Hypothesis test for a population proportion $H_0: $ the probability that a person has the characteristic A = $0.05$
$H_1: $ the probability that a person has the characteristic A $­\neq 0.05$
I take a sample of size $500$ and find that $40$ people ($0.08$) have the characteristic A. I want to test the null hypothesis with $\alpha = 0.01$.
First, I would have to find the sample's variance, which is $\dfrac{1}{500-1}(0.08-0.05)^2 = 0.0000018$, so the standard deviation is $0.00134$.
The discriminating function would be Student's T fonction. The observed T is $\dfrac{0.08-0.05}{0.00134}*\sqrt{500} = 499$. That means that the critical $\alpha$ value would be incredibly small. That seems a bit extreme, did I go wrong somewhere?
 A: I don't see how the t distribution is appropriate here.
Test using normal approximation. Using the normal approximation to the binomial, we have the
test statistic
$$Z = \frac{\hat \theta - \theta_0}{\sigma_{\hat \theta}},$$
where $\hat\theta = 40/500 = 0.08$ is the estimate of the
true population proportion $\theta$ having the characteristic.
Also $\theta_0 = 0.05$ is the hypothetical population proportion.
And, according to $H_0$, $\sigma_{\hat \theta} = SD(\hat \theta) = 
\sqrt{\frac{\theta_0(1-\theta_0)}{n}}.$
Then, assuming $H_0$ to be true, $Z \stackrel{aprx}{\sim} Norm(0,1)$
and we reject $H_0$ at the 1% level if $|Z| > 2.576.$
Because $\theta_0 = 0.05,$ we get 
$SD(\hat \theta) = \sqrt{.05(.95)/500} =  0.009745.$
So $$Z = (.08 - .05)/.009745 = 3.08$$ and We reject $H_0$ because
$|Z| = 3.08 > 2.576.$
The P-value in this situation is $P(|Z| > 3.08) =  0.0021.$
Reference: Except for notation, the exposition above follows very closely 
the explanation of the 'binomial test' in Ott & Longnecker, 7e, pp 488-489.
Perhaps your textbook contains something similar. 
Esact binomial test. Here is printout from Minitab's 'exact' test procedure for this test,
which does not use the normal approximation. $H_0$ is rejected at
the 1% level because the P-value is less than 0.01. Also, notice
that the 99% CI for $\theta$ does not include 0.05.
 Test and CI for One Proportion 

 Test of p = 0.05 vs p ≠ 0.05

                                                    Exact
 Sample   X    N  Sample p         99% CI         P-Value
 1       40  500  0.080000  (0.051879, 0.116412)    0.005

Note: Requesting the normal approximation instead of the exact test
in Minitab, one gets P-value 0.002, which agrees with our result from
the normal approximation.
Plot of BINOM(500, .05). Perhaps more intuitively, here is a
plot of your hypothetical binomial distribution with $n = 500$ and $\theta = .05$ over the span of $x$-values where it has substantial probability.
Notice that your observed value $x = 40$ is very far out into the right
tail of this distribution. The exact P-value from Minitab is essentially
double the (very small) probability in this distribution to the right of $x = 40.$  

Addendum (5 hrs later): Some users seem to be offended by any computer code at all on this site (perhaps
preferring duck quill pens and oxblood/charcoal ink on papyrus).
I try to keep blood pressures low by omitting detailed code for my graphs. But @Brandon has requested code for the plot above, so here it is.
 x = 10:45;  pdf=dbinom(x, 500, .05)
 plot(x, pdf, lwd=2, col="blue", type="h", ylab="PDF", 
     main="PDF of BINOM(500, .05)")
   abline(h=0, col="green2")
   abline(v=40, col="red", lty="dashed", lwd=2)

