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Ch6 Sec4: Problem 3 PreviousProblem List Next PREVIEW ONLY- -ANSWERS NOT RECORDE Attempt Results Entered Answer Preview 0.0019 0.0019 (1 point) Determinetthe probability of making a Type Il error for the following hypothesis test, given that 939 :H>910 For this test, take σ-46, n-24 , and α 0.09. P(Type II Error)0019 Preview My Answers Submit Answers You have attempted this problem 4 times. Your overall recorded score is 0%. You have unlimited attempts remaining
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Given Ho : μ-910 H1 H>910 α-0.09 σ: 46 n=24 We know P(Type l error)-P(Reject Ho | Ho is true) = α P(Reject Ho | Ho is true)-α P(X> 910 | μ = 910) = 0.09 Converting to z-scores P(Z>z) -0.09 where (X-μ) o/ n (x - 910) 46 v24 46 V24 1 - P(Z <z) 0.09 P(Z < z) = 0.91 +2+910 = From Excel function, NORM.S.INV we find Z z NORM.S.INV(0.91) 1.34076 Using z to find X 46 46 -*Z 910 - * (1.34076)910 V24 V24 又=922.5893 Thus we reject Ho when X> 922.5893 That is Do not reject Ho when X s 922.589:3 To find P(Type Il error)- P(Do not reject Ho | H1 is true) P(Type l I error) = P(X 922.5893 | μ = 939) Converting to z-scores = P(Z Z | μ = 939) (x-u) (922.5893- 939 46 /24 where Z-1.74773 From Excel function, NORM.S.DIST we find P(Z s Z) -NORM.S.DIST(-1.74773, TRUE) P(Z Z) = 0.0403 P(Type II Error) 0.0403

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