For a Dicotomous IV, the OR against DV is lesser then 0.
Does it means a protective effect.
decrease in the IV will increase DV.
Probably your results are being presented as coefficients, not odds ratio.
The coefficients can be negative, but when you apply the exponential function to convert to odds ration, the previous negative values will be between zero and one.
What software are you using? Some already present the results in odds ratio, like Stata and SPSS. But for other, like R regression packages, you need to manually apply the exponential function on the coefficients.
And yes, if the coefficient is negative (so that, after exponentiation, the odds ratio is a decimal fraction between zero and 1.0), that means the relationship is inverse -- increases in the IV are associated with lower odds on the DV.
Marcus Vinicius Veber Lopes i m using stata.
Case control study.
Vit D is the independent variable and is divided into 10.
10 the OR is 0.04.
if it is a protective factor,how to explain it.
what i think is,vit D >10 is protective for disease.
Your OR of .04 says the odds of having the disease are multiplied by a factor of .04 if vit D > 10, compared to the odds if vit D < 10. In other words, the odds are about 96% lower when vit D is above the reference value.
When you have OR < 1.0, you might find interpretation easier if you take the reciprocal: 1/.04 = 25. Then you can say, the odds of the disease are 25 times as large if vit D < 10, compared to the odds if vit D > 10. Mathematically, it makes no difference which way you say it, but "25 times as large below the reference value" might be a little easier to understand than ".04 times as large above the reference value).
Burke D. Grandjean reference value =10, Exposure to disease is 96% reduced.
Correct me if i am wrong.
That's not exactly the way to say it. If your DV is "do you have this disease, yes or no?" then EXPOSURE to the disease is not 96% reduced. Some people might have been "exposed" to the disease without contracting it.
The ODDS of actually having the disease are reduced 96%. "Odds" is the ratio of the proportion (or the number) who have it, divided by the proportion (or number) who don't. If the disease is quite rare, the odds will be close in numerical value to the probability, but odds and probabilities are two different things. Probability, of course, is the number who have it divided by the total number in the sample. In other words, the probability is the proportion who have it, not the ratio of two proportions.
But if your DV is "have you been exposed to this disease, yes or no?" then in that case your way of saying it would be correct.
Nasira Munawar Lone,
The "logit" command for binary outcomes in stata present results as coefficients that can be negative. You can use the same sintax with the command "logistic" to get results already in odds ratio.
"logistic DV i.IV"
In regard to the interpretation of your output, I agree with Burke. The "The odds of actually having the disease are reduced 96%" its a safe presentation. Some could call the "probability of having the disease" to make it easier to understand, but the term "probability" is not correct in terms of the calculation procedures, as Burk have already explained.
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