Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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 $X\gt k$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.