A survey of $61, 647$ people including questions about office relationships. Of the respondents, $26$% reported that bosses scream at employees. Use a $.05$ significance level to test the claim that 1/4 of the people say that bosses scream at employees.
Please keep things as simple as possible.
My work: $H_{0}: p=.25$
$H_{1}: p \gt.25$
$\hat{P}$ is given as $.26$
Using the formula $z=\dfrac {\hat{P}}{\sqrt{\frac {pq}{n}}}$ with $q=.75$, I get a $z$-score of around $5.73$. However, this is an extreme value but $26$ percent is very close to $25$ percent., so I was expecting a $z$ score of almost zero. What did I do wrong?
Under the null hypothesis, $p=0.25$. The test you are using (appropriately, I might add), relies on modeling the sampling distribution of $\hat P$ as $\mathcal{N}\left(\mu=p,\sigma=\sqrt{\frac{pq}{n}}\right)$.
The test you are referring to is not correctly stated in your post, it should be $z=\frac{\hat P - 0.25}{\sqrt{\frac{pq}{n}}}$,which does in fact give you $5.73$. As Henry has noted, this results from your very large sample size, which will have a very high power. Thus, it can detect even very small deviations from $25\%$.
The 95% confidence interval for your true $p$ would be approx. $0.26\pm1.96(.002)\approx 0.26\pm 003$. So you can be quite confident that $p=0.26$ (to two significant figures.) So yeah, its different from $0.25$ but not by very much.