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**

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