Problem
===
$e_i = np$

$\chi^2_{(n-1)} = \Sigma_i{(f_i-e_i)^2 \over e_i}$[^1]
$\Rightarrow \Sigma_i{(n \bar p - np)^2 \over np}$

And $\chi^2_{(n-1)} = \Sigma_i({\bar p - p \over \sqrt{ pq \over n}})^2$ [^2]
$= \Sigma_i{n(\bar p - p)^2 \over pq}$
$= \Sigma_i{n^2(\bar p - p)^2 \over npq}$
$= \Sigma_i{(n \bar p - np)^2 \over npq}$
$= \Sigma_i{(f_i - e_i)^2 \over e_iq}$

So $q=1$ ?

[^1]: book key formula 12.6

[^2]: book key formala 9.4

---

$b_i=a_ip+x_i$, and $\forall i \in \mathbb{N}$, and $x_i<p$
$\{\Sigma_{\forall i} b_i\}\ (mod\ p)\\
\equiv\{\Sigma_{\forall i} (a_ip+x_i)\}\ (mod\ p) \\
\equiv\{\Sigma_{\forall i} (x_i)\}\ (mod\ p)\\
\equiv\{\Sigma_{\forall i} (x_i\ mod\ p)\}\ (mod\ p)$