2008hw6_solutions

2008hw6_solutions - ST 314 1. Text, p. 235, exercise 4.39....

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ST 314 HW #6 Solutions 1. Text, p. 235, exercise 4.39. a. Parallel boxplots are: 1 2 Box-and-Whisker Plot 0.7 1 1.3 1.6 1.9 2.2 2.5 VISCOSIT GROUP b. Let μ H = true average thickness for high viscosity and μ L = true average thickness for low viscosity. Then the appropriate test is: (1) H 0 : μ H - μ L = 0 H a : μ H - μ L 0 (looking to see if higher viscosity paint leads to thicker coatings) (2) Use a test statistic of t = ( f8e5 y H - f8e5 y L )/s p {1/n H + 1/n L } 1/2 (3) With α = .05 and n H + n L – 2 = 16 + 16 – 2 = 30, we find t 30, .05 = 1.697 so we reject H 0 if t > 1.697. (4) Calculating gives f8e5 y L = 1.348, s L 2 = .1146, f8e5 y H = 1.487 and s H 2 = .2474, so s 2 p = 15(.1146) + 15(.2474)/30 = 0.181 and s p = .4254. Thus, t = ( f8e5 y H - f8e5 y L )/s p {1/n H + 1/n L } 1/2 = [(1.487 – 1.348)]/(.4254){1/16 + 1/16} 1/2 = .924. Since .924 is not > 1.697, we do not reject H 0 . (5) At α = .05, there is insufficient evidence to conclude that the average thickness for high viscosity exceeds the average thickness for low velocity. c. With α = .05, t 30, .025 = 2.042 so a 95% confidence interval for the true mean difference in coating thickness between the high viscosity and low viscosity is: [( f8e5 y H - f8e5 y L ) ± t 30, .025 s p {1/n H + 1/n L } 1/2 ] = (1.487 – 1.348) ± 2.042 (.4254) {1/16 + 1/16} 1/2 = [.139 ± .307] or [-.168, .446]. d. We assumed the two samples were independent of one another. We also assumed the two populations of viscosities have a common variance. Normal probability plots of the two data sets would show some concern with the normality assumption in the thickness from the high viscosity. e.
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This note was uploaded on 05/27/2008 for the course ST 314 taught by Professor Arthur during the Spring '08 term at Oregon State.

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2008hw6_solutions - ST 314 1. Text, p. 235, exercise 4.39....

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