Suppose we have a baseline exposure group and 2 other exposure groups for a case control study. Suppose the odds ratio for the first exposure is $1.5$ and the odds ratio for the second exposure is $1.8$. Does this mean that cases are $1.5$ times as likely to have exposure 1 than the controls? Also does this mean that cases are $1.8$ times as likely to have exposure 2 than the controls?
Solved – How to interpret odds ratio
case-control-studyepidemiologyodds-ratio
Related Solutions
I did the following in Stata, the first is fixed effect and the second is random effect. I got different answers than you did.
Study | ES [95% Conf. Interval] % Weight
---------------------+---------------------------------------------------
1 | 2.700 1.800 4.000 63.47
2 | 1.300 0.500 3.400 36.53
---------------------+---------------------------------------------------
I-V pooled ES | 2.189 1.312 3.065 100.00
---------------------+---------------------------------------------------
Heterogeneity calculated by formula
Q = SIGMA_i{ (1/variance_i)*(effect_i - effect_pooled)^2 }
where variance_i = ((upper limit - lower limit)/(2*z))^2
Heterogeneity chi-squared = 2.27 (d.f. = 1) p = 0.132
I-squared (variation in ES attributable to heterogeneity) = 56.0%
Test of ES=0 : z= 4.89 p = 0.000
. metan or ll ul, effect(Odds Ratio) null(1) lcols(trialname) texts(200) random
Study | ES [95% Conf. Interval] % Weight
---------------------+---------------------------------------------------
1 | 2.700 1.800 4.000 55.93
2 | 1.300 0.500 3.400 44.07
---------------------+---------------------------------------------------
D+L pooled ES | 2.083 0.721 3.445 100.00
---------------------+---------------------------------------------------
Heterogeneity calculated by formula
Q = SIGMA_i{ (1/variance_i)*(effect_i - effect_pooled)^2 }
where variance_i = ((upper limit - lower limit)/(2*z))^2
Heterogeneity chi-squared = 2.27 (d.f. = 1) p = 0.132
I-squared (variation in ES attributable to heterogeneity) = 56.0%
Estimate of between-study variance Tau-squared = 0.5488
Test of ES=0 : z= 3.00 p = 0.003
It is not true in all situations. The odds ratio only gives an estimate of the relative risk if the outcome is a low probability outcome. (Same insight as Poisson approximation to the binomial distribution).
Imagine a case-control study for lung cancer, then we check the number of smokers in both groups. Technically, the only thing we can test is given that an individual has lung cancer, what is the probability that they smoke. We can do the same for the non cancer group, and obtain a ratio of the both probabilities. This would be the relative risk. But we do not really care about this quantity. We actually want, given that an individual smokes, what is the probability that they have lung cancer divided by the same probability for non-smokers.
The nice thing about the odds ratio is that it is bi-directional. So: the odds of smoking given lung cancer divided by the odds of smoking given control is actually equivalent to the odds of lung cancer given smoking divided by the odds of lung cancer without smoking. This bi-directionality of the odds ratio allows us to obtain the comparison we want from case-control studies.
Now, if we know the outcome to have a low rate in both groups $-$ the proportion of individuals with lung cancer is small among smokers and non-smokers $-$ then the odds ratio approximates the relative risk. This is the one time we can use the odds ratio to approximate the relative risk in case-control studies. I'm assuming that in case-control studies, the cases are rare events. So certain persons may skip this caveat and state what the OP stated.
The best resource I've found for questions like the one here is Agresti's book on Categorical Data Analysis.
Best Answer
It means that the odds of a case having had exposure #1 are 1.5 times the odds of its having the baseline exposure. This is not the same as being 1.5 times as probable: odds are not the same as probability (odds of 2:1 against means a probability of $\frac{1}{3}$). So it comes down to what you mean by 'likely'.