Solution to the fraction of defective items in a large lot is ‘p’. to test the null … - Sikademy
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Peace Weguma

the fraction of defective items in a large lot is ‘p’. to test the null hypothesis h0: p=0.2, one consider the number ‘f’ of defectives in a sample of 8 items and accept the hypothesis if f≤6, and reject the hypothesis otherwise. what is the probability of type-1 error of the test? what is the probability of type-2 error of the test?

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The number of defectives (f) in the sample is a random variable which follows a Binomial distribution with parameters n=8 and p.

The probability of x defectives is given as,

\binom{8}{x}p^x(1-p)^{8-x}  


a)

The probability of type 1 error.

From definition, probability of type 1 error is he probability of rejecting H_0 when it is true. So, p=0.2 and we find the probability of 7 or 8 defectives.

So, when the number of defectives is  7, its probability is, \binom{8}{7}0.2^7(0.8)^{1}=0.00008192  and when the number of defectives is 8, its probability is, \binom{8}{9}0.2^88^{8-8}=0.2^8=0.00000256

The probability of committing type 1 error is 0.00008192+0.00000256=0.00008448


b)

The probability of type 2 error.

From definition, type 2 error is the probability of accepting H_0 when H_1 is true. So we determine the probability of 6 or less defectives when p=0.1.

p(x\le6)=1-p(x\gt 6)=1-\{\binom{8}{7}0.1^70.9+0.1^8\}=1-(7.3e-7)=0.99999927Therefore, the probability of committing type 2 error is 0.99999927

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