I am running into an issue with a question that I am working on. I want to find the $p$-value of a $z$-score by hand. Here is the exact question and the work that I have done so far on it.:
A controversy has arisen in the mathematics department at a large university over the proportion of freshman who had AP statistics in high school. The department chair insists that exactly $70\%$ of freshman had AP statistics in HS, but the other department member suspect that the proportion may be different. To resolve this issue, the department surveys $55$ freshman finding that $32$ had AP statistics in high school. Using level $0.05$, test for evidence that the "other department members" are right. Give the p-value.
The work that I have done so far is as follows: $H_0$: $\pi=0.7$ vs HA: $\pi \ne 0.7$.
This is a two sided test. $\hat\pi=32/55=.5818$ about. $\alpha=0.05$
Working the formulas I obtain $z=-1.91$. The rejection region is $|z|>1.96$. We conclude that we fail to reject $H_0$ as $1.91 \ngtr 1.96$.
My issue now is figuring out how to find the $p$-value without using a calculator. Any advise would be helpful.
Actually I think I answered this question myself after a little playing around. Someone please correct me if I am wrong.
I simply found the value on the table that corresponded to my test statistic $-1.91$ and doubled it for the 2-sidedness. This would give me the probability of having a score that is more extreme than what I actually obtained. Therefore the p-value would be $2(0.0287)=0.0574$. This would also confirm that we are unable to reject $H_0$.