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:
With P as the sample p, 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
Two times computing:
being the true p
(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.