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I have a survey data set with missing values and I generated 10 multiple imputations in which the missing values were imputed. There are several categorical variables in the data sets and I'd like to see if there's any association among these variables using chi-squared test.

However, I've been searching how to run this analysis in either SAS or Stata but didn't find a solution. It seems that mi estimate doesn't support svy: tab in Stata. I can run chi-squared test on each imputed data set using in SAS using proc surveyfreq, however, I don't know how to combine the results. Can anyone please guide me through the steps on how to test the associations between categorical variables?

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How to do this has been elaborated by Rubin (1988, p.87) and Li et al. (1991).

First, you take mean $\chi^2$ across multiply imputed data sets.

$$\chi^2=\frac{1}{m}\sum _{l=1}^{m}\chi^2_l$$

You estimate the relative increase in variance:

$$r=\left(1+\frac{1}{m}\right)\frac{1}{m-1}\sum _{l=1}^{m}\left(\sqrt{\chi^2_l}-\overline{\sqrt{\chi^2}}\right)^2$$ where

$$\overline{\sqrt{\chi^2}} = \frac{1}{m}\sum _{l=1}^{m}\sqrt{\chi^2_l}$$ And the test statistic:

$$D_x=\frac{\frac{\chi^2}{k}-\frac{m+1}{m-1}r}{1+r}$$

where $k$ is the degrees of freedom of $\chi^2$. Now, $D_x$ is F-distributed:

$$P_x = Pr(F_{k,v} \gt D_x)$$

where

$$v=k^{-3/m}(m-1)\left(1+\frac{1}{r}\right)^2$$

which gives you the Chi-square test across multiply imputed data sets.

Li, K.-H., Meng, X.-L., Raghunathan, T. E., & Rubin, D. B. (1991). Significance levels from repeated p-values with multiply-imputed data. Statistica Sinica, 1(1), 65–92.

Rubin, D. B. (1988). Multiple Imputation for Nonresponse in Surveys. John Wiley & Sons.

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  • $\begingroup$ @user32145 was this any useful? $\endgroup$
    – tomka
    Commented Dec 4, 2013 at 13:02
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    $\begingroup$ I found it useful, but there is an error: In the expression for $r$, the $1/(m-1)$summation term is the variance of the square roots of the test statistics from each imputed data set. As such, the second term should be the mean of the square roots of all test statistics, not the square root of the mean test statistic. $\endgroup$
    – Meg
    Commented Mar 8, 2014 at 23:06

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