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

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 .
Image of page 2
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.
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern