# Why is confidence width narrower, when sample proportions are smaller?

I am using a binomial distribution to calculate confidence intervals around a proportion.

I noticed that for a fixed sample size, the width between the confidence intervals gets narrower as the proportion goes down (this is true also if you use a normal distribution).

Someone has asked my why, if the proportion is lower, we need fewer samples to have the same degree of confidence. I can see the maths of why the CI get narrower, but am struggling to think of a good lay-explanation.

• Hint: What is variance of number of successes, and therefore of sample proportion, in a Binomial as a function of p? How does that variance change as a function of p? Apr 26, 2016 at 13:53
• Variance is np(1-p), so as p goes down, variance goes down, so confidence intervals are narrower?
– rw2
Apr 26, 2016 at 14:03
• So are you saying that the confidence intervals are lower for a smaller observed proportion because the variance is lower... why is this? Is it because there's less possible variance as you move away from 50%? Thanks
– rw2
Apr 26, 2016 at 15:13
• Note that the width in absolute terms goes down but not the relative width (the ratio between the upper and lower limit) Apr 26, 2016 at 15:30
• Thanks, but I'm not sure what the relative width tells us? So does it stay the same?
– rw2
Apr 26, 2016 at 19:09

I would say the interval gets narrower as the success parameter moves away from 0.5 because you have more information about the process.

Think about your ability to predict the next toss of a coin. If the coin is fair, you are in the worst possible situation - either guess is equally valuable or accurate (in the long run of having several tosses to predict).

But if the coin is biased in either direction you can better predict its outcome. And as the success probability approaches 0 or 1 you begin to have high accuracy in your predictions. So we see the amount of uncertainty in the process is a maximum at success probability of 0.5, but increasingly shrinks as we move away from that point.

• Thanks. It seems to come down to variance getting smaller as the sample proportion moves away from 50%. Hopefully this is a good explanation for why.
– rw2
Apr 26, 2016 at 19:07

The comments above are correct - since the variance of a Binomial distribution with parameters $n$ and $p$ is $np(1-p)$, with fixed $n$, as $p$ gets farther from 0.5, the variance gets smaller. With my students, I showed a graph of $p(1-p)$ and showed that it's biggest at $p=0.5$ and gets smaller everywhere else.

An explanation that I felt was far more intuitive was that, if the confidence intervals were as wide when $p$ was close to 0 or 1 as it was when $p\approx 0.5$, then your interval would often go below 0 or above 1. We don't want that, so the interval adjusts the closer we get to the boundaries of $[0,1]$. (There are some fringe cases where you might end up with a confidence interval that extends beyond the $[0,1]$ boundaries but these aren't very common and the explanation of shrinking the confidence interval to better fit inside [0,1] is something most people can get their heads around.)

• Thanks for your answer. I'm using a binomial distribution, which avoids confidence intervals extending beyond [0,1]. However, even if the sample size is large enough that the confidence intervals never get near zero, you still see a narrowing of the confidence width as your sample proportion moves away form 50%.
– rw2
Apr 26, 2016 at 19:05