chapter 1

Chapter 7: Statistical Intervals Based on a Single Sample50. =229.764 + 233.502x the middle of the interval = = 231.633.2⎛ s ⎞width = 2 ( t.025,4) ⎜ ⎟ , and solve for s. Here, n = 5, t. 025,4= 2. 776⎝ n ⎠s 5( 3.738)3 .738 = 2 2776 ⇒ s = = 1.50555 2( 2.776)t 4. , and the interval islimit – lower limit = 3.738. ( )a 99% C.I., 604. 005,4=1.5055231 .633 4.604 = 213.633 ± 3.100 =5( 228.533,234.733)± .To find s we use, and width = upper. So for51.136a. ˆp = = . 680 ⇒ a 90% C.I. isb.2001.645.680 +2n =2( 200)± 1.6451.6451 +200(.680)(.320)22001.645+42( 200)2.6868 ± .0547==1.01353(.624,.732)22 242( ) (.25) − ( 1.645) (.05) ± 4( 1.645) (.25)( .25−.0025) + .05 ( 1.645)2 1.6451.3462 ± 1.3530= = 1079. 7 ⇒.0025.0025use n = 10804c. No, it gives a 95% upper bound.52.a. Assuming normality, t 1. 753 , do s 95% C.I. for µ is. 05,15=.036. 214 ± 1.753 = .214 ± .016 =16(.198,.230)2b. A 90% upper bound for σ , with χ 1. 341 , is. 10 ,15 =(.036)151.3412=.0145 = .120c. A 95% prediction interval, with t 2. 131 , is. 025 , 15 =(.036) 1 +1 = .214 ± .0791 = (.1349 ,.2931 ). 214 ± 2.131.16233