chapter 1

Chapter 9: Inferences Based on Two Samples84. Let p 1 = true survival rate at ο CH p − p 0 vs. p − p 00:1 2=11 ; p 2 = true survival rate at ο CH a:1 2≠. The test statistic is30 ; The hypotheses are73102175p ˆ1= = .802 , and p ˆ 2= = . 927 , p ˆ = = . 871, ˆ = . 12991110201.802 − .927 − .125z == = −3.91. The p-value =1 1.871 .129 + .0320Φ( )( )( )91 110( −3 .91) < Φ( − 3.49) = . 0003rates appear to differ.zpˆ− pˆ1 2= . Withpq ˆ ˆq .1 1( + ), so reject H o at any reasonable level. The two survivalmn85.a. We test µ − µ 0H vs. H µ − µ 00:1 2=a:1 2≠. Assuming both populations havenormal distributions, the two-sample t test is appropriate. The approximate degrees of2(.042721)2(.0325125) (.0102083)freedom ν == 11.4 , so we use df = 11.+711t 4.437 , so we reject H o if t ≥ 4. 437 or ≤ −4. 437. 0005,11=.68t = ≈ 3.3 , which is not ≥ 4. 437.0427212t The test statistic is, so we cannot reject H o . At significancelevel .001, the data does not indicate a difference in true average insulin-binding capacitydue to the dosage level.b. P-value = 2P( t > 3.3) = 2 (.004) = .008 which is > .001.86.σ ˆ2=2( ˆ )2222[( n −1) S + ( n −1) S + ( n −1) S + ( n −1)S ]112n + n2+ n3+ n− 41 2 3 42222[( n1− 1) σ1+ ( n2−1) σ2+ ( n3−1) σ3+ ( n4−1)σ4]2E σ == σn1+ n2+ n3+ n4− 4[ 15( .4096) + 17( .6561) + 7( .2601) + 11 (.1225)]== .40950the given data is344. The estimate for291