1
$\begingroup$

I do have the following data

TABLE A
Population Size | Type    | Value
340             | Control | .7+03[SD]
350             | Test    | .015+.25[SD]

Can I convert this to something like this? How ?

TABLE B
Population Size | Type    | Value
340             | Control | 25
350             | Test    | 33

If not how to I convert TABLE A into a Odds Ratio (based on Mean and SD only)

$\endgroup$
1
$\begingroup$

It would be possible if the mean is the mean of a binary variable coded using the values 0 or 1. In that case, the mean would actually be the proportion of 1s. In that case the odds of a 1 would be $\frac{0.7}{1-0.7}\approx 2.3$, i.e. within the control group we expect to find 2.3 1s for ever 0. The odds of a 1 for the test group would be $\frac{0.015}{1-0.015}\approx 0.015$. The odds ratio would than be just the ratio of those odds: $\frac{\frac{.7}{1-.7}}{\frac{0.015}{1-0.015}} = \frac{0.7}{1-0.7} \times \frac{1-0.015}{0.015}= \frac{0.7(1-0.015)}{(1-0.7)0.015}\approx153$, that is the odds of a 1 in the control group is a 153 times larger than the odds of a 1 in the test group.

However, your means are not proportions, so this method won't work for you. I know this because if the mean for the control group were a proportion, then the standard deviation would have been $\sqrt{0.7*(1-0.7)}\approx0.46$, which is completely different from the standard deviation you report in the table.

If you want to find out if there is anything possible, then we need to know more about your problem: In particular what is your dependent variable and how is it measured?

$\endgroup$
  • $\begingroup$ Thanks Maarten, actually the value that I have is cholesterol level for each of the population (with drug and control sets). so, yes the means are not proportions. Here is $\endgroup$ – rlpatrao Jun 1 '17 at 9:15
  • $\begingroup$ Here are actual values <pre> 348 Placebo 0.07+0.15 [SD] 339 Ra20mg 0.17+0.2 [SD] 114 Placebo 0.3+0.68 [SD] 114 Oxxxx20mg -0.05+0.76 [SD] </pre> $\endgroup$ – rlpatrao Jun 1 '17 at 9:24
  • $\begingroup$ So the dependent variable is not binary? $\endgroup$ – Maarten Buis Jun 1 '17 at 9:50
  • $\begingroup$ No, they are not. $\endgroup$ – rlpatrao Jun 1 '17 at 10:47
  • $\begingroup$ Then the whole concept of Odds Ratio does not make sense for your problem. So it cannot be computed. $\endgroup$ – Maarten Buis Jun 1 '17 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.