Consider the problem of testing H0: μ = 10 versus H1: μ > 10 with n = 64, σ = 2 (known), and α = .025. The rejection region of this test is given by

Suppose we wish to calculate the power of this test at the alternative μ1 = 11. Power = the probability of rejecting the null hypothesis when the alternative is true. Since our test rejects the null hypothesis when > 10.49, its power at μ1 = 11 is the probability
P[ > 10.49 when the true mean μ1 = 11]
If the population mean is 11, we know that X has the normal distribution with mean 11 and sd = σ/ˆšn = 2//64 = .25. The standardized variable is

and we calculate

Following the above steps, calculate the power of this test at the alternative:
(a) μ1 = 10.5
(b) μ1 = 10.8

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X-10 27、 1.96 2/64 R: X 10 + 1.96 = 10.49 ー 64 X-11 25 Power = P [ X 1049 when μ.-11 ] 10.49 11 .25 = P [Z > -2.04] = .9793 (using normal table )