Actually i just wanna assure about what i'm thinking now.

Given Problem :

The proportion of adults living in a small town who are college graduates is estimated to be $p =0.6$. To test this hypothesis, a random sample of $15$ adults is selected. If the number of college graduates in the sample is anywhere from $6$ to $12$, we shall not reject the null hypothesis that $p =0.6$; otherwise, we shall conclude that $p\ne0.6$.

Here is what i'm thinking about:

If given problem show that $\alpha$ (type I error, reject $H_0$ when it's true) greater than $\beta$ (type II error, receive $H_0$ when it's false), is that mean the hypothesis is good?


If given problem show that $\alpha$ (type I error, reject $H_0$ when it's true) less than $\beta$ (type II error, receive $H_0$ when it's false), is that mean the hypothesis is bad?

But i'm not sure with my thinking... anyway, how to determine those hypothesis are bad or good? Is there another way to determine it?

Just for an additional..

When i have to use the binomial distrobution and when i have to use the normal distribution (for calculate the error $\alpha$ and $\beta$)?

Bcz when i use excell to calculate binomial, and i put $x>15$ it'll error. So, i don't know how am i supposed to be. Then i try use the normal dist, and sometimes it works. Is that mean, normal distribution is used when $x>15$?


1 Answer 1


You do not need to use a normal approximation to calculate the Type I and II errors of the test.

Your hypothesis is $$H_0 : p = 0.6 \quad \text{vs.} \quad H_a : p \ne 0.6.$$ Your sample size is $n = 15$. Your test statistic is $X$, the number of college graduates in the sample of adults. Your criterion to reject $H_0$ in favor of $H_a$ is $$R = (X < 6) \cup (X > 12).$$ Therefore, under the assumption that $H_0$ is true, $$X \mid H_0 \sim \operatorname{Binomial}(n = 15, p = 0.6),$$ and $$\alpha = \Pr[R \mid H_0] = 1 - \sum_{x=6}^{12} \binom{15}{x} (0.6)^x (1 - 0.6)^{15-x} \approx 0.0609473.$$ This is your Type I error, because it is the probability of rejecting the null hypothesis when it is true. The Type II error is $$\beta = \Pr[\bar R \mid H_a],$$ the probability of failing to reject $H_0$ when $H_a$ is true. But $H_a$ does not provide sufficient information about the sampling distribution: you don't know, for example, if $H_a$ means that $p = 0.9$, or $p = 0.1$, or $p = 0.60001$. All of these are possibilities under the alternative hypothesis. You can certainly calculate a Type II error function that depends on the value of $p$; e.g., $$\beta(p_a) = \Pr[\bar R \mid p = p_a] = \sum_{x=6}^{12} \binom{15}{x} p_a^x (1 - p_a)^{15-x}.$$ This is a polynomial of degree $15$ in the variable $p_a$, and on the interval $p_a \in [0,1]$, we can plot this curve:

enter image description here

As expected, the closer $p_a$ is to $0.6$, the higher the Type II error $\beta$ of the test, because it becomes harder for the test to distinguish between the null and the alternative hypotheses. For example, if choice were between deciding whether $p = 0.6$ versus $p = 0.60001$, the distribution of the test statistic would look almost exactly the same under either of these hypotheses. You'd need an incredibly large sample size to be able to begin to tell the difference between these cases.

As we have seen above, you lack sufficient information to use the Type II error as a measure of the suitability of the test. You can use the Type I error $\alpha$, which depends only on the null hypothesis and the choice of rejection region. If you want the $\alpha$ to be smaller, you would choose a more strict criterion--make it harder to reject $H_0$ when it is true. For example, if you chose $$R = (X < 5) \cup (X > 13),$$ what would be the resulting Type I error of such a test? The tradeoff is that when you make type Type I error smaller, it makes the Type II error larger (for the same sample size $n$). This again makes intuitive sense, because reducing the size of the rejection region $R$ makes it more difficult to reject even when $H_0$ is false--this leads to a reduction in the power of the test.

  • $\begingroup$ But if i have interval of sample that greater than $15$, like $100$, $200$. The binomial way will fails right? What i mean here is when i input $x$ greater than $15$ for example $48\leq x \leq 72$, my book suggests to use the normal approximation, cz excell can't find the value when $x$ greater than $15$. And actually i know how to calculate $\alpha, \, \beta$, but i don't know when they are said to be good testing and when they are said to be bad testing. And this two statements is the main of my question. Sorry if my english is difficult to understand . I'm not good at it $\endgroup$
    – user516076
    Commented Oct 4, 2019 at 21:41
  • $\begingroup$ @user516076 When $n$ is large, you can still use a computer for calculating the exact $\alpha$, but a normal approximation is also possible. You would perform the approximation with the statistic $$Z \mid H_0 = \frac{X/n - p_0}{\sqrt{p_0 (1-p_0)/n}}$$ where $p_0 = 0.6$ in your case and $Z$ is approximately standard normal. $\endgroup$
    – heropup
    Commented Oct 4, 2019 at 22:16
  • 1
    $\begingroup$ As for when a test is "good" or "bad," that is subjective. I cannot answer that question because it requires more information than you have provided. $\endgroup$
    – heropup
    Commented Oct 4, 2019 at 22:17

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