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H 0 : H 1 : α =

.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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  1. .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.

  2. 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

  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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