chapter 1

Chapter 7: Statistical Intervals Based on a Single Sample38. N = 25, x = . 0635 , s = .00650635 2.064a. 95% P.I. : (.0065) 1+1 = .0635±.0137⇒ (.0498,.0772).25± .b. 99% Tolerance Interval, with k = 95, critical value 2.972 (table A.6):. 0635 ± 2.972 .0065 ⇒ .0442,.0828 .( ) ( )39.a.Normal Probability PlotProbability.999.99.95.80.50.20.05.01.001Average: 52.2308StDev: 14.8557N: 1330 50 70volumeAnderson-Darling Normality TestA-Squared: 0.360P-Value: 0.392Based on the above plot, generated by Minitab, it is plausible that the populationdistribution is normal.b. We require a tolerance interval. (from table A6, with 95% confidence, k = 95, and n=13,the tcv = 3.081.( tcv) = 52.231±3.08114.856 ( ) = 52.231±45.771⇒( 6.460,98.002)x ± sc. A prediction interval, with t 2. 179 :. 025,12=( ) 1+1 = 52.231±33.593 ⇒ ( 18.638,85.824)52.231±2.179 14.85613229