Question

Consider the accompanying observations on bond strength. 11.7 12.3 9.8 9.2 7.7 6.4 6.7 6.8 13.2 17.2 9.2 5.7 5.5 5.3 5.4 5.0 4.7 10.6 15.3 8.7 4.3 4.1 3.8 4.0 3.5 3.6 20.4 25.4 13.7 12.7 13.3 9.0 8.3 10.8 14.3 7.4 5.1 5.6 5.2 4.9 5.3 5.0 3.9 3.9 3.5 3.8 3.5 3.8 (a) Estimate true average bond strength in a way that conveys information about precision and reliability. [Hint: < x; = 388.5 and Σκαι x2 = 4,253.41.] (Round your answers to three decimal places.) A 95% CI for u = the true average strength is (b) Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10. (Round your answers to three decimal places.)

Answer 1

الم clata from the giren member of bende Strength n =48 EU) - 338.5 <(xi) - 4253.41 V то find the mean (x1, given formula 388.5 48 → To find the Standoond deviation X = 8.09375 given formula SE (n-1) 4253.41 - (328.5)°/48 48-1 4253.41-3144.42 41 1108.99 41 23.596 S: 4.857 =X & text S a) 95 % Confidence Pntewal for H= the true avereco Strength then confidence interval (CI) then t critical value tew= there , n-1) then a 1-C1 : 1-951100 : 1-0.95 10.05 tev t 10.0612 ,48-1)

. t (201174) Internal then 2.01174 X 0.70104 confidence CI = x + tev x 4.857 8.09375 I 2.01174 x 148 8.09375 8.09375 + 1,4103 : (8.0935 -1.4103), (8.0935 +1.4103) (6.6832 ), (9.5038) A 95% confidence interval for 4 - the true average Strength 's (5.683), (1.509 >b) the number of rabie Enceeds loin(x) =13 Ộ - 0.271 then confidence interval era på 2/2 13 48 p+2212/20 120 IZH2 + 40² 1 + Z2 2/2 1+ 2222 n 2 0.271 +(1.967/2X48 + $1.96 I+ (1.96/48 0.27180.729. 1.982 48 4x48 1.969 It 48

=(бая19-б.1924), (О.8919+Oіч) 3) е. 87 +0.0цоој6 t 1.96 0.00 19 to: 00043 - (+0,08003 ТО.03 О.3101 6 1.08003 + 1.96 No.00ч 55 1. 08002 О.Ә819 + 1.96 x 0.0614 1.68003 6. Ә879 4 146 х Оло524 О. ax19 4 о 1924 О165ч. ОЧ103 б.165 [б.io, 2