chapter 1

Chapter 8: Tests of Hypotheses Based on a Single Sample14.a. = . 04σx, so P ( x ≥ 10.1004or≤ 9. 8940 whenµ = 10 ) = P(z ≥ 2.51or≤ −2.65)= .006 + .004 = .01b. ( 10 .1) = P(9.8940< x < 10. 1004β whenµ = 10 .1) = P(−5.15< z < .01) = .5040 , whereas( 9 .9) = P(−.15< z < 5.01) = .5596.β Since µ = 9. 9 and = 10. 1µ representequally serious departures from H o , one would probably want to use a test procedure forwhich ( 9.9) β ( 10.1)β = . A similar result and comment apply to any other pair ofalternative values symmetrically placed about 10.Section 8.215.α when z has a standard normal distribution) = 1 − Φ( 1.88) = . 0301a. = P( z ≥ 1. 88α when z ~ N(0, 1) = Φ( − 2 .75) = . 003b. = P( z ≤ −2. 75α Φ ( − 2 .88) + ( 1− Φ( 2.88) = . 004c. =16.a. = P( t ≥ 3. 733α when t has a t distribution with 15 d.f.) =.001, because the 15 d.f.row of Table A.5 shows that t .001,15 = .3733b. d.f. = n – 1 = 23, so α = P( t ≤ −2.500)= . 01c. d.f. = 30, and α = P( t ≥ 1.697) + P(t ≤ −1.697)= .05 + .05 = . 10242