Question

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range. 413 421 421 422 425 427 431 430 437 439 446 447 448 453 460 463 465 (a) Calculate a two-sided 95% confidence interval for true average degree of polymerization (as did the authors of the article). (Give answer accurate to 1 decimal place.) (430.1 446.2 X) Does the interval suggest that 440 is a plausible value for true average degree of polymerization? yes no Does the interval suggest that 450 is a plausible value? O yes O no (b) What is the correct interpretation of this interval? 95% of the data lies within this interval. 0 95% of all possible values of are contained in this sample. We are 95% confident that is within this interval. There is a 95% chance that this interval includes u. c) Calculate a two-sided 98% confidence interval for true average degree of polymerization. (Give answer accurate to 1 decimal place.) 428.3 X 447.9 X)

Answer 1

A B С 6 S.NO (X) (X-X BAR)^2 630.8961938 7 1 413 00 2 421 9 3 421 LO 4 422 293.0138408 293.0138408 259.7785467 172.0726644 123.6020761 50.66089965 11 5 425 12 6 427 13 7 431 14 00 430 65.89619377 15 9 437 1.249134948 16 10 439 0.778546713 17 11 18 12 19 13 20 14 21 15 446 62.13148789 447 78.89619377 448 97.66089965 453 221.4844291 460 478.8373702 463 619.1314879 465 722.6608997 7448 4171.764706 438.1176 16.1473 STDEV(B7:B23) 22 16 23 17 24 SUM 25 MEAN 26 ST. DEV

n=1 w = 438.12 SS 4 171.7647 Ja 4171:7647 11-1 167473 a) 95% CI = 0.95 af = no= 17-1 Q=1-0.95 -0.05 te zite,025, 16 = -16 16= 2.20 95%. CID tz cx blue 438:12+ 2.120 x 16.1473 V7 - 438.12 + 8.3 429.82 , 446.42 95% CI = H2968 44614 = 0.98 c) 98%, CI Oc=O q8 -0.02 te= tapdf tool, 16 = 2.584 (From t table) 98% CI = I tex te verho 438.12 + 2.584 x ! 6-1473.. 438.12 + 10.12 42810, 4482 98% CI - H28. O, 4482