Probability of accepting null hypotheses given p value

Question

Hi i'm currently struggling to answer part b and c of this question:

For part (a) this is just regarding the significance level used for the hypotheses test

for part (b) Type 1 error denotes rejecting a correct null hypotheses when x(bar) = 1 this takes probability of p^3 with H0 being correct when p=1 so the probability of this not coming to pass ought to be (1-p^3)

for part (c) my only current idea is either substituting p=0.9 into p^3 or summing all values 0 through 2/3 but am very confused

Any insight is appreciated and thanks in advance

Answer 1

for part (b) Type 1 error denotes rejecting a correct null hypotheses when x(bar) = 1

No, part (b) said explicitly that the rule is: "accept the null hypothesis if $\bar x = 1,$ otherwise reject." So they will reject the null hypothesis only if $\bar x \neq 1.$

Type 1 error is simply when we reject a correct null hypothesis. There's no additional "if" condition. The thing to figure out is, what has to happen for the researcher to reject a correct null hypothesis? To begin with, this can only happen if the null hypothesis is correct. In this experiment, that means $p = 1.$ Secondly, in order for a Type 1 error to occur, the researcher has to reject the null hypothesis (that is, reject the possibility that $p = 1$) despite the fact that the null hypothesis is correct (that is despite the fact that it is true that $p = 1$). What event would cause the researcher to reject the null hypothesis, and given that $p = 1,$ how likely is that event to occur?

Similar reasoning applies in part (c), except that now we're considering the case where $p = 0.9$ and we're asking for the probability of accepting the null hypothesis rather than the probability of rejecting it.

Answer 2

Careful on part (b), you want $\alpha = P(\text{Type I Error}) = P(\text{reject true null}) = P(\bar{X} \neq 1 | p = 1) = 0$ according to the probabilities of the values of $\bar{X}$. Which makes sense because if $p = 1$ then with probability one, $\bar{X} = 1$ according to the given distribution.

Part (c) is asking you to compute the probability of a type II error, or $\beta$:

The decision rule in part (b) is to reject the null ($H_0: p = 1$) if $\bar{x} \neq 1$. A type II error occurs if we fail to reject (or accept) a null that is false. You can think of $\beta$ as function of $p$ and as a conditional probability dependent on the true value of $p$:

$$\beta(p) = P(\text{Type II Error}|p) = P(\text{fail to reject false null} | p) = P(\bar{X} = 1 |p) = p^3$$

We need to compute $\beta(.9)$

$$\beta(.9) = .9^3 = .729$$

In the context of this problem this also makes sense to have such a high value of $\beta(.9)$ because if $p = .9$ and you are checking if $p = 1$ then the odds of you picking up that the null is false are small based on the possible values of $\bar{X}$. But if you are checking to see if $p = .1$ then the probability of a type II error goes down by a lot.

Answer 3

The three parts are asking, respectively,

(a) find a set $R$ such that $\Pr[\bar X \in R \mid p = 1]$ is minimized.

(b) $\Pr[\bar X \ne 1 \mid p = 1]$

(c) $\Pr[\bar X = 1 \mid p = 0.9].$

It is important to understand that the distribution of $\bar X$ is degenerate under $H_0 : p = 1$, because in such a case $(\bar X \mid H_0) = 1$ almost surely. Then, the first part can be answered with almost any choice of $R$, since for example $R = \{0\}$ gives $$\Pr[\bar X = 0 \mid p = 1] = 0,$$ and the choice $R = \{0, 1/3, 2/3\}$ also gives $$\Pr[\bar X \in \{0, 1/3, 2/3\} \mid p = 1] = 0.$$ $R$, the "rejection region" of the test, is not uniquely determined, since there is more than one set for which the test never rejects the null hypothesis when it is true.

Indeed, you can argue that for any hypothesis test, you can minimize the Type I error by choosing a rejection region that amounts to the empty set; i.e., you can assure your Type I error is $0$ if you just say "never reject $H_0$." But this is not an interesting test for obvious reasons.

The above also answers part (b) of your question.

The last part of the question is a straightforward calculation: set $p = 0.9$ in the above table, and read out the entry corresponding to the outcome $\bar X = 1$.

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From a pedagogical standpoint I think this question is poorly written. First, it uses the common but misleading phrase "accept the null hypothesis" which obfuscates the asymmetric nature of the null and alternative hypotheses. Second, the answer to the first part is not unique as I have explained, but the wording suggests it is. Moreover, there is a philosophical issue with part (a) as I have also explained.