chapter 1

Chapter 8: Tests of Hypotheses Based on a Single Sample36.a.1 p = true proportion of all nickel plates that blister under the givencircumstances.2 H o : p = .103 H a : p > .104z =popˆ− p5 Reject H o if z ≥ 1.6456z =opˆ− .10( 1 − p )/n .10(.90)/n(.90)o=14/100 −.10= 1.33.10 /1007 Fail to Reject H o . The data does not give compelling evidence forconcluding that more than 10% of all plates blister under thecircumstances.The possible error we could have made is a Type II error: Failing to reject the nullhypothesis when it is actually true.b. ( 15)(.90)⎡ .10 − .15+1.645 .10 /100 ⎤. = Φ⎢⎥ = Φ − =⎢⎣.15(.85)/100⎥⎦( .02) . 4920β . When n =200, β ( 15)(.90)⎡ .10 − .15 + 1.645 .10 / 200⎤. = Φ⎢⎥ = Φ − =⎢⎣.15(.85)/200 ⎥⎦( .60) . 2743c.( ) + 1.28 .15(.85)⎡1.645.10 .90⎤2n = ⎢⎥ = 19.01 = 361.4 , so use n = 362⎢⎣.15 −.10⎥⎦237.1 p = true proportion of the population with type A blood2 H o : p = .403 H a : p ≠ .404z =popˆ− popˆ− .40( 1 − p )/n .40(.60)/n5 Reject H o if z ≥ 2.58 or z ≤ -2.586z =(.60)o82/150 − .40 .147= = 3.667.40 /150 .04=7 Reject H o . The data does suggest that the percentage of the population with type Ablood differs from 40%. (at the .01 significance level). Since the z critical value fora significance level of .05 is less than that of .01, the conclusion would not change.248