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(10 By the central limit theorem, when - is large, the distribution of ;-- Ny. , Hence, to test the null hypothesis, Mg:p= po against the alternative p> po an approximate GLR test is to reject if ; > c for some constant c which fulfills the constraint of the specified level of significancea. For -s this approximate test may be written as reject n, if ; > pom where pot ™ po + L6a o Show that an appronimate power function of the test is SPSL tio For a specified power at to be as and significance level of a » os show that the sample size that must be taken is The following "answers" have been proposed. tao For part , by the definition of the power function, and the central limit theorem, where z- Na. . For part dio we need to find - that satisfies both of the constraints nd Solving for - gives the specified formula for bo For part dit, by the definition of the power function, and the central limit theorem, where z- Na, 1). For part ti we need to find - that satisfies both of the constraints A- 1645 .1.262 nd Solving for - gives the specified formula for » i For part , by the definition of the power function, and the central limit theorem, where z-Na D. For part di we need to find. that satisfies both of the constraints nd Solving for - gives the specified formula for ido Both parts ta and hi are correct. te None of the above. The correct answer is (a) (b) (c) (d) (e) N/A (Select One)

(10 By the central limit theorem, when - is large, the distribution of ;-- Ny. , Hence, to test the null hypothesis, Mg:p= po against the alternative p> po an approximate GLR test is to reject if ; > c for some constant c which fulfills the constraint of the specified level of significancea. For -s this approximate test may be written as reject n, if ; > pom where pot ™ po + L6a o Show that an appronimate power function of the test is SPSL tio For a specified power at to be as and significance level of a » os show that the sample size that must be taken is The following "answers" have been proposed. tao For part , by the definition of the power function, and the central limit theorem, where z- Na. . For part dio we need to find - that satisfies both of the constraints nd Solving for - gives the specified formula for bo For part dit, by the definition of the power function, and the central limit theorem, where z- Na, 1). For part ti we need to find - that satisfies both of the constraints A- 1645 .1.262 nd Solving for - gives the specified formula for » i For part , by the definition of the power function, and the central limit theorem, where z-Na D. For part di we need to find. that satisfies both of the constraints nd Solving for - gives the specified formula for ido Both parts ta and hi are correct. te None of the above. The correct answer is (a) (b) (c) (d) (e) N/A (Select One)

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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(10) By the central limit theorem, when n is large, the distribution of = - N(p, P-P). Hence, to test the null hypothesis, Ho : p = po against the alternative H1 : p > po, an approximate GLR test is to reject Ho if p > c for some constant c which fulfills the constraint of the specified level of significance a. For a = 0.05 this approximate test may be written PO(1-po) as reject Ho if p > pcrit where pcrit := po+1.645V (i) Show that an approximate power function of the test is Pcrit-P A(p) = P|Z > PO SPS1. VP(1-p) (ii) For a specified power at p = pi to be 0.9 and significance level of a = 0.05 show that the sample size that must be taken is 1.645 Po(1 – po) + 1.282 /p1(1 – p1) PO-P1 The following ^answers" have been proposed. (a) For part (i), by the definition of the power function, and the central limit theorem, Pcrit-P T(p) = Pp (p s perit) = P|Z< VP(1-p) where z - N(0, 1). For part (ii) we need to find n that satisfies both of the constraints perit- PO Perit-P1 = -1.645, and = -1,282, VPO(1-PO) VPI(1-P1) Solving for n gives the specified formula for n. (b) For part (i), by the definition of the power function, and the central limit theorem, Pcrit-P T(p) = Pp (p s perit) * PZs VP(1-p) where z - N(0, 1). For part (ii) we need to find n that satisfies both of the constraints Pcrit- PO Pcrit- P1 = -1.645, and = 1.282. VPO(1-Po) VP1(1-P) Solving for n gives the specified formula for n. (c) For part (i), by the definition of the power function, and the central limit theorem, Perit-P T(p) = Pp (p > Perit) a P|Z> VP(1-p) where z - N(0, 1). For part (ii) we need to find n that satisfies both of the constraints Pcrit- PO Pcrit- P1 = 1.645, and = -1.282, VPO(1-Po) VPI(1-p1) Solving for n gives the specified formula for n. (d) Both parts (a) and (b) are correct. (e) None of the above. The correct answer is (a) (b) (d) N/A (Select One)
Transcribed Image Text:(10) By the central limit theorem, when n is large, the distribution of = - N(p, P-P). Hence, to test the null hypothesis, Ho : p = po against the alternative H1 : p > po, an approximate GLR test is to reject Ho if p > c for some constant c which fulfills the constraint of the specified level of significance a. For a = 0.05 this approximate test may be written PO(1-po) as reject Ho if p > pcrit where pcrit := po+1.645V (i) Show that an approximate power function of the test is Pcrit-P A(p) = P|Z > PO SPS1. VP(1-p) (ii) For a specified power at p = pi to be 0.9 and significance level of a = 0.05 show that the sample size that must be taken is 1.645 Po(1 – po) + 1.282 /p1(1 – p1) PO-P1 The following ^answers" have been proposed. (a) For part (i), by the definition of the power function, and the central limit theorem, Pcrit-P T(p) = Pp (p s perit) = P|Z< VP(1-p) where z - N(0, 1). For part (ii) we need to find n that satisfies both of the constraints perit- PO Perit-P1 = -1.645, and = -1,282, VPO(1-PO) VPI(1-P1) Solving for n gives the specified formula for n. (b) For part (i), by the definition of the power function, and the central limit theorem, Pcrit-P T(p) = Pp (p s perit) * PZs VP(1-p) where z - N(0, 1). For part (ii) we need to find n that satisfies both of the constraints Pcrit- PO Pcrit- P1 = -1.645, and = 1.282. VPO(1-Po) VP1(1-P) Solving for n gives the specified formula for n. (c) For part (i), by the definition of the power function, and the central limit theorem, Perit-P T(p) = Pp (p > Perit) a P|Z> VP(1-p) where z - N(0, 1). For part (ii) we need to find n that satisfies both of the constraints Pcrit- PO Pcrit- P1 = 1.645, and = -1.282, VPO(1-Po) VPI(1-p1) Solving for n gives the specified formula for n. (d) Both parts (a) and (b) are correct. (e) None of the above. The correct answer is (a) (b) (d) N/A (Select One)
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