Test a proportion with limited information I was reading this article, and wanted to check whether 203/2000 was statistically different from 1/12. Is this a comparison between two means? How can I do this without knowing the standard deviations?
 A: The key is to ask the right question.  The value of 203 is the maximum observed number of astrological signs among the "almost 2000" people arrested in southwestern Ontario in 2011.  It is not legitimate to treat it as if it were a single, isolated count, for that would mistakenly (and greatly) exaggerate the apparent significance of this value (the "Jelly beans cause acne" phenomenon).  One way to assess whether this value is consistent with a hypothesis of no association between astrological sign and being arrested in 2011 is to compute the chance of observing 203 or more people of some common astrological sign when 2000 people are selected independently and randomly from a large general population.
Obviously we cannot compute that chance exactly: we don't know whether "almost 2000" is 1999 or 1 or anything in between.  Let's take it to be 2000.  We also don't know the proportions of the various astrological signs in the general population, but it's a good approximation to take each of the 12 signs to be equally distributed.  (Assuming 2000 will slightly overestimate the answer and assuming equal distributions will slightly underestimate it; the two effects--hopefully--should come close to canceling each other out.)  Under these assumptions, the distribution of the maximum can be closely approximated as the maximum of 12 independent draws from a Binomial(2000, 1/12) distribution.  (The 12 counts, one for each sign, are not truly independent--they must sum to 2000--but they are close enough to being independent that this simple Binomial calculation will be close enough.) That's enough to compute the probability of interest; its calculation is routine.

This figure plots the probability of the maximum of 12 independent Binomial(2000, 1/12) variates.  The shaded area depicts the chance that the maximum will be 203 or greater.  This chance closely approximates the chance in which we are interested.
There is a 2.8% chance that the maximum will be 203 or greater: this is a mild indicator that the hypothesis of no association may be incorrect, but it is not conclusive.  In most technical publications this chance would not be small enough to be considered "significant" (because, arguably, a "two-tailed" analysis is warranted--an unusually low count would have drawn attention, too--requiring us to double the chance to 5.6%, above the traditional threshold of 5%).
Such a result has to be looked at sceptically: the police may have broken down their arrest records in many different ways before noting this particular number.  If they did, we shouldn't be in the least surprised they found an event with a 2.8% chance of happening.
By the way, because 203 is fairly well out in the tail of a skewed distribution, a Normal approximation is not trustworthy for this calculation.  I therefore obtained this value using the Binomial distribution itself.
A: The main problem here is that months of birth are not equiprobable; there is a seasonality, so you don’t expect 1/12 of Aries.
If it was the case, the number of Aries would follow a binomial law $B(n=2000,p=1/12)$, the expected number of Aries would be $np = 2000/12 = 166.7$, and its standard deviation $\sqrt{np(1-p)} = 12.36$. You make a normal approximations, you have much more than two standard deviations so this is significant... (you have 2.94 standard deviations and the one-sided p value is 0.0016; two-sided p = 0.0032.). Alternatively you could do a $\chi^2$ test with 1 df, comparing the observed values 203, 1797 to the expected values 166.7, 1833.3. 
However, other important issue, the hypothesis is not stated beforehand, and you should correct for multiple testing... with which multiplicity? the number of variables you could test in such data is surely too big to be surprised by the observation of a p value of 0.001. So you would need to replicate this important result in some other study! edit A simple correction for 12 independent tests (considering the 12 astrological signs as independent variables) leads to $1-(1-0.016)^{12} = 0.019$.  The difference with the $p$-value computed by whuber is due to the normal approximation.
