Question

2. Suppose that we have 9 independent observations from a normal distribution with standard deviation 10. We wish to test Ho : μ-150 vs. H A : μ 150 The best test with level a- 0.05 uses the test statistic T1 =1元-1501 and has a critical value of c 6.53. The test rejects the null hypothesis when T> c (a) Calculate the power of this test against the alternative μ-151. (b) Calculate the power of this test against the alternative 11-48. (c) Alternatively, consider the test statistic T2 = IZ-152 and critical value c 7.55 i. Calculate the level of this second test. ii. Calculate the power against the alternatives μ-151. iii. Calculate the power of this test against the alternative μ = 148. (d) Which of the two tests do you think is better? Why?

Answer 1

$\bar{X}\sim N\left(\mu,~10^2/9 \right )\\ Power~at~\mu_1=P(|\bar{X}-150|>6.53)=1-P\left(-6.53\leq \bar{X}-150\leq 6.53 \right )\\ =1-P\left(\frac{3(-6.53-\mu_1+150)}{10}\leq \frac{3(\bar{X}-\mu_1)}{10} \leq \frac{3(6.53-\mu_1+150)}{10}\right )\\ =1-\Phi\left(\frac{3(156.53-\mu_1)}{10} \right )+\Phi\left(\frac{3(143.47-\mu_1)}{10} \right ) \\(a)~When~\mu=151,~Power=1-\Phi(1.659)+\Phi( -2.259)=0.0605\\ (b)~When~\mu=148,~Power=1-\Phi(2.559)+\Phi( -1.359)=0.0923\\ (c)~\bar{X}\sim N\left(\mu,~10^2/9 \right )\\ Power~at~\mu_1=P(|\bar{X}-152|>7.55)=1-P\left(-7.55\leq \bar{X}-152\leq 7.55 \right )\\ =1-P\left(\frac{3(-7.55-\mu_1+152)}{10}\leq \frac{3(\bar{X}-\mu_1)}{10} \leq \frac{3(7.55-\mu_1+152)}{10}\right )\\ =1-\Phi\left(\frac{3(159.55-\mu_1)}{10} \right )+\Phi\left(\frac{3(144.45-\mu_1)}{10} \right )\\ i.~Level=Power~at~\mu_1=152=0.0235 \\ ii.~Power~at~\mu_1=151=0.0299\\ iii.~Power~at~\mu_1=148=0.1437\\ (d)~First~test~is~better~since~Power~at~\mu=151~for~second~test~is~less~than~\\ 0.05.$