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There are 7 balls in urn. $Q$ of them are white and the rest are black. We have hypothesis $H_0:Q=3$ and $H_1:Q=5$. To test this we draw 2 balls (balls don't come back to the urn - i.e. they are drawn without replacement).

We reject $H_0$ when both balls drawn are white. Calculate probability of type I and type II errors.

MY PROBLEM - what are probabilities of $H_0$ and $H_1$? Are they actually needed?

SOLUTION SKETCH:

Let $x$ be number of drawn white balls.

$\alpha=\mathbb{P}(x>1|H_0)=\frac{\frac{3}{7}\cdot\frac{2}{6}}{\frac{3}{7}}=\frac{2}{6}$

or it is

$\alpha=\mathbb{P}(x>1|H_0)=\frac{3}{7}\cdot\frac{2}{6}=\frac{1}{7}$

and for II type error: $\beta=\mathbb{P}(x<2|H_1)=\frac{\frac{5}{7}\cdot\frac{2}{6}+\frac{2}{7}\cdot\frac{1}{6}+\frac{2}{7}\cdot\frac{5}{6}}{\frac{5}{7}}=\frac{11}{15}$

Or is it wrong?

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  • $\begingroup$ I'm not sure what role H1 plays here. Normally, in a problem like this you either reject the null hypothesis (H0) or determine that you can not reject the null hypotheses. The only reasonably H1 would be Q >= 4 (ie, the actual number of white balls is more than 3), not actually having Q set to a specific number. $\endgroup$
    – user1566
    Jan 15, 2016 at 2:06

2 Answers 2

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I believe you need an a priori estimate for P(Q3) and P(Q5)

If you assume these two events to be equally likely (before pulling balls), then let each be 0.5

From Bayes $P(x=2)*P(Q_3|x=2) = P(x=2|Q_3)*P(Q_3)$

$P(Q_3|x=2) = \frac{P(x=2|Q_3)*P(Q_3)}{P(x=2)}$

$P(x=2)=P(x=2|Q_3)*P(Q_3)+P(x=2|Q_5)*P(Q_5)$

$P(x=2|Q_3)=\frac{3}{7}*\frac{2}{6}=\frac{1}{7}$

$P(x=2|Q_5)=\frac{5}{7}*\frac{4}{6}=\frac{10}{21}$

$P(x=2)=\frac{1}{7}*\frac{1}{2}+\frac{10}{21}*\frac{1}{2}=\frac{13}{42}$

$P(Q_3|x=2) = \frac{P(x=2|Q_3)*P(Q_3)}{P(x=2)}=\frac{\frac{1}{7}*\frac{1}{2}}{\frac{13}{42}}=\frac{3}{13}$

$P(Q_5|x=2) = \frac{P(x=2|Q_5)*P(Q_5)}{P(x=2)}=\frac{\frac{10}{21}*\frac{1}{2}}{\frac{13}{42}}=\frac{10}{13}$

Thus a Type I is $\frac{3}{13}$

A Type II error you have to complete a similar analysis for each of the cases (one of each and zero white) which would not have rejected H0 in favor of H1

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Your question seems to be incorrect regarding the hypotheses "H0:Q=3 and H1:Q=5"

Null and alternative hypotheses are mutually exclusive and exhaustive statements.

It can be of the form:

H0:Q is equals to 3 and H1:Q not equals to 3

Please edit your question to get a sensible answer.

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  • $\begingroup$ I may be wrong, but I think you're talking about Fisher hypothesis testing, while this question is about Neyman-Pearson hypothesis testing. Thus the question is sensible. $\endgroup$
    – beginner
    Jan 15, 2016 at 9:46

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