2
$\begingroup$

Im doing a multiple imputation of a dataset using R's MICE package.

imp <- mice(nhanes, m=5, print = FALSE, seed = 55152)

I figured out that to pool regression coefficients you really only need to get the mean of the 5 regression coefficients for the 5 datasets.

But now i need to pool means, confidence intervals and standard deviation using Rubin's rules.

How do i do that?

/Kind regards

$\endgroup$
1
$\begingroup$

You can use the pool function that comes along with mice. In ist helpsite you will find the following example:

imp <- mice(nhanes, m=5, print = FALSE, seed = 55152)
fit <- with(data=imp,exp=lm(bmi~hyp+chl))
summary(pool(fit))

what gives

                    est         se          t       df     Pr(>|t|)       lo 95       hi 95 nmis       fmi     lambda
(Intercept) 19.63676903 4.33084987  4.5341606 15.73596 0.0003524824 10.44324131 28.83029675   NA 0.2394702 0.14858446
hyp         -0.43069297 2.07375135 -0.2076879 18.41666 0.8377520514 -4.78042837  3.91904242    8 0.1563331 0.06943169
chl          0.03803107 0.02241891  1.6963831 14.99264 0.1104714393 -0.00975576  0.08581789   10 0.2619634 0.16966640

Unfortunately, at the moment, I'm too blind to see/find the smart solution. So let's do the pooling according to Rubin's rules by hand. If we assume $x_i, ~ \ldots, x_n$ to be iid samples from $N(\mu, \sigma^2)$, then $\bar{x} \sim N(\mu, \sigma^2/n)$. We will need $\sigma^2/n$ for the within variance:

n <- nrow(nhanes)
m <- 5
Q <- array(dim = m)
U <- array(dim = m)

for (i in 1:m){
  Q[i] <- mean(complete(imp, i)$bmi)
  U[i] <- var(complete(imp, i)$bmi)/n
}
B <-  var(Q)
mean_of_means <- mean(Q)
total_variance_of_means <- mean(U) + (1 + 1/m) * B
$\endgroup$
  • $\begingroup$ This answers how i get the regression coefficient and the associated confidence interval. But im interested in the mean. $\endgroup$ – Simon Aug 30 '16 at 5:16
  • $\begingroup$ Looks like I didn't get your question right. So, you are interested in the mean of what? The variables (like imp$bmi)? $\endgroup$ – Qaswed Aug 30 '16 at 12:12
  • $\begingroup$ Exactly, it could be the mean and standard deviation/confidence interval of imp$bmi. Thanks for your patience. $\endgroup$ – Simon Aug 31 '16 at 15:42

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.