Originally Posted by

**Pim** The distribution that fits with this problem is a binomial one, you get a visitor (trial), who will (or will not) click your ad. (Success/failure)

There is a formula, which can give you confidence intervals for proportions:

$\displaystyle

P \pm z_c \sqrt{\frac{pq}{n}}

$

With P as the sample p, $\displaystyle z_c$ as your confidence level, p as the actual success, which in this case will be approximated by P, q as p -1, failure chance and n as the sample size.

We'll take 95% as your confidence level, which converts to $\displaystyle z_c = 1.96$

Two times computing:

$\displaystyle p_{actual}$ being the true p

$\displaystyle

n = 203

$

$\displaystyle

p = 0.0197

$

$\displaystyle 0.0197 \pm 1.96 \sqrt{\frac{0.0197 * 0.9803}{203}}$

$\displaystyle 0.0197 \pm 0.01911704079$

$\displaystyle 0.0005829592104 \leq p_{actual} \leq 0.03881704079$

$\displaystyle

n = 191

$

$\displaystyle

p = 0.0052

$

$\displaystyle 0.0052 \pm 1.96 \sqrt{\frac{0.0052*0.9948}{191}}$

$\displaystyle 0.0052 \pm 0.0102002061$

$\displaystyle 0 \leq p_{actual} \leq 0.01540020612$

(The first part would be smaller than 0, but per defenition, chances cannot be smaller than zero)

Now, as you can see those intervals overlap for quite a large part, however they are indeed significantly different. What this means is that it's not unlikely for them to remain different. However, if you want more certainity, I suggest you compute this again once you have quite a bit more views, as that means there are more trials and therefore your intervals will get shorter and you will see how much so.