※ 引述《jusimi (Jusimi)》之銘言：
: A manufacturers’s quality control department is testing whether
: the probability to have defective products is 2% or more
: (i.e. H0: p=0.02 ; Ha: P>0.02).
: The department randomly checks 100 products and records the number
: of defective products (Y).
: The decision rule to reject H0 is Y>2.
: Use Poisson distribution to approximate the Type I error (α), which is :
: a. 7exp(-2)
: b. 1-7exp(-2)
: c. 5exp(-2)
: d. 1-5exp(-2)
: e. None of above
: 希望有高手可以幫我回答~~~先謝謝你們了!!(跪)
α= P(reject H0│H0 is true) = P(Y>2│p=0.02)
符合大樣本實務法則 n 大於等於 100
np ≦ 10
所以用Poisson近似
np=100*0.02= 2 → λ= 2
α = P(Y>2) = 1-P(Y≦2)
= 1-{ P(Y=0) + P(Y=1) + P(Y=2) }
= 1 - { e^(-2) + 2e^(-2) + 2e^(-2) }
= 1 - 5e^(-2)
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※ 編輯: royalwind 來自: 219.69.113.88 (03/16 05:52)