A preliminary sample of $1200$ engineers has been taken, out of which $260$ are females. The sample is used to estimate the proportion of female engineers denoted by $pF$ .

Estimate the $95$% confidence interval for $pF$.

I know how to calculate CI with known variance or standard deviation, but I can't seem to find a way to do it with unknown variance or standard deviation.


There is a quick little approximation for CIs for sequences of bernoulli trials.

$$ \dfrac{n_F}{n} \pm z_{1-\alpha/2}\sqrt{\dfrac{n_Mn_F}{n}}$$

Here, $z_{1-\alpha/2}$ is the $1-\alpha/2$ quantile from the standard normal distribution, $n_F$ is the number of observed females, $n_M$ is the number of observed males, and $n$ is the sum of both males and females.

So, doing some plugging and chugging yields...

$$ \dfrac{260}{1200} \pm \dfrac{1.96}{1200} \sqrt{\dfrac{260\cdot (1200-260)}{1200}} $$

So the CI is approximately $[0.19,0.24]$

It may be worth noting that this estimate is called a Wald Interval and is known to be a bad estimate over all. Agresti-Coulli intervals for binomial proportions are better, but I doubt that is what your instructor is asking.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.