Finding z-value and p-value given a proportion A news company claims that $90\%$ of their readers like to read their paper. In a random sample of $65$ people, $9$ said that they like to read their paper.
How do I find the z-score and p-value from this information? I thought to do a 1-Prop Z-Test, however, this gave me that $p = 0$, and $p = 1$. I do not think that this is correct.
 A: Let, $$X_i = \left\{ \begin{array}{c}
 1&  \text{if } i^{th} \text{person reads a newspaper}   \\
0 &  \text{otherwise}
\end{array} \right.$$
then, if $Pr[X_i=1]=P$
$$\sum_{i=1}^{n}{X_i}$$ will follow Binomial(n,P) 
I hope this helps!!
A: To find the z-value:
From the given, we see that $ $ $\hat{p} = \frac{9}{65} \approx 0.138 $.
We use that to find the z-value using this formula:$$z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}$$
Where $p=0.90$, $q=0.10$, and $n=65$
Plugging in shows that the z-value is about $-20.478$, which is very abnormal for a z-value. But seeing as how the claim is $90\%$ of people like the paper and only $9$ out of $65$ sampled like it, it makes sense that the z-score will be very large in magnitude.
You can normally find the p-value using the normal distribution table. However in this case, no matter which test you use, this p-value is going to be ridiculously small, as the z-value $-20.478$ isn't on the standard tables. The best we can get through using a table is that the p-value $\lt.0002$ which would be good enough to reject the claim under the standard levels of significance.
