On 2004-11-02, at 19.02, Leonelo Bautista wrote:
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,
This is a very neat solution. I will try it today and see how it
performs. To me it looks like a feasible "bootstrap" approach. But
then, I'm not a statistician. I think though that the code needs to be
slightly modified to describe risks:
a)
gen random_a = log_odds_se * invnorm(uniform())
gen random_b = log_odds_se * invnorm(uniform())
gen risk_a = exp(log_odds_a + random_a) / (1 + exp(log_odds_a +
random_a))
gen risk_b = exp(log_odds_b + random_b) / (1 + exp(log_odds_b +
random_b))
Michael
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