Proper Interpretation of odds ratio less than 1 for my study

#1
Hello everyone,

This is for a PhD, research in psycholinguistics.

I have done logistic regression analyses where some chosen predictor predicts the likelihood of the event happening. That is my rudimentary summary of what I am doing. Let me quickly define my event and non-event. These consist of difference scores that have been made binary. So if a difference score was positive, I coded it as 1; if it was negative, I coded it as 0. In my study this difference is very important. 0 does not mean simply NOT 1. It actually means that my participants were doing something else entirely. Interpreting the odds ratio correctly is crucial.

I am getting several odds ratios less than 1. For example, 0.881. I am told that this means: For every one unit increase in the independent variable, there is an 11.9% decreased likelihood of the event happening. If this is correct, it only partially answers what I need to know.

My question, how much does this tell me about the likelihood of the non-event happening? And how can I calculate that? I've read up quite a bit online about calculating reciprocals (because the odds ratio is invertible) but I get stuck and confused every time. And my Andy Field textbook doesn't address this in the detail that I need.

I've tried to provide as much detail as I think you need but not so much that it is superfluous and boring. If you can at all help me understand this, I would really appreciate it.

MariaF.
 

noetsi

Fortran must die
#2
In terms of the logistic regression "not one" and "doing something entirely different from one" are actually the same thing. You are either 1 or 0 - what being in one or zero means substantively is not pertinant to the method.

It is traditional, because the meaning of odds ratios less than one are not intuitively obvious, to reverse the event you are maximising when you have odds ratios less than one. That is maximize 1 rather than 0 if you were initiall maximising 0 (the software have different defaults of which is being maximised). This will cause odds ratios less than one to now be greater than one. You then interpret the odds ratio in terms of what is being maximized (which of course is the opposite of what had been maximized).

Say you were initially maximising 0 and you get a odds ratio of .75. So you change the coding to maximize 1 instead. The software will automatically generate the new odds ratio for you which will be the reciprocal of the previous finding - you don't have to do this yourself. You would then say the odds of being in 1 for a one unit change in the predictor are....(whatever the new odd ratio is, I think 1.33).:p

What software are you using.
 
#3
Hi noetsi,

Thank you very much for the reply. I am using SPSS and I believe it maximised 1 but I can't say for sure, as I am still learning all this. I have a pretty equal proportion of odds ratios greater than 1 and less than 1. I've interpreted the greater than 1s but clearly got stuck when I got to the less than 1s. So are you saying that the reciprocal of an odds ratio less than 1 (1/odds ratio) gives me the odds ratio from which I can calculate the increased likelihood of non-event happening (rather than the decreased likelihood of the event happening)? For example, in the case of what you wrote above, with an odds ratio .75 there is a 25% decreased likelihood of the event happening. Calculating the reciprocal: 1/.75 = 1.33. Therefore, the change in odds is 1.33-1=.33. Therefore there is a 33% greater likelihood of the nonevent happening. Am I correct?

Thank you so much for taking the time to read and answer me. Much appreciated!
Maria
 

noetsi

Fortran must die
#4
Yes SPSS maximizes 1

So are you saying that the reciprocal of an odds ratio less than 1 (1/odds ratio) gives me the odds ratio from which I can calculate the increased likelihood of non-event happening (rather than the decreased likelihood of the event happening)?
You can calculate the odds ratio of the non-event (if by that you mean the state opposite the one you were calculating 0 rather than 1 in SPSS normally). But the percent change point you discussed is something I am unclear on. Odds Ratios are about odds not liklihood (which I believe are relative risk). The relative risk is not the same as the odds ratio at all (except in very rare cases).

Whether you can do what you did in the calculations above I don't know. I have never seen anyone do that. Normally when you use odds ratios you just discuss the odds ratio not liklihood (which again I think may be relative risk).

You might want to look at this and see if relative risk is what you are interested in. I am really not clear again about change in the odds, something I have never seen discussed. I do not know in honesty if you can even calculate that.

http://www.broadinstitute.org/~eliana/Papers/when_to_use_odds_ratio_relative_riskfulltext.pdf
 
#5
As noetsi said, odds ratio means the ratio of odds of some outcome with and without some treatment (or risk factor, or in general an independent variable). So if OR is greater than 1, it shows that the odds of that outcome happening in the presence of the independent variable is greater than the odds of the same outcome in the absence of that independent variable. We have an OR = 2.5 for smoking's effect on a cancer. We can say: Cancer is 2.5 times more likely to occur in smokers. Or, we can say smoking might increase the odds of cancer for 1.5 times (1.5+1 = 2.5).

For ORs less than 1, all the above holds, except that all the words "increase" and "more" would be replaced with "decrease" and "less' etc., AND we would use the reciprocal of OR for our interpretation, instead. For example, OR = 0.5 for a treatment of blood pressure. We should say "High blood pressure is two times (two as the reciprocal of 0.5) less likely in the presence of the treatment", or we also can say "the odds of high blood pressure in the presence of this treatment is half the odds of the same disease without treatment".

When you want to discuss ORs in terms of changes in percentages, every unit of OR means a 100% change, when the change is positive. When the change is negative (OR less than 1), we can take the reciprocal of OR. Then subtract it by 1. For example, OR drops from 1 to 0.1. It means that the disease is 10 times less likely to occur with treatment. So we have 10 (reciprocal of 0.1) - 1 (baseline OR) = 9 = 900%. So in this case, our treatment reduced the odds of that disease for 9 reciprocal OR units, or for 900%.

So are you saying that the reciprocal of an odds ratio less than 1 (1/odds ratio) gives me the odds ratio from which I can calculate the increased likelihood of non-event happening (rather than the decreased likelihood of the event happening)? For example, in the case of what you wrote above, with an odds ratio .75 there is a 25% decreased likelihood of the event happening. Calculating the reciprocal: 1/.75 = 1.33. Therefore, the change in odds is 1.33-1=.33. Therefore there is a 33% greater likelihood of the nonevent happening. Am I correct?
Maria
I think you are.
 
#6
Dear noetsi and victorxstc,

Thank you both very much for your replies. It has really helped a lot. I had also found this website that had sort of lead me to believe there had to be at least some component of truth in what I was saying: http://www.ats.ucla.edu/stat/mult_pkg/faq/general/odds_ratio.htm
If the word "likelihood" is problematic and not used in discussions of odds ratios, I'll just change it to "increased/decreased odds."
And thank you, victorxstc, for confirming the possibility of discussing it in percentages.

You've both helped me a lot with a 2.5 year-old dissertation on the verge of getting finished! I appreciate it so much!

Maria
 

noetsi

Fortran must die
#7
Congradulations on the disertation. It is always a great relief to be done with those:)

Mine took a lot longer than 2 and a half years....
 
#8
Sorry I never acknowledged this, noetsi. Thank you for congratulating me. I've been getting tons of pressure to finish; they wanted me to finish in one year after my qualifying exams. When I didn't make it, my funding ran out at the university and I had to get a job. So I really can't wait to get this done. One more chapter to go...

:)
 
#10
Hi all,

Thanks for all the explanations, but after reading this post over and over, I am still a bit confused. Also because my situation is somewhat different:

Model 1:
I have a dichotomous outcome: yes/no diagnosis
I have a continuous predictor: score A

Model 2:
I have a dichotomous outcome: yes/no diagnosis
I have a continuous predictor: score B

For model 1 I have an OR above 1 (Exp(B), thus the OR, =1.17), I interpreted this as:
"the odds of having a diagnosis was 1.17 times greater with one point increase in score A"
I think this is correct isn't it?

For model 2 I have an OR below 1 (Exp(B), thus OR, =0.91). Based on the previous posts I figured I had to take the reciprocal of 0.91 (being 1.10), so I interpreted this as:
"the odds of having a diagnosis was 1.10 times smaller with one point increase score B"
Is this correct?
I started doubting because somewhere else I read that I should not take the reciprocal but instead do 1-0.91=0.09 and use this number for my interpretation. Although I am not sure how exactly I would use this number in my interpretation...

I hope you can shed some light on the different numbers and ways of interpreting my OR's!
Thanks!
 

noetsi

Fortran must die
#11
For model one that looks right although its more common to say a one unit increase than a one point increase.
The same for 2.

I have seen the reciprocal used many times so I am sure it is right. I don't know the second method you mention.

This comes from a pretty good source

Thus the final recommendation for those of you engaged in logistic regression and similar analyses using odds or probability ratios-- when possible, refrain from reporting RRs or ORs less than 1.0. It would make sense to standardize the reporting of this effect size so that all ratios be reported as >1.0. Analyses that result in ratios less than 1.0 would take the inverse of the RR or OR, and reverse the categories or the description of the results to keep the conclusion consistent.
http://pareonline.net/getvn.asp?v=11&n=7
 

noetsi

Fortran must die
#12
This contradicts much of what I have read in the past and I wondered what others thought about it.

Using the data above, if one wants to know if boys are at greater risk of being recommended to remedial reading then girls, we can calculate a relative risk by dividing the probability for boys by the probability for girls (.35/.10) which yields a relative risk of 3.50. In other words, boys are 3.50 times more likely to be recommended for remedial reading than girls. This is intuitive, yet this statistic is rarely the one reported in research. Odds ratios are much more common, partly because many popular software packages readily report ORs. The odds ratio for these data is the odds for boys divided by the odds for girls (.54/.11) which yields an odds ratio of 4.91. In this case, the odds for boys are 4.91 that of girls.
However, that does not mean one can say that boys are 4.91 times as likely, or 4.91 times more likely to be recommended to remedial reading than girls. Technically, the odds of being assigned are 4.91 times greater for boys relative to girls. But since odds are tricky to understand, the meaning of this is less clear. Technically, it means that for every boy not recommended to remedial reading, 4.91 times as many boys will be recommended for remediation (0.54) than the number of girls recommended for every girl not recommended.
http://pareonline.net/getvn.asp?v=11&n=7

I was wondering if anyone could say how one gets P0 here from standard statistical output....ideally SAS :)

RR = OR / [(1-P0)+ (P0 x OR)] (1)
where RR=relative risk, OR= calculated odds ratio, and P0 = the proportion of non-exposed individuals that experience the outcome in question. In the case of our example, P0 would be .10, the probability that girls would be referred to remedial reading, and the OR is 4.91. Completing the calculations we end up with an estimated RR of 3.53, a very close approximation to the actual RR of 3.50.
 
#13
Also to throw in the mix, when the outcome is rare, typically construed as < 10%, the OR and RR are fairly comparable.
 
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#15
Outcome death and <10% of people die.


I believe what goes down is in odds you divide p/q and in risk you p/p+q. If p is small then the denominators start to become comparable.
 
#16
I am sure that I have used genmod, but not regularly. I think of it more like "lm" in R. It is a platform where you then specify the family of regression you are interested in.


It comes back to, can you use the phrase risk with retrospective data, since it implies the exposure is established before outcome.
 

noetsi

Fortran must die
#17
death isn't always the outcome in a model, at least not mine :p

Genmode has lots of limitations notably for diagnostics. You would only use it to do RR.
 
#18
Using OR is fine. The general public shouldn't be reading actual journal article anyways. It is too easy for them to get misguided. It is better, for them to read synopses designed in layman speak. Unfortunately, some believe the same thing about doctors. There are many publications on how easily results can be misinterpreted.


A last note, relative risks are not invertible. Thus you can't use the reciprocal method like in OR.
 
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noetsi

Fortran must die
#19
The general public shouldn't be reading actual journal article anyways.
Elitist :p

In honesty reading that article I was not clear substantively what OR meant any more so I could not explain it to others at a substantive level. Technically I know it means the odds of something are X greater than the odds of something else, but that has little intuitive meaning to me. On the other hand what the probability of something is relative to the probability of something else is fairly straightforward.

I know I have told people routinely that a specific OR means that something is more likely than something else. And in fact that is not correct, it is the odds of something that is more or less likely using the OR.

Technically, the odds of being assigned are 4.91 times greater for boys relative to girls. But since odds are tricky to understand, the meaning of this is less clear. Technically, it means that for every boy not recommended to remedial reading, 4.91 times as many boys will be recommended for remediation (0.54) than the number of girls recommended for every girl not recommended.