# Calculate confidence interval from point estimate and standard error

Im having a problem with calculating a confidence interval for an incidence rate ratio between two populations. Using PROC STDRATE from SAS, I've obtained the Incidence Rate Ratio (the difference in incidence rate between two time periods), the log rate ratio, the standard error and the z-value.

I wonder (1) why on earth did SAS not provide confidence intervals for the rate ratio? (2) How can I myself calculate these?

But this is not a mean, which is why the simple formula: SE = SD/√(sample size), probably isnt correct.

Many thanks in advance for help!

## 2 Answers

They do give you a standard error and a $z$ value, which suggests that the SAS developers thought it reasonable to use the normal distribution to approximate the sampling distribution of the statistic involved. So, if you can confirm that your $z$ statistic is equal to the estimated log rate ratio (call it $LRR$) divided by the standard error, then I would be comfortable in using $LRR \pm 2\times SE$ as an approximate 95% CI for the true log rate ratio. Supposing that this results in an interval $(\ell,u)$, then the interval $(e^\ell, e^u)$ provides a 95% CI for the incidence rate ratio itself.

• Dear Professor Lenth, Many thanks for Your answer, it surely was helpful. Have I understood it correctly, as follows: to calculate upper reference limit, I use the formula: exp(LRR + 2*SE) to calculate lower reference limit, I use: exp(LRR - 2*SE) I've tried it; the estimates land on each side of the point estimate, but the ones I've tried return a lower estimate quiet close to the point estimate. Many thanks for Your time – Frank49 Jul 30 '14 at 10:26
• The lower confidence limit will be closer to the point estimate than the upper one. But "quite close" has me concerned if you mean it's practically the same value. BTW, I'd say "confidence limit", not "reference limit". – rvl Jul 30 '14 at 13:26
• The majority of my events have a reasonable upper and lower limit compared to the point estimate (although one event has a lower limit of 1.14, point estimate of 1.17, and upper limit 1.57). Theres one more thing thats troubling me. In some events 2 *SE is greater then the log rate ratio (LRR). So if i have LRR +/- 2*SE with a negative lower limit and use that estimate in the exp function, i receive a value below zero for the lower limit, the point estimate and upper limit for the same event is above zero (e.g stroke has a point estimate of 1.17, upperlim 1.40, lower limit0.89). continues -> – Frank49 Jul 30 '14 at 16:11
• -> -> Now what I’ve done to solve this is, if LRR-2*SE = <0 then I’ve simply multiplied this value with (-1) before using the exp function. In these cases i obtain a reasonable upper and lower limit compared to the point estimate, is this incorrect and should i then simply accept a lower limit below zero when point estimate and upper limit is above zero in those events. – Frank49 Jul 30 '14 at 16:12
• NO - leave it negative! $e^x$ is still positive when $x<0$. – rvl Jul 30 '14 at 16:31
• It is not reasonable to multiply (or perform any other operation) with certain values that appears odd.
• Just like hazard ratio, odds ratio, relative risk etc, an incidence rate ratio can never be negative (as far as I know...). Higher risk will be expressed as risk >1.0, while lower risk will be expressed as ratio <1.0. I cannot recall that there exists negative relative risks.
• But it's log can be negative. That's the context – rvl Jul 31 '14 at 13:16