Suppose that we have 9 independent observations from a normal distribution with

Question

Suppose that we have 9 independent observations from a normal distribution with standard deviation 10. We wish to test H_0:mu = 150 vs. H_A:mu notequalto 150 The GLRT with level alpha = 0.05 uses the test statistic T_1 = |x- 150| and has a critical value of c = 6.53. The test rejects the null hypothesis when T > c. Calculate the power of this test against the alternative mu = 151. Calculate the power of this test against the alternative mu = 149. Alternatively, consider the test statistic T_2 = |x = 152| and critical value c = 7.55 Calculate the level of this second test. Calculate the power against the alternatives mu = 151. Calculate the power of this test against the alternative mu = 149. Which of the two tests do you think is better? Why?

Explanation / Answer

Solution

Back-up Theory

1. Average of a sample of size n from a normal distribution with mean m and standard deviation s, is also Normally distributed with mean m and standard deviation s/n

Part (a)

P(H0 is accepted) = P(|MeanX – 150| 6.53) = P(143.47 MeanX 156.53)

Now, power of the test = 1 – P(accepting the null hypothesis when alternative is true)

= 1- P(143.47 MeanX 156.53/µ = 151)

= 1- P[3{(143.47 – 151)/10} 3{(MeanX – 151)/10} 3{(156.53 – 151)/10}]

= 1- P(- 2.289 Z 1.659), where Z is the Standard Normal Variate

= 1- [P(Z 1.659) – P(Z - 2.289)]

= 1- (0.9506 – 0.0110)

= 0.0604 ANSWER

Part (b)

P(H0 is accepted) = P(|MeanX – 150| 6.53) = P(143.47 MeanX 156.53)

Now, power of the test = 1 – P(accepting the null hypothesis when alternative is true)

= 1- P(143.47 MeanX 156.53/µ = 149)

= 1- P[3{(143.47 – 149)/10} 3{(MeanX – 149)/10} 3{(156.53 – 149)/10}]

= 1- P(- 1.659 Z 2.259), where Z is the Standard Normal Variate

= 1- [P(Z 2.259) – P(Z - 1.659)]

= 1- (0.9889 – 0.0484)

= 0.0595 ANSWER

Part (a)(i)

Level of the test = P(H0 is rejected when H0 is true) = P(|MeanX – 152| > 7.55)

= P(MeanX > 159.55 or MeanX < 144.45/ µ = 150)

= P[Z > {3(159.55 – 150)/10} or Z < {3(144.45 – 150)/10}]

= P(Z > 2.985) + P(Z < - 1.665)

= 0.0044 + 0.0480 = 0.0524

So, level of the test is 5.24% ANSWER

Part (c)(ii)

Now, power of the test = 1 – P(accepting the null hypothesis when alternative is true)

= 1- P(|MeanX – 152| 7.55/ µ = 151)

= 1- P[{3(144.45 – 151)/10} Z {3(159.55 – 151)/10}]

= 1- P(- 1.965 Z 2.565)

= 1- [P(Z 2.565) – P(Z - 1.965)]

= 1- (0.99485 – 0.0247)

= 0.02985 ANSWER

Part (c)(iii)

Now, power of the test = 1 – P(accepting the null hypothesis when alternative is true)

= 1- P(|MeanX – 152| 7.55/ µ = 149)

= 1- P[{3(144.45 – 149)/10} Z {3(159.55 – 149)/10}]

= 1- P(- 1.365 Z 3.165)

= 1- [P(Z 3.165) – P(Z - 1.365)]

= 1- (0.9996 – 0.0861)

= 0.0865 ANSWER

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