chapter 1

Chapter 9: Inferences Based on Two Samplesd. A 95% confidence interval for the difference between the true average strengths for thetwo types of joints is ( 80.95 − 63.23)± t2( 9.59) ( 5.96). 025, ν+1091.9681 35.5216 2( + )91.9681235.5216( ) ( )102. The10 10approximate degrees of freedom is ν == 15.05 , so we use 15d.f., and t 2. 131. The interval is , then,. 025, 15=( 2.131)( 3.57) = 17.72 ± 7.61 ( 10.11,25.33)17 .72=109± . With 95% confidence, wecan say that the true average strength for joints without side coating exceeds that of jointswith side coating by between 10.11 and 25.33 lb-in./in.+109271. m = n = 40, x = 3975. 0 , s 1 = 245.1, = 2795. 0confidence interval for1µ2( 1180.0) 1560.5 ≈ ( 1024, 1336)y , s 2 = 293.7. The large sample 99%µ − is ( 3975.0 − 2795.0)± 2.58245.1402293.7+40± . The value 0 is not contained in this interval so we canstate that, with very high confidence, the value of2concluding that the population means are not equal.µ − is not 0, which is equivalent to1µ272. This exercise calls for a paired analysis. First compute the difference between the amount ofcone penetration for commutator and pinion bearings for each of the 17 motors. These 17differences are summarized as follows: n = 17, d = −4. 18 , s = 35. 85 , where d =(commutator value – pinion value). Then t 2. 120 , and the 95% confidence interval. 025,16=for the population mean difference between penetration for the commutator armature bearingand penetration for the pinion bearing is:⎛35.85⎞− 4.18±−⎝ 17 ⎠( 2.120) ⎜ ⎟ = −4.18± 18.43 = ( 22.61,14.25)d. We would have to saythat the population mean difference has not been precisely estimated. The bound on the errorof estimation is quite large. In addition, the confidence interval spans zero. Because of this,we have insufficient evidence to claim that the population mean penetration differs for thetwo types of bearings.286