# Hypothesis test: Type I error

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.

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The use of $109.5$ and $130.5$ is called the continuity correction when used, as you have done, in treating a discrete random variable (the binomial random variable) as a continuous random variable (the normal random variable). – Dilip Sarwate Nov 21 '12 at 1:10
@Dilip Sarwate Thank you very much. – Hiperion Nov 21 '12 at 1:25

Probably there is currently a typo, you meant $130.5$, not $130.0$. For one thing, our expression should be symmetric about $120$.
We reject if we get a result of $131$ or bigger. The probability of this is $1$ minus the probability that the result is $130$ or less. Let $X$ be the number of college graduates. Then $X$ takes on consecutive integer values $0,1,2,\dots,200$. When we approximate $\Pr(X \le k)$ by the normal, one usually gets a better approximation by calculating the probability that the normal is $\le k+\frac{1}{2}$. So we want $1$ minus the probability that the normal is $\le 130.5$. That is exactly the probability that the normal is $\ge 130.5$.
Remark: Minor tip for the continuity correction: It is best to only try to remember the method for $\Pr(X\le k)$. and use this to figure out each time what to do when we have a situation like $X \lt k$, or $X\ge k$, or $X\gt k$.