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Suppose two numbers are drawn one by one from the set ${1,3}$ and it is not known whether the numbers are drawn with replacement or without replacement. Suppose you set up the following test:

H0: Numbers are drawn without replacement

HA: Numbers are drawn with replacement

And the Null is not rejected if the sum of the draws is equal to 4 and rejected otherwise. Find the probability of type I error and type II error.

Prob(Type 1 error)=P(X+Y=4|with replacement)=$\frac{2}{3}*\frac{1}{3}$=$\frac{2}{9}$

Is this answer correct? Any help will be appreciated!

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  • $\begingroup$ Is this question from a course or homework? If so please consider adding the self-study tag. $\endgroup$
    – Wes
    Mar 26, 2017 at 7:55

2 Answers 2

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If $H_0$ is true:

You will alway get $4$ and not reject null hypothsis $H_0$. So type 1 error probility is 0.

If $H_A$ is true:

You will get $Prob(X+Y=2)=0.25,Prob(X+Y=4)=0.5,Prob(X+Y=6)=0.25$. Since if $X+Y = 4$ then you will not reject null hypothesis that should be rejected, So type 2 error probility is $Prob(X+Y=4) = 0.5$ .

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Type I error is the incorrect rejection of a true null hypothesis (a "false positive").

In this case, it would be considering the numbers to be drawn with replacement, when they are actually drawn without replacement. The null is rejected if the sum of both numbers do not equal 4, so we want the probability that the numbers drawn without replacement do not equal 4.

There is only 1 possible set of numbers drawing both 1 and 3 without replacement. As both of these numbers sum to 4, and there are no other possibilities, the Type 1 error rate is 0%

I hope this helps in explaining exactly what the type 1 error is, as it can be easy to get muddled up and forget what the focus of a question or concept.

All the best :)

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