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:

415	420	421	423	425	428	432	435	437
438	445	446	449	451	457	461	464

(c) Calculate a two-sided 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.)

Answer 1

∑x =    7447  

∑x² =    3265875  

n =    17  

Mean , x̅ = Ʃx/n =    7447/17 =    438.0588

Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] =    √[(3265875-(7447)²/17)/(17-1)] =    15.1058

95% Confidence interval :  

At α = 0.05 and df = n-1 = 16, two tailed critical value, t-crit = T.INV.2T(0.05, 16) =    2.120

Lower Bound = x̅ - t-crit*s/√n = 438.0588 - 2.12 * 15.1058/√17 =    430.29

Upper Bound = x̅ + t-crit*s/√n = 438.0588 + 2.12 * 15.1058/√17 =    445.83

430.29 < µ < 445.83