Question

is drawn from a Normal population with variance ?^2-5. Consider the z test with hypotheses A simple random sample of size n-1 H0: ?: 14 H1: ? > 14 with significance level 0.05 If it were true that ?-157, what would be the power of this test? (Give answers correct to 3 decimal places)

Answer 1

Solution :

This is right tailed test .

First we compute the rejection region (in terms of   $\bar{x}$ )

Critical value z(0.05) = 1.645, So Rejection region = {z / z > 1.645}

Now test statistics z = (  $\bar{x}$ – 14 )/ 2.24/?(10) =    ($\bar{x}$ – 14)/ 0.71

Then Rejection region = { ($\bar{x}$ – 14)/ 0.71    >1.645} = {  $\bar{x}$   >15.168}

Ho is rejected when    $\bar{x}$ > 15.168

Ho is not rejected when   $\bar{x}$   < 15.168

? = P (type II error) = P (     $\bar{x}$ < 15.168 / µ = 15.7) = P ( z < (15.168-15.7)/ 0.71)

          = P ( z <-0.75)

          = 0.227

Power of test = 1 - ? = 1-0.227 = 0.773

Power of the test = 1 – 0.088 = 0.912