I understand you can get the log of the risk for each group (lnR1 and lnR2,
for example) and their corresponding standard errors (se_lnR1 and se_lnR2).
Consider the following approach:
a) Get a large number of values from the distribution of the risk in each
group using the -uniform- function:
gen A = lnR1 + se_lnR1 * invnorm(uniform())
gen B = lnR2 + se_lnR2 * invnorm(uniform())
b) Calculate the relative risk using A and B
gen rrisk=A/B
c) get the values corresponding to the 0.025 and 0.975 percentiles of the
distribution of "rrisk"
local l025 = .025 * _N
local u025 = .975 * _N
These two values should correspond to a 95% confidence interval for the
relative risk. Maybe other members of the list could comment on the
appropriateness of this approach.
I hope this help,
Leonelo Bautista
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Michael Ingre
Sent: Tuesday, November 02, 2004 11:43 AM
To: [email protected]
Subject: Re: st: RRR with CI from logit model
On 2004-11-02, at 18.19, Constantine Daskalakis wrote:
> Look at the estimates and estimated standard errors for the different
> situations in your example. You'll probably find that the estimated
> RRR increases but its estimated standard error goes to hell (increases
> much more). This is a problem with Wald-type tests that has been
> pointed out before (eg, see Hauck & Donner, JASA 1977, w/ corrrection
> in 1980). The delta method (especially for anti-log functions of
> coefficients) seems to exacerbate that.
Thank you, I think you got it. But what is the solution to the problem?
Am I stuck with just OR? Is there a way to calculate RRRs with CIs
directly from predicted probabilities with CIs instead?
Michael
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