The power of the test is probability of Type II error. IN this case
it is probability of type II error when x-bar = 23.6. This is the
probability of Ho not being rejected when it is false because x-bar =
23.6

B(23.6) = 1 - z(23.6 - 25.1)(14/sqrt(100))

1- z(-1.07)

Finr z(-1.07) on standard normal distribution chart to get .1423

1- .1423 = .8577

b) we know the power of the test is .9

so 1 - z(23.6 - 25.1)/(14/sqrt(n)) = .9

therefore z(-1.5/914/sqrt(n)) = .1

looking at the z-chart we know z(-1.28) = .1

so -1.28 = -1.5(14/sqrt(n))

-17.92/sqrt(n) = -1.5

cross multiply and square both sides and solve for n and you get n = 143.

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