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.4600. Do not reject H 0. Reject H 0. .04. Test Statistic:. 0. Answer to Question 3. H 0 : H 1 : α =. μ = .25. μ > .25. .04. σ = .004. Z α = 1.75. The picture above shows a Type I situation. Here we are assuming that

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## H 0 : H 1 : α =

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**.4600**Do not reject H0 Reject H0 .04 Test Statistic: 0 Answer to Question 3 H0: H1: α = μ = .25 μ > .25 .04 σ = .004 Zα = 1.75 The picture above shows a Type I situation. Here we are assuming that H0 is true. To work this problem we have to examine a Type II situation where the H0 is false (1 = .251). The “trick” is to determine where 1 = .251 is with respect to the Reject H0 region.**If Zcomputed is less than 1.75, we do not reject H0.**If Zcomputed is greater than or equal to 1.75, we reject H0. 1.75 is the critical value that separates the Reject part from the Do not reject part. To work a power(beta) problem that deals with a hypothesis test of the mean you have to convert the critical value (1.75 on the Z scale) to the equivalent value of the sample mean (the scale). To do this you solve the test statistic for . Solving the test statistic for , we obtain: Answer to Question 3**.4600**Do not reject H0 Reject H0 .04 Zα = 1.75 0 .5 .2734 area = .2734 Do not reject H0 Reject H0 Power = .5 + .2734 = .7734 0 Answer to Question 3 Reject H0 (Power) Do not reject H0 (Beta) Z = (.2507-.251)/(.004/10) Z = -.75

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