# 420Hw03ans - STAT 420 Spring 2010 Homework#3(12 points(due...

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STAT 420 Spring 2010 Homework #3 (12 points) (due Friday, February 12, by 4:00 p.m.) 1. a) 4.6 The American Car Company is interested in the average number of cars coming off three assembly lines with obvious defects (determined by visual inspection). The number of defects per week for each of the assembly lines is recorded for several weeks. The data follow. Assembly Line 1 2 3 49 61 55 67 48 50 68 57 64 73 66 56 61 44 53 47 58 49 Given the observations above, test for differences in the mean number of visually defective cars coming off the assembly lines using the F test. Use α = 0.10. > Y = c(49,61,55,67,48, 50,68,57,64,73,66,56, 61,44,53,47,58,49) > A = c( 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3) > summary(aov(glm(Y ~ factor(A)))) Df Sum Sq Mean Sq F value Pr(>F) factor(A) 2 330.0 165.0 2.8846 0.0871 . Residuals 15 858.0 57.2 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 p-value = 0.0871 < 0.10. Reject H 0 : μ A = μ B = μ C at α = 0.10.

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OR A n = 5 , A x = 56 , 2 A s = 65 , B n = 7 , B x = 62 , 2 B s = 63.66667 , C n = 6 , C x = 52 , 2 C s = 43.2 , J = 3 . N = n A + n B + n C = 5 + 7 + 6 = 18 . 18 1026 18 52 6 62 7 56 5 ... 2 2 1 1 = + + = + + + = N n n n J J y y y x = 57 . SSB = ( ( ( 2 2 2 2 2 1 1 ... y y y y y y J J n n n - + + - + - = 5 ( 56 – 57 ) 2 + 7 ( 62 – 57 ) 2 + 6 ( 52 – 57 ) 2 = 5 + 175 + 150 = 330 . MSB = 2 330 1 SSB = - J = 165 . SSW = ( ( ( 2 2 2 2 2 1 1 1 ... 1 1 J J s n s n s n - + + - + - = 4 65 + 6 63.66667 + 5 43.2 = 260 + 382 + 216 = 858 . MSW = 15 858 SSW = - J N = 57.2 . SSTot = SSB + SSW = 330 + 858 = 1188 . F = 2 . 57 165 MSW MSB = 2.8846 . ANOVA table: Source SS DF MS F Between 330 2 165 2.8846 Within 858 15 57.2 Total 1188 17 F > F 0.10 ( 2, 15 ) = 2.70 . Reject H 0 : μ A = μ B = μ C at α = 0.10. p-value = [ = FDIST ( 2.8846 , 2 , 15 ) ] = 0.0871 .
b) Construct three 90% ( individual ) confidence intervals for the differences μ 1 μ 2 , μ 1 μ 3 , μ 2 μ 3 . ( 29 j i J N j i n n t 1 1 MSW d.f. Y Y 2 γ + - ± - t 0.10 / 2 ( 15 ) = t 0.05 ( 15 ) = 1.753 . μ A μ B 56 – 62 ± 1.753 2 . 57 7 1 5 1 + 6 ± 7.763 μ A μ C 56 – 52 ± 1.753 2 . 57 6 1 5 1 + 4 ± 8.028 μ B μ C 62 – 52 ± 1.753 2 . 57 6 1 7 1 + 10 ± 7.376 c) Use Bonferroni method to construct three simultaneous 90% confidence intervals for the pairwise differences of means.

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