How can I determine a statistically significant sample size in a yes/no test with no control. First up, I'm not a mathematician, I'm just a digital marketer trying to do the right thing. I apologise in advance if my use of terminology is wrong here.
I'm trying to learn how to solve a problem that I see marketers doing wrong all of the time, but I don't have the requisite math experience to know where to begin.
The problem
Suppose you have a landing page, or an ad. You show the add to a hundred people (100 impressions) and 10 people click (10 conversions). You draw the conclusion that the ad has a click-through-rate (CTR) of 10%.
It's an initial ad or landing page, so the is no control group to test against:
I remember from university biology that a sample of 100 people from a population of potentially millions (we call that target market size) is not significant. And that if the same test was applied to a much larger sample, the result would likely be very different).
My question then, given that the possible variance is yes, or no (is this a binomial distribution?), how large a sample would I need to have before I can rely on the CTR value of my ad? (Let's say I want a confidence level of 95%)
 A: The good news is this: 
You don't need a control group. You want to know the CTR for people who actually come to this page, so looking at what those people do is what's needed. 
Now in talking about a CTR, there's a hidden assumption: the folks visiting the page are divided into two groups, the Clickers and the Others. Some fraction $f$ are clickers, and a fraction $1-f$ and Others. And the assumption is that people visiting the page do so in a random order, i.e., any given visitor has a probability of $f$ of being a Clicker, and $1-f$ of being an Other. 
This seems completely reasonable, until you realize that for a page about "breakfast spots in Seattle", for instance, views arriving at 4PM or 2AM might well come from bots, while those arriving at 830 AM are likely to be from humans hoping to find breakfast. It would be no surprise if the CTR for bots was very different from that of humans, and the assumption above would be false. 
Still, without that assumption, there's nothing you can really do (except test both day and night, and on rainy and sunny days, and on weekdays and weekends, and ... well, you get the idea). So we make that assumption. 
Now the question is "If I get $n$ clickthroughs from $N$ visits, I can estimate that $ f = n / N$ (as in "20 clicks in 100 visits: I'm guessing clickthrough rate is $0.2$), and you want to know whether that's a reasonable thing to do. 
There's a clever trick to answer this, which is to change the question: instead, you say, "Given that the 'sample CTR' was 0.2, what are the chances that the TRUE CTR is, say, between $0.15$ and $0.25$?" If you have 10 samples...the chances are relatively small; if you have 1000000 samples, probably pretty good.  
Even better, you might ask: for $N = 200$ samples, I computed a sample CTR of 0.2; how large an interval around 0.2 must I take to be 95% sure that it contains the true CTR? Would $[1.99, 2.01]$ be enough? Would $[1.92, 2.08]$ be enough? 
Pause Why "95%"? Tradition. You can ask the same question for 80% or 30%, but people seem to like "95% confidence intervals," so we often use those. 
I'm hoping that this setup -- the transformation from your problem to a general one of "can I find an interval $[a, b]$ around the number $s = n / N$ that's large enough that's it's 95% sure to contain the number $f$ -- makes sense. Because if it does, then you're ready to read a nice Wikipedia article about doing just this thing. 
Binomial Proportion Confidence Interval
The unknown fraction $f$ (the percentage of Clickers) is the "binomial proportion" of the title. If I were suggesting one of the many choices to use, I'd say "try Agresti-Coull", on the grounds that it's pretty simple. 
