If either x1 1 or x2 1 we should2reject 2h0 because if

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Unformatted text preview: 1 (θ) = Pθ (X1 > .95) = θ + .05 if −.05 < θ ≤ .95 1 if .95 < θ. Using the distribution of Y = X1 + X2 , given by fY (y |θ) = ￿ y − 2θ 2θ + 2 − y 0 if 2θ ≤ y < 2θ + 1 if 2θ+1 ≤ y < 2θ + 2 otherwise, we obtain the power function for the second test as 8-8 Solutions Manual for Statistical Inference if θ ≤ (C/2) − 1 0 d. If either X1 ≥ 1 or X2 ≥ 1, we should2reject 2H0 , because if θ = θ ≤ (C i < 1) = 1. Thus, (2θ + − C ) /2 if (C/2) − 1 < 0, P (X − 1)/2 β2 (θ) = Pθ (Y > C ) = consider the rejection region given − (C − 2θ)2 /2 if (C − 1)/2 < θ ≤ C/2 1 by 1￿ if C/2 ￿ θ. < {(x1 , x2 ) : x1 + x2 > C } {(x1 , x2 ) : x1 > 1} {(x1 , x2 ) : x2 > 1}. c. From the graph it is clear that φ1 is more powerful for θ near 0, but φ2 is more powerful for The θs. set is the rejection more for φ2 . than φ1 largerfirst φ2 is not uniformly regionpowerfulThe test .with this rejection region has the same Problem 3[C&B] 8.14the last two sets both have probability 0 if θ = 0. But for 0 < θ < C − 1, size as φ2 because The power function of this test is strictly larger than β2 (θ). If C − 1 ≤ θ, this test and φ2 have the same power. ￿ ￿ 8.14 The CLT tells us￿ Z = ( i Xi − np)/ np(1 − p) is approximately n(0, 1). For a test that that rejects H0 when i Xi > c, we need to find c and n to satisfy ￿ ￿ ￿ ￿ c−n(.51) c−n(.49) = .01 and P Z > ￿ = .99. P Z>￿ n(.49)(.51) n(.51)(.49) We thus want c−n(.49) ￿ = 2.33 n(.49)(.51) and c−n(.51)...
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This note was uploaded on 08/09/2013 for the course STATS 10-705 taught by Professor Wasserman during the Fall '12 term at Carnegie Mellon.

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