Calculating number needed to treat with confidence intervalls when using multiple imputation

[Christian Meyer]

Dear Statalist users!
I am going to analyze data from an RCT. To handle "lost to follow-up" I imputed missing outcome data (dichotomous measure) with the ice procedure. Now I would like to compute the “number needed to treat” (NNT) with confidence Intervals (ci). I got the estimate for the proportion, se and ci for those getting cured by "mi estimate: proportion cured , over(group). So I can calculate the risk difference and NNT by hand. But I don´t know how to calculate the ci.

I would be very grateful if anyone could give advice.
Christian Meyer

[Clyde Schechter]

The NNT is just the reciprocal of the risk difference. So all you need to do is take the confidence interval for the risk difference and take the reciprocals of both endpoints (and reverse the order). The resulting interval necessarily has exactly the same coverage probability.

[Christian Meyer]

Sorry for asking again, but i also do not know how to calculate the ci for the RD (across all imputated data sets)

[Clyde Schechter]

So, instead of using -mi estimate: proportion cured, over(group)-, do -mi estimate: regress cured i.group-. The coefficient of the group variable will be the estimated risk difference, and the CI for that will be the CI for risk difference.

[Rich Goldstein]

be careful if the CI apparently includes 0; note that I am not a fan of NNT and some comments by me can be found at http://biostat.mc.vanderbilt.edu/wiki/Main/NntProblems

[Christian Meyer]

Thank you very much Clyde, I will try this. Never thought about OLS regression with a dichotomous dependent variable before.

[Svend Juul]

I am a bit worried about Clyde's suggestion to use regress. I would be less worried to use binreg with the rd option. The estimated risk difference is the same, but the confidence interval is different. Try these:

webuse lbw.dta , clear
regress low smoke
binreg low smoke , rd

I agree with Rich that NNT can become messy and can complicate communications more than helping it.

[Christian Meyer]

Is there any option to calculate the ci based on the Wilson score method when using mi data?

Thank you Rich for good advice, so I may drop ci for non-significant differences. Or may I try the reporting proposed by Altmann BMJ, 1998; 317(7168): 1309–1312, e.g. NNTH 3.7 to ∞ to NNTB 9.1 or is this outdated?

[Christian Meyer]

Just to complete the thread. The calculation via regess as suggested by Clyde exactly replicated the (asymptotic) ci generated via epitab command ci. The binreg solution yield ci close to but not identical with results calculated by bcii under the label "Bender´s 95% CI" (which is known to be more accurate and conservative compared to the asymptotic one). I assume the binreg solution yields "exact confidence intervals".

[Svend Juul]

I guess that Christian is talking about the epitab command cs. I find:

webuse lbw.dta

                         RD             95% CI
regress low smoke      0.1532       0.0181, 0.2884
binreg low smoke, rd   0.1532       0.0161, 0.2904
cs low smoke           0.1532       0.0161, 0.2904

that is, the same point estimate. binreg and cs also give the same CI while regress give a slightly different CI.