Question 730580: I am trying to calculate the 95% confidence interval for a given odds ratio (OR) using the formulas
a = exposure and outcome present
b = exposure and outcome absent
c = no exposure and outcome present
d = no exposure and outcome absent
OR = (a/c)/(b/d) which can be rewritten as OR = (a)x(d)/(b)x(c)
95% CI for the OR = [1n (OR) +/- 1.96√1/a + 1/b + 1/c + 1/d)]
OR = (285)x(111)/(98)x(268) = 1.21
So I used a natural logarithm table to find the ln for the OR I calculated. ln for 1.21 = 0.1906
95% CI = [1n (OR) +/- 1.96√(1/285 + 1/98 + 1/268 + 1/111)] = 0.1906 +/- 0.1567
= 0.0339 to 0.3473 = exp(0.0330) to exp(0.3473)
So,
OR (1.21) [95% CI = 1.03 to 1.42]
I think that is correct. My problem is, how do I know what the natural log is for an OR that is less than 1.0? I have an OR = 0.95 and I don't know how to calculate the 95%CI for that OR. The natural logarithm tables are for numbers 1.0 and higher.
Thanks you! Answer by lynnlo(4176) (Show Source):
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