Question

1. It is desired to test the null hypothesis u = 40 against the alternative hypothesis u < 40 on the basis of a random sample from a population with standard deviation 4. If the probability of a Type I error is to be 0.04 and the probability of Type II error is to be 0.09 for u = 38, find the required size of the sample.

Answer 1

Given

$Ho:\mu=40$

$Ha:\mu<40$

$\sigma = 4$

P(Type I error) = 0.04 = significance level

=> Reject Ho if P(Z < z) = 0.04

=> z < -1.751

Which means we fail to reject Ho if z > -1.751

=> X - $_\mu$ / $_\sigma$ > -1.751

=> X > -1.751*4/$_\sqrt n$ + 40

P(Type II error) = 0.09 for $_{\mu = 38}$

=> P(fail to reject Ho | Ho is false) = P(X > -1.751*4/$_\sqrt n$ + 40 | $_{\mu = 38}$ )

=> P(Z >  ( -1.751*4/$_\sqrt n$ + 40 - 38) / (4/$_\sqrt n$ ) ) = 0.09

=> ( -1.751*4/$_\sqrt n$ + 2) / (4/$_\sqrt n$ ) = 1.341 , from Inverse Normal table

=> On solving n = 38.24

But n is a whole number

So sample size = 39