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acf_2
Fluorite | Level 6

Hello Everyone, I have this code:

 

proc logistic data=B_FINAL PLOTS = (ROC) Descending ;
	class _genderD _childrenD _educlevelD _trainingD 
			_amountinvestedD / param=ref;
	model _approvalD (event='1') = _genderD _childrenD _educlevelD _trainingD 
			_amountinvestedD 
		   
/ link=logit technique=fisher ctable pprob=0.25 pprob=0.5 pprob=0.75 rsquare expb lackfit;
	
run;

The odds ratio table shows as result:  _genderD 0 vs 1 = 0.862 (point estimate). How can I interpret this result for the gender variable (the variable is coded '1' for female and '0' for male) Thanks for any help 

10 REPLIES 10
Reeza
Super User

It means a man is 0.86 times as likely to experience the event/outcome as a female. 

 

acf_2
Fluorite | Level 6

Hello Everyone,

I have this code:

proc logistic data=B_FINAL PLOTS = (ROC) Descending ;
	class _genderD _childrenD _educlevelD _trainingD 
			_amountinvestedD / param=ref;
	model _approvalD (event='1') = _genderD _childrenD _educlevelD _trainingD 
			_amountinvestedD
		   
/ link=logit technique=fisher ctable pprob=0.25 pprob=0.5 pprob=0.75 rsquare expb lackfit;
	
run;

The odds ratio table shows as result: _genderD 0 vs 1 = 0.862 (point estimate).

How can I interpret this result for the gender variable (the variable is coded '1' for female and '0' for male)


Thanks for any help

 

Ksharp
Super User
That means when sex from Male (0) to Female (1) , the Prob of _approvalD=1 is creasing 0.862 times .
a.k.a
Prob of _approvalD=1 for Female is 0.862 times of Prob of _approvalD=1 for Male.
Female have lower Prob than Male.
acf_2
Fluorite | Level 6

Thanks for helping guys, but I am still confusing, both answers seems to me different. In the first answer, female has greater chance, in the second answer seens the opposite. 

Sorry, but I am a first time SAS user....

Reeza
Super User

I'm going to assume that the event is loan approval. 

The odds ratio is stating that a Sex=0 is 86% times as likely to get a loan compared to a Sex=1 or 14% less likely. Assuming a Sex=1 chance of getting a loan is 100% (not true) relatively, the Sex=0 would be 86% instead, or 14% less likely. 

 

Looking at it in reverse, 

 

1 vs 0 is 1/0.86 = 1.160 => a Sex=1 is 16% more likely to get a loan compared to a women. 

 

This has nothing to do with understanding SAS and has to do with understanding logistical regression models. 

It's a statistical concept that is the same regardless of whatever language you use. 

Here's a tutorial on Odd's Ratio specifically.

 

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2938757/

 

https://s4be.cochrane.org/blog/2013/08/13/a-beginners-guide-to-interpreting-odds-ratios-confidence-i...

 

Odds ratios are hard to understand the first time but make sense once you get the hang of it.

 

EDIT: Replaced male/female with SEX=# to help clarify. 

RosieSAS
Obsidian | Level 7
1 is for female, not male. So I would say a female is 16% more likely to get a loan compared to a male.
Reeza
Super User

@RosieSAS wrote:
1 is for female, not male. So I would say a female is 16% more likely to get a loan compared to a male.

Not what the original post says so I would clarify that for sure. I've updated my previous answer to refer to it generically, ie Sex=1 compared to Sex=0.

StatDave
SAS Super FREQ
If the odds ratio is 0.862, then mathematically Odds(male)/Odds(female) = 0.862. Rewriting that gives Odds(male)*0.862 = Odds(female). Expressed in words, this means that the odds of the event for females is 0.862 times the odds of the event for males. Meaning also that the odds of the event for males is less than the odds of the event for females, where according to your code, the event is _approvalD = 1. Note that the odds of the event is not the same as the probability of the event. The odds of the event is the ratio of the probability of the event over the probability of the nonevent: Pr(event)/Pr(nonevent).
RosieSAS
Obsidian | Level 7
I think "Odds(male)*0.862 = Odds(female)" should be a typo, that Odds(male)/0.862 = Odds(female). So Odds(male)<Odds(female), which means male is less likely to cause event=1 compared to female.
StatDave
SAS Super FREQ
Yes... sorry. Obviously Odds(female)*.862=Odds(male), so females have higher odds. And note that the OP states that the odds ratio is labeled "_genderD 0 vs 1 = 0.862" with males=0, meaning that the odds ratio has Odds(males) in the numerator.

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