Question

A manufacturing firm needs to test the null hypothesis Ho that the probability p of a defective item is 0.2 or less, against the alternative hypothesis H,: p 0.2. The procedure is to select four items at random. If all four items are defective, Ho is rejected; otherwise, a fifth item is selected. If the fifth item is defective, Ho is rejected. In all other cases, Ho is accepted. What is the power of this test in terms ofp? What is the value of a, the probability of a Type I error?

help with this hypothesis testing procedure?

Answer 1

$Power=\beta(p)=P(H_0\;is\;rejected|p)\\ =P(all\;4\;items\;are\;defective|p)+P(fifth\;one\;is\;defective\;and\;not\;all\;preceeding\;4|p)\\ =p^4+\sum_{k=0}^{3}\binom{4}{k}(1-p)^{k+1}p^{3-k}p\\ =p^4+(1-(1-p)^4)(1-p)$

$P(type I error)=P(reject H_0|p=0.2)\\ =P(first\;4\;defective|p=0.2)\\ +P(fifth\;is\;defective\;but\;not\;all\;the\;preceedings|p=0.2)\\ =0.2^4+\sum_{k=0}^{3}\binom{4}{k}0.8^{k+1}0.2^{3-k}0.2\\ =0.4739$