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A casino dealer suspects that his die isn't a fair die. He thinks that the die gives more chances to the even numbers (the probability for an even number is larger than $0.5$). To test his suspects he decides to have an experiment, he writes down the outcome of $60$ independent die rolls. The dealer would say that the die is unfair if the number of even numbers in those $60$ rolls will be larger than $35$.

I'm asked to:

  1. Formalize the dealer's assumption.
  2. What is the rejection region ($C$)?
  3. What is the probability for both mistakes ($\alpha$ and $\beta$)?

I started by saying that $X$, the number of even results of a fair cube $X\sim B(\frac{1}{2}, 60)$, and thus $X \sim N (30, 15)$

$H_0: \mu = 30$
$H_1: \mu > 30$

So this is the dealer's assumption. The rejection area should be derived from the fact that the dealer will reject $H_0$ if and only if the even numbers on the experiment will be less or equal to $35$. I'm having a hard time finding $C$ and calculating $\alpha$ and $\beta$.

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  • $\begingroup$ Nicely stated. You might be over-thinking part of this. I don't believe you need to find $C$ at all: it is given to you explicitly in the problem statement. Question (2) is there merely to confirm you understand what this terminology means. $\endgroup$ – whuber Jan 28 '14 at 20:48
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    $\begingroup$ @whuber, So you mean that $C: x > 35$, since in this case we reject $H_0$. $\endgroup$ – Quaker Jan 28 '14 at 20:58
  • $\begingroup$ @whuber, I'm thankful for the upvote but I still don't know how to calculate $\alpha$ and $\beta$ since I need to use the mean of $H_1$ in order to calculate it, but all I know is that $\mu > 30$ when it's $H_1$. $\endgroup$ – Quaker Jan 28 '14 at 22:06
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    $\begingroup$ The upvote means you asked a good question and asked it well. As far as answers go, I am confident you will see some good ones soon. In the meantime, it often is helpful to return to the definitions (and you might even want to state your definitions in the question, because not everybody will have the same understanding of your symbols). What precisely does $\alpha$ mean? What does $\beta$ mean? How do those definitions translate into the terms of this particular question (now that you know the distribution, the hypotheses, and the rejection region)? Go on from there... . $\endgroup$ – whuber Jan 28 '14 at 22:09
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    $\begingroup$ Since you haven't been given a specific $p$, I was suggesting you consider $\beta$ as a function of $p$ *. $$\quad$$ *(or perhaps more traditionally, $1-\beta$ as a function of $p$ - i.e. the power function) $\endgroup$ – Glen_b -Reinstate Monica Jan 29 '14 at 8:37
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Let $X$ be the number of even numbers observed in 60 rolls.

  1. Formalize the dealer's assumption.

What exactly is required here would depend on what you've been taught. The fact that it says "assumption" rather than "assumptions" leaves me in some doubt as to which assumption is sought and the degree of formalism required. I presume it's the assumption of symmetry on the die, but it might be a formal statement relating to the resulting probability. But then again it might be about the Bernoulli assumptions (constant probability and independence across trials). Or it might be as you suggested (and you're likely to know better than I what is sought).

2. What is the rejection region ($C$)?

How you'll write the relevant subset will depend on how you've been taught, but it will be some way of expressing the set of values where $X>35$.

3. What is the probability for both mistakes ($\alpha$ and $\beta$)?

$\alpha$ is straightforward. It's $P(\text{rejection}|H_0\text{ is true})$, which is the probability of being in the rejection region when $p=0.5$.

Under $H_0$, $X\sim \text{Binomial}(60,.5)$. So $P(X>35|n=60,p=0.5)$ is a simple binomial distribution calculation. I make that 0.0775 using the binomial distribution function in R. You were doing it using a normal approximation and that may be what is required instead. The question is then whether to use the continuity correction or not. I'll use it here:

$P(X> 35|p=0.5)\approx P(\frac{X-60p}{\sqrt{60p(1-p)}}>\frac{35.5-60p}{\sqrt{60p(1-p)}}|p=0.5)$

$= P(Z>\frac{35.5-30}{\sqrt{15}}) \approx 0.0778$

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$\beta$ depends on the true value of $p$. You can write it as a binomial tail area that is a function of $p$, but it seems that you can write it as a normal probability which simplifies things a little.

$P(X\leq 35|p)= P(X<35.5|p)$

$\approx P(\frac{X-60p}{\sqrt{60p(1-p)}}<\frac{35.5-60p}{\sqrt{60p(1-p)}}|p)$

$= P(Z<\frac{35.5-60p}{\sqrt{60p(1-p)}}|p)$

$= \Phi(\frac{35.5-60p}{\sqrt{60p(1-p)}})$

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