Pooling Estimates from Multiple Imputation Logistic Regression I'm currently attempting to pool multiple logistic regression results together. I used SPSS to perform the logistic regressions using complex weights. The option to pool the results together is not available when using complex weights in SPSS. Unfortunately, the resources available to me are SPSS and Excel. I'm attempting to pool in the results for the variables (significance, Odds ratio, Confidence interval) but was not able to find how to do so. 
How would I be able to combine the results in those multiple logistic regressions?   
 A: Overview
The rules/formulae themselves have been published in several locations. Here's an accessible version (framed in terms of SAS, but the equations are under the heading "Combining Inferences from Imputed Data Sets" on page 5.)
https://stats.idre.ucla.edu/wp-content/uploads/2016/02/multipleimputation.pdf
Caveats
As noted by Amparo, for logistic regression the calculations are all done on the log scale, then exponentiated to give odds ratios.
As a reminder, one key point of multiple imputation (over single imputation) is to properly reflect the between-imputation variability into the overall precision of estimates (e.g. confidence intervals). This combination process is more complex than suggested in Amparo's answer.
Annotation:
$m$ is the number of imputed datasets;
$\bar{Q_i}$ is the point estimate for a given parameter in imputation set i
$\bar{U_i}$ is the variance for that estimate in imputation set i
Calculations for point estimates and variance
$\bar{Q}$ gives the point estimate across imputations:
$$
\bar{Q} = \frac{1}{m} \sum_{i=1}^{m}\hat{Q_i}
$$
Then $\bar{U}$ is the average variation in each imputation (called within-imputation variance)
$$
\bar{U} = \frac{1}{m} \sum_{i=1}^{m}\hat{U_i}
$$
And $B$ is the between-imputation variance:
$$
B = \frac{1}{m-1}\sum_{i=1}^{m}(\hat{Q_i} - \bar{Q})^2
$$
And then $T$ is the total variance (combined $\bar{U}$ and $B$)
$$
T = \bar{U} + (1 + \frac{1}{m})B 
$$
One can then take $\sqrt{T}$ as the standard error, and derive a Wald confidence interval as follows:
$$
Lower 95\% CI = \bar{Q} - 1.96 \times \sqrt{T}
$$
$$
Upper 95\% CI = \bar{Q} - 1.96 \times \sqrt{T}
$$
Final caveat on derivation of confidence interval
Note that the 1.96 above is useful for large-scale datasets but approximate elsewhere -- this is ideally derived with a t-distribution based on degrees of freedom (see page 5 of the linked paper, as this gets a bit more complex... I may add the formulae later but have run out of time to typeset now!.)
The recommended calculations for the degrees of freedom for this t-statistic depend on the number of imputations $m$, the average within-imputation variation, $\bar{U}$ and the between imputation variance, $B$.
A: I have exactly the same problem. Rubin's rule tells us to average the estimates (and something a little more complex to calculate the pooled variance). The estimates in a logistic model are the betas, not the exponent of the betas (odds ratios), although these are the ones we are usually interested in. My conclusion, therefore, is that I should average the betas (which are the log of the odds ratios) and their confidence interval limits; these estimates will follow Rubin's rule. Once we have averaged the betas (that is, the log of the odds ratios) and their confidence limits (again, the log of the confidence limits of the odds ratios) then we can transform them back again into odds ratios (and their confidence limits) by exponentiating them. 
With a little bit of math we can safe ourselves all the back and forth: unless I got this wrong, the exponent of the average of the betas (our pooled odds ratio of say, m imputed datasets) is equal to the (product of the m odds ratios)^(1/m). Just multiply all the odds ratios you want to pool, and calculate the m-root.
I have just looked into this solution this afternoon, and I was trying to find out what others say, so if I got this wrong I will appreciate some kind feedback. Thanks!  
