference in NED [i.e., Normal equivalent deviate] by dividing by pi / SQRT(3), which is 1.81 to 2 decimal places." I /suspect/ that the R function I mentioned in the other thread uses this method. R B wrote > > *Correction: > > sd = pi / sqrt(3) ~ 1.81 > > Ryan > > On Jun 30, 2012, at 4:38 PM, R B <ryan.andrew.black@> wrote: > >> Rich, >> >> I haven't read the article, but my guess is that they suggested >> dividing by 1.81 since that value approximates the standard deviation >> of the standard logistic distribution, which is pi/3. >> >> Ryan >> >> On Sat, Jun 30, 2012 at 3:46 PM, Rich Ulrich <rich-ulrich@> wrote: >> >> That's an interesting note. I'm not sure why it makes good sense, to >> divide the ln(OR) by 1.81. But the writer does make good sense in >> suggesting that a good measure of effect size is log(OR), and what >> you need for further information is not the N, but the standard >> error. (I think he is suggesting that.) >> >> The original question was about converting an OR to a chi-squared >> test, in the context of meta-analysis. >> I doubt why anyone should want to do that, except as an intermediate >> step to finding the error term -- the chi-squared statistic itself is >> a *test* statistic, and is poorly suited for "effect size". Where it >> is appropriate, the OR is a fine measure of effect, and log(OR) is >> the version that serves as an interval-scaled measure. >> >> -- >> Rich Ulrich >> >> >> Date: Sat, 30 Jun 2012 07:44:43 -0700 >> From: srmillis@ >> Subject: Re: odds ratio to chi square conversion >> To: SPSSX-L@.UGA >> >> FYI: >> >> http://www.ncbi.nlm.nih.gov/pubmed/11113947 >> >> ~~~~~~~~~~~ >> Scott R Millis, PhD, ABPP, CStat, PStat® ... >> > ----- -- Bruce Weaver [hidden email] http://sites.google.com/a/lakeheadu.ca/bweaver/ "When all else fails, RTFM." NOTE: My Hotmail account is not monitored regularly. To send me an e-mail, please use the address shown above. -- View this message in context: http://spssx-discussion.1045642.n5.nabble.com/odds-ratio-to-chi-square-conversion-tp5713894p5714260.html Sent from the SPSSX Discussion mailing list archive at Nabble.com. ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD |
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