Answer to Question #293367 in Statistics and Probability for Shahnaz

Question

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?

Answer 1

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 $p=0.1$.

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