I've been trying to understand the solution to this exercise:
It is estimated that the proportion of adults living in a small town who are college graduates is $p = 0.6$. To test this hypothesis, we selected a random sample of $200$ adults. If the number of graduates in our sample is any number between $110\leq x\leq 130$, accept the null hypothesis that $p = 0.6$, otherwise, we conclude that $ p\neq 0.6$.
Evaluate $\alpha$ (Type I error) with the assumption that $p = 0.6$. Use the normal distribution.
My anwers is:
\begin{align*} \alpha&= P(\text{Type I error})\\ &=P\left(z\leq \frac{109,5-200(0,6)}{\sqrt{200(0,6)(0,4)}}\right)+P\left(z\geq \frac{130,0-200(0,6)}{\sqrt{200(0,6)(0,4)}}\right)\\ &\approx2\cdot(0,0655)\\ &=0,131 \end{align*}
Is this correct? I do not know why (and when) I have to use the values $109.5$ and $130.5$, because the theorem of normal approximation to the binomial does not say anything about it. Can anyone help?
Thank you very much.