Which statistical test to use for A/B/C test with proportions I face the following problem: For my startup I came up with 3 different marketing messages. I will display each of the messages to 1000 different people, so 3000 people in total. I will then measure click-through rate.
Now my question is: Which test should I use to test for statistical significance? I want to know which message performs best (if any).
 A: Given that your objective is to determine which, if any, of the three messages is best, I'd skip the $\chi^2$, which merely tests for differences.  If two messages are equally good and substantially better than the third, the $\chi^2$ will (hopefully) return a significant result, but you won't have learned all that you want to.
An alternative is the parametric bootstrap.  Using the example in the comments above, we have three samples of size $1000$ which we can model as drawn from Binomial distributions with $n=1000$ and unknown probabilities.  We estimate the probabilities by the three observed frequencies and generate a large number ($B = 10,000$, for example) of draws from each of the three Binomial distributions.  We then compare the three draws for $b=1, 2, \dots, B$, identifying which is the largest, and report the resulting frequencies:
# Observed data
n <- 1000
observed_counts <- c(75, 50, 50)

# Estimate probabilities
p <- observed_counts / n

# Generate 10,000 samples for each message
x1 <- rbinom(10000, n, p[1])
x2 <- rbinom(10000, n, p[2])
x3 <- rbinom(10000, n, p[3])

# Count the frequency with which each is best
best <- ifelse(x1 > x2, ifelse(x1 > x3, 1, 3),
                        ifelse(x2 > x3, 2, 3))

with the result:
> table(best)
best
   1    2    3 
9778  106  116 

Message 1 was best 97.78% of the time, corresponding to a p-value of 0.0222, roughly the same as the p-value of the $\chi^2$ test given in comments above (0.028).
However, consider a situation with observed frequencies of 7.5%, 7.5%, and 5%.  The bootstrap returns:
    # Observed data
n <- 1000
observed_counts <- c(75, 75, 50)

...

> table(best)

best
   1    2    3 
4823 5157   20 

which makes it quite clear that, although message 3 is worse, messages 1 and 2 are not significantly different.  The $\chi^2$ test, on the other hand, returns a p-value of 0.0439, not as helpful a result!
A: When testing proportions between multiple groups/bins of data, I usually go with the $\chi^2$ Goodness-of-Fit Test.  The NIST Engineering Statistics Handbook and Minitab Online help have great information on this topic.
The simple version is that you will have your observed counts in each category ($O_{A},O_{B},O_{C}$). The expected value ($E$) in this situation is the sum of all the people talked to divided by the product of the click-through rate for the group and the total click through rate.  This gives you ($E_{A},E_{B},E_{C}$).
The $\chi^2$ for each category is calculated as: $$\chi^2=\frac{\left(O-E\right)^2}{E}$$ and results in ($\chi^2_A,\chi^2_B,\chi^2_C$).
The analysis then takes the sum of the chi-squared values as the hypothesis.  The critical threshold of this value is defined by $\chi^2_{\alpha,df}$ and a p-value can be calculated with a chi-test comparing the observed and expected values.
A: Tavrock mentions the Chi-squared test as a solution.  I disagree that this would be an appropriate test for a few reasons.  First of all, I imagine you are not interested in knowing if differences exist between messages, you're more interested in knowing which message leads to largest click through.  This is an altogether different hypothesis than the one the chi-squared is designed to test.  Additionally, even if the chi-squared was an appropriate test, the test does not easily offer estimates of click through rate along with their uncertainty.
A better (and not much more complex approach) is to use a Bayesian decision making framework, as I do here.  This is approach is easily extended to $n$ marketing messages.
