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Exercise :

Let $p$ be the probability of having Heads during a coin toss. For the hypothesis testing $H_0 : p = 0.5$ against the alternative $H_1 : p > 0.5$ we count the number of independent trials $X$ needed until having Heads for the first time. If the critical region of the test is $K = \{ X > 3 \}$, then :

i) Calculate the importance level $a$ of the test given.

ii) Calculate the probability of the type II error for the test above, if $p=0.7$.

Attempt :

Since we are interested for an experiment regarding trials until a first success, we have a Geometric Distribution. Thus :

$$\mathbb{P}(X=k) = (1-p)^{k-1}p$$

For part (i), we have :

$$a = \mathbb{P}(\text{Reject} \; H_0|H_0) = \mathbb{P}(X>3|p=0.5)=1-\mathbb{P}(X\leq 3 | p=0.5)$$ $$=$$ $$1-\mathbb{P}(X=3|p=0.5)=1-\sum_{i=1}^3 (1-0.5)^{i-1}\cdot 0.5$$

For part (ii), we have :

$$\text{Type II Error}=\mathbb{P}(\text{Accept} \; H_0|H_1) = \mathbb{P}(X\leq3|p=0.7)$$ $$=$$ $$\sum_{i=1}^3(1-0.7)^{i-1}\cdot 0.7$$

Question : Is my reasoning and my solution correct ? I am worried a bit about the probability calculating part.

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    $\begingroup$ The critical region looks more suitable for an alternative hypothesis that $H_1 : p \lt 0.5$ if $p$ is the probability of heads $\endgroup$ – Henry Jun 6 '18 at 23:55
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"a" is significance not importance although in this case they mean the same thing :) Ok, the actual probability of rejecting the null given it's a fair coin is getting 4 tails which is $0.5^4 = .0625$.

Are you sure you have stated the question correctly? This test is set up to make a type II error an almost certain outcome. For a type II error, this would result in getting a head in 4 flips $= 0.7 + 0.7\cdot 0.3 + 0.7\cdot 0.3^2 + 0.7\cdot 0.3^3 = .9919$.

In most schools, accepting the null isn't an option. It's failure to reject the null.

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  • $\begingroup$ Hi and thanks for your answer. Well, regarding $a$, I think you are mistaken. The number of trials is represented by $X$. Now, the critical region is the region of $\mathbb{R}^n$ for which we reject the $H_0$. Truly, for $X=4$ we would reject the $H_0$ but we would also reject it for any other $X>3$. Thus, I think that $a$ isn't simply the $\mathbb{P} = 4$ but generally $\mathbb{P}(X>3|p=0.5)$ as described in my solution and found by the geometric distribution samle (it's still a small number). For type II error, I think you have to sum up until $X=3$ and not $X=4$. $\endgroup$ – Rebellos Jun 7 '18 at 7:00
  • $\begingroup$ Good points of discussion. But "a" is a significance level not a summation of all the probabilities for rejecting the null. That is, any probability less than .0625 is reason to reject the null. For the type II error, you have to go to 4 trials as it would take 4 tails to reject the null for K = {X > 3}. Type II error questions are purely academic, one has to know the actual population parameter so why bother with a test. $\endgroup$ – Phil H Jun 7 '18 at 13:26

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