HW13

ISyE 3030 Homework 13 Hannah McClure 12/1/2021 Exercise 13-8 a) a = 3 , n = 5 , α = 0 . 01 , H 0 : μ 1 = μ 2 = μ 3 , H 1 : at least one mean differs from the others. y1 = c( 19 , 22 , 20 , 18 , 25 ) y2 = c( 20 , 21 , 33 , 27 , 40 ) y3 = c( 16 , 15 , 18 , 26 , 17 ) y = c(y1,y2,y3) circuit = rep( 1 : 3 , rep( 5 , 3 )) model1 = lm(y~factor(circuit)) anova(model1) ## Analysis of Variance Table ## ## Response: y ## Df Sum Sq Mean Sq F value Pr(>F) ## factor(circuit) 2 260.93 130.467 4.0061 0.04648 * ## Residuals 12 390.80 32.567 ## --- ## Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 qf( 0.99 , 2 , 12 ) ## [1] 6.926608 f 0 = 4 . 0061 f α,a - 1 ,a ( n - 1) = f 0 . 01 , 2 , 12 = 6 . 9266 f 0 . 01 , 2 , 12 = 6 . 9266 > f 0 = 4 . 0061 → Fail to Reject H 0 pval = 0 . 04648 > α = 0 . 01 → Fail to Reject H 0 ∴ We can conclude the three circuit types have the same response time. c) 95% confidence interval on the response time for circuit 3. sum(y3)/ 5 ## [1] 18.4 qt( 0.975 , 12 ) ## [1] 2.178813 1

¯ y i . - t α 2 ,a ( n - 1) MSE n ≤ μ i ≤ ¯ y i . + t α 2 ,a ( n - 1) MSE n MSE = 32 . 567 (from ANOVA chart in (a)) ¯ y 3 . = 18 . 4 (calculated in R above) 18 . 4 - t 0 . 025 , 12 32 . 567 5 ≤ μ 3 ≤ 18 . 4 + t 0 . 025 , 12 32 . 567 5 → 18 . 4 - 2 . 1788 32 . 567 5 ≤ μ 3 ≤ 18 . 4 + 2 . 1788 32 . 567 5 → 12 . 8394 ≤ μ 3 ≤ 23 . 9606 Exercise 13.2.8 a) a = 4 , n = 5 , α = 0 . 05 , H 0 : μ 1 = μ 2 = μ 3 = μ 4 , H 1 : at least one mean differs from the others. m = read.table( "13-12.txt" , header = TRUE) attach(m) model2 = lm(Temperature~factor(Method)) anova(model2) ## Analysis of Variance Table ## ## Response: Temperature ## Df Sum Sq Mean Sq F value Pr(>F) ## factor(Method) 3 22.124 7.3747 14.846 6.952e-05 *** ## Residuals 16 7.948 0.4968 ## --- ## Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 qf( 0.95 , 3 , 16 ) ## [1] 3.238872 f 0 = 14 . 846 f α,a - 1 ,a ( n - 1) = f 0 . 05 , 3 , 16 = 3 . 2389 f 0 . 05 , 3 , 16 = 3 . 2389 < f 0 = 14 . 846 → Reject H 0 ∴ We can conclude the presence of oxygen during preparation affects the mean transition temperature. b) pval = 6 . 952 × 10 - 5 < α = 0 . 05 → Reject H 0 ∴ We can conclude the presence of oxygen during preparation affects the mean transition temperature. c) 95% confidence interval on mean Tc when method 1 is used splittemp = split(Temperature, Method) m1 = splittemp$ 1 sum(m1)/ 5 ## [1] 14.8 qt( 0.975 , 16 ) ## [1] 2.119905 2

LSD = 0 . 94496 | ¯ y 1 . - ¯ y 3 . | = 2 . 50 > LSD = 0 . 94496 → μ 1 and μ 3 are significantly different | ¯ y 1 . - ¯ y 4 . | = 1 . 42 > LSD = 0 . 94496 → μ 1 and μ 4 are significantly different | ¯ y 2 . - ¯ y 3 . | = 2 . 50 > LSD = 0 . 94496 → μ 2 and μ 3 are significantly different | ¯ y 2 . - ¯ y 4 . | = 1 . 42 > LSD = 0 . 94496 → μ 2 and μ 4 are significantly different | ¯ y 3 . - ¯ y 4 . | = 1 . 08 > LSD = 0 . 94496 → μ 3 and μ 4 are significantly different Exercise 13.4.3 a) n = read.table( "13-47.txt" , header = TRUE) attach(n) model3 = lm(Shape~factor(NozzleType)+factor(Velocity)) anova(model3) ## Analysis of Variance Table ## ## Response: Shape ## Df Sum Sq Mean Sq F value Pr(>F) ## factor(NozzleType) 4 0.102180 0.025545 8.9162 0.0002655 *** ## factor(Velocity) 5 0.062867 0.012573 4.3886 0.0073642 ** ## Residuals 20 0.057300 0.002865 ## --- ## Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 qf( 0.95 , 3 , 16 ) ## [1] 3.238872 boxplot(Shape~NozzleType, col = rainbow( 5 )) 1 2 3 4 5 0.8 0.9 1.0 1.1 NozzleType Shape 4

boxplot(Shape~Velocity, col = rainbow( 5 )) 11.73 14.37 16.59 20.43 23.46 28.74 0.8 0.9 1.0 1.1 Velocity Shape f 0 = 8 . 9162 → By looking at the box plots, we can Reject H 0 . b) library(agricolae) comp2 = LSD.test(model3, "factor(NozzleType)" , group = FALSE) comp2 ## $statistics ## MSerror Df Mean CV t.value LSD ## 0.002865 20 0.8586667 6.233582 2.085963 0.06446268 ## ## $parameters ## test p.ajusted name.t ntr alpha ## Fisher-LSD none factor(NozzleType) 5 0.05 ## ## $means ## Shape std r LCL UCL Min Max Q25 Q50 Q75 ## 1 0.7816667 0.02136976 6 0.7360847 0.8272487 0.75 0.81 0.7725 0.780 0.7950 ## 2 0.8533333 0.03723797 6 0.8077513 0.8989153 0.81 0.92 0.8350 0.850 0.8575 ## 3 0.9016667 0.04215052 6 0.8560847 0.9472487 0.83 0.95 0.8900 0.905 0.9275 ## 4 0.9433333 0.11360751 6 0.8977513 0.9889153 0.83 1.14 0.8650 0.925 0.9775 ## 5 0.8133333 0.08664102 6 0.7677513 0.8589153 0.75 0.97 0.7600 0.770 0.8400 ## ## $comparison ## difference pvalue signif. LCL UCL ## 1 - 2 -0.07166667 0.0311 * -0.13612935 -0.007203983 ## 1 - 3 -0.12000000 0.0009 *** -0.18446268 -0.055537317 ## 1 - 4 -0.16166667 0.0000 *** -0.22612935 -0.097203983 ## 1 - 5 -0.03166667 0.3177 -0.09612935 0.032796017 ## 2 - 3 -0.04833333 0.1335 -0.11279602 0.016129350 5