chapter 1

Chapter 8: Tests of Hypotheses Based on a Single Sample38.a. We wish to test H o : p = .02 vs H a : p < .02; only if H o can be rejected will the inventory be15ˆ =1000postponed. The lower-tailed test rejects H o if z ≤ -1.645. With p = . 015 , z =-1.01, which is not ≤ -1.645. Thus, H o cannot be rejected, so the inventory should becarried out.b. β ( 01)c. ( 05)(.98)⎡ .02 − .01+1.645 .02 /1000 ⎤. = Φ⎢⎥ = Φ ≈⎢⎣.01(.99)/1000⎥⎦(.98)( 5.49) 1⎡ .02 − .05 + 1.645 .02 /1000 ⎤. = Φ⎢⎥ = Φ − =⎢⎣.05(.95)/1000⎥⎦( 3.30) . 0005β , so is p =.05 it is highly unlikely that H o will be rejected and the inventory will almost surely becarried out.39. Let p denote the true proportion of those called to appear for service who are black. We wish totest H o : p = .25 vs H a : p < .25. We usez .01 = -2.33. We calculate = . 1686zpˆ −.25= , with the rejection region z ≤ -.25(.75)/n177.1686 − .25p ˆ = , and = = −6.11050.0134z . Because –6.1 < -2.33, H o is rejected. A conclusion that discrimination exists is very compelling.40.a. P = true proportion of current customers who qualify. H o : p = .05 vs H a : p ≠ .05,pˆ − .05z = , reject H o if z ≥ 2.58 or z ≤ -2.58. p ˆ = . 08.05(.95)/n.03z = = 3.07 ≥ 2.58.00975, so, so H o is rejected. The company’s premise is not correct.b. (.10)(.95)⎡ .05−.10 + 2.58 .05 /500 ⎤= Φ⎢⎥ = Φ − 85⎢⎣.10(.90)/500 ⎥⎦( 1. ) =β .0332249