For a method, I'm calculating it's sensitivity and specificity. I also want to calculate standard errors, but I'm unsure how. I don't have a sample to calculate it from. All I have are 50 iterations where I applied the method to a different dataset each time. So that, I have 50 values for the number of TPs, FPs, TNs, FNs. From this, I calculate the sensitivity and specificity by summing all TPs, FPs, TNs, FNs.

I haven't been able to find much information online, aside from a wikihow article, which states that it can be calculated as:

$\sqrt{\frac{(1-Sensitivity)*Sensitivity}{n_p}} * 1.96,$

where $n_p$ is the number of positives in the dataset. Is this correct?


1 Answer 1


Let's define some notation.

$y\in\{-1,+1\}$ is the true state.

$\hat y\in\{-1,+1\}$ is the predicted state.


$$ \text{Sensitivity} = \mathbb{P}\big(\hat y = +1 \vert y = +1)\\ \text{Specificity} = \mathbb{P}\big(\hat y = -1 \vert y = -1) $$

So both of these are just proportions. Treat them like you would any other proportion parameter of a Bernoulli distribution. That's where you get the standard error formula you gave. The common way to calculate standard error (SD) is the following: The outcome of each experiment is Bernoulli distributed with parameter $p$ $X_1,\dots,X_n \overset{iid}{\sim} Bern(p)$. The number of successes is then binomially distributed (since order does not matter) and we need to deal with the https://en.m.wikipedia.org/wiki/Binomial_proportion_confidence_interval The standard error for the proportion is then $SD(\hat{p})=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$. Other possible ways to compute a confidence interval are highlighted in Wikipedia.

(I dispute that the $1.96$ should be included. Standard errors should not be tied to particular confidence levels. What if I want a $90\%$ confidence interval or a $99\%$ confidence interval?))

To get from there to your formula, the $\hat{p}$ is the sensitivity, and the $n$ is the number of positive cases, which you are calling $n_p$. Likewise, the standard error for specificity would use $\hat{p} = \text{specificity}$ and $n$ equal to the number of true negative cases.

  • $\begingroup$ I don’t see a nice way to make the prose flow, but $\hat p$ is the estimate of $p$, calculated as the proportion of $1$s in the data. $\endgroup$
    – Dave
    Aug 28, 2021 at 4:39

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