What does a t-statistic measure and how is this related to the student t distribution? Set Up
Let's suppose I am surveying $n$ democratic citizens about whether they will vote for Clinton or Sanders. Let's suppose my survey yields $x$ for Clinton and $n-x$ for Sanders, $x \in \Bbb Z_+$. Let $p$ denote the fraction of voters who prefer Clinton. Let's suppose I estimate $p$ by using $\hat{p} = \frac{x}{n}$. Suppose my variance is $\hat{p}(1-\hat{p})$. Suppose $H_0 : p = 0.5$ and $H_A : p \ne 0.5$. 
picking a t-statistic
My t statistic might be then 
$$t_{\hat{p}} = \frac{\frac{x}{n}-0.5}{\sqrt{\frac{x}{n}\left(1-\frac{x}{n}\right)\frac{1}{n}}}$$
From what I understand, this is essentially a ratio 
$$\frac{ \text{difference between estimator and hypothesized value}}{\text{standard deviation}} $$
My Question:


*

*What purpose does a t-statistic serve? Specifically, what useful information does this ratio provide? 

*What relation does a t-statistic and its purpose have to the student t distribution? 


MORE INFO
In the problem I was working with, $n= 400$ and $x = 215$. The question asks me to find the p value for a hypothesis as stated above. The solution read 

Source: For Students
Solutions to Odd-Numbered End-of-Chapter Exercises, Author Unknown, Pages 8-9 
Source: Introduction To Econometrics, Stock and Watson, 3rd edition, pages 75,97
 A: That statistic does not have a t-distribution.
In the context of a hypothesis test against a hypothesized population proportion, it doesn't make a lot of sense to me to divide by the sample estimate of the standard deviation, since the true standard deviation is readily available under the null. (For confidence intervals we don't have a null, however.)
There is an asymptotic argument that as $n\to\infty$ that a correctly standardized proportion will have a standard normal distribution, but I don't see any argument that indicates that in small samples the $t$ is a better approximation. [It is possible that with the sample proportion in the denominator the statistic is better approximated by a $t$ distribution, but I haven't seen an argument to that effect.]
So I don't see any good reason to call that a t-statistic, at least not if the intent is to suggest that it has a t-distribution, and it seems that's what it suggested to you (which is not surprising). 
One possibility that has occurred to me is that the authors may conflate the concept of studentizing - dividing a statistic by a sample estimate of its standard deviation - with the idea of an actual t-statistic. 
The benefit of studentizing a statistic depends on what you're using the statistic to do


*

*as a measure of deviation from some hypothesized value, it estimates how many standard deviations of that statistic (standard errors) that are apart.

*used for a test statistic it allows us to deal with the fact that we don't know the standard deviation; we replace it by an estimate. If the estimate is consistent, then (by Slutsky's theorem) in the limit as $n\to\infty$ the distribution of the ratio will go to the distribution of the numerator scaled by the unknown population standard deviation. This means that when the standardized statistic is asymptotically normal the studentized statistic will also be standardized normal.
However, as long as they don't claim that the standardized statistic has a t-distribution there's not any problem with calling it by that name rather than calling it a Studentized statistic. 
--
Comments relating to the extract in the exercise:
$n=400$ with population $\pi=\frac12$ is plenty large enough to use the normal approximation (especially if you use the hypothesized proportion to calculate the standard error, but even without that it should be fine). 
With the more usual statistic where the hypothesized proportion is used for the standard error, I'd happily work with a sample size a sixteenth as large.
However it's not necessary to use any approximation at all, since packages with the binomial cdf built in are readily available, so exact tests are easy to do.
