# 420Hw06ans - STAT 420 Homework #6 (due Friday, October 12,...

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STAT 420 Fall 2007 Homework #6 (due Friday, October 12, by 3:00 p.m.) 1. Recall Problems #1 and #2 from Homework 2: It has been proposed that the brightness measured in some unit of color or a commercial product is proportional to the time it is in a certain chemical reaction during the production process, or Y i = β x i + ε i , i = 1, 2, … , n , where ε i ’s are i.i.d. N ( 0, σ 2 ), where Y i measures brightness, x i measures time, and β is a parameter. The following data on x and Y are available: x 1.0 1.2 1.4 1.6 1.8 2.0 Y 3.4 7.0 6.0 9.0 8.0 11.0 x y y ˆ = 5 x e = y y ˆ 1.0 3.4 5 1.6 1.2 7.0 6 1.0 1.4 6.0 7 1.0 1.6 9.0 8 1.0 1.8 8.0 9 1.0 Then ˆ = = = n i i n i i i x y x 1 2 1 = 5. 2.0 11.0 10 1.0 Without β 0 in the model, SSTotal = Y T Y = Σ y 2 . Without β 0 in the model, dfTotal = n . RSS = Σ e 2 . Without β 0 in the model, SS R = Y Y ˆ ˆ T = 2 y ˆ = Σ 2 ˆ x 2 . Set up the ANOVA table. Does β appears to be different from zero? State the null and the alternative hypotheses, report the value of the test statistic, the critical value, and the decision. Use a 5% level of significance.

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> x <- c(1.0, 1.2, 1.4, 1.6, 1.8, 2.0) > y <- c(3.4, 7.0, 6.0, 9.0, 8.0, 11.0) > fit <- lm(y ~ x + 0 ) or > fit <- lm(y ~ x - 1 ) > summary(fit) Call: lm(formula = y ~ x - 1) Residuals: 1 2 3 4 5 6 -1.6 1.0 -1.0 1.0 -1.0 1.0 Coefficients: Estimate Std. Error t value Pr(>|t|) x 5.0000 0.3263 15.32 2.15e-05 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 1.23 on 5 degrees of freedom Multiple R-Squared: 0.9791, Adjusted R-squared: 0.975 F-statistic: 234.8 on 1 and 5 DF, p-value: 2.148e-05 < 0.05 = α Reject H 0 : β = 0 at α = 0.05. OR > sum(y^2) [1] 362.56 > sum(fit\$residuals^2) [1] 7.56 > sum(fit\$fitted.values^2) [1] 355 SSTotal = Σ y 2 = 362.56. RSS = Σ e 2 = 7.56. SS R = 2 ˆ y = 355. Source SS df MS F Regression 355 1 355 234.7884 Error (Residual) 7.56 5 1.512 Total 362.56 6 Critical Value: F 0.05 ( 1 , 5 ) = 6.61 . Reject H 0 : β = 0 at α = 0.05.
OR > anova(fit) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) x 1 355.00 355.00 234.79 2.148e-05 *** Residuals 5 7.56 1.51 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Critical Value: F 0.05 ( 1 , 5 ) = 6.61 . Reject H 0 : β = 0 at α = 0.05. 2. The data in the table below came from a recent time study of a sample of 15 employees performing a particular task on an automobile assembly line. Time to Assemble,

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## This note was uploaded on 04/29/2010 for the course STAT stat 420 taught by Professor Stepanov during the Spring '07 term at University of Illinois at Urbana–Champaign.

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420Hw06ans - STAT 420 Homework #6 (due Friday, October 12,...

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