mont4e_sm_ch13_sec03.pdf

# mont4e_sm_ch13_sec03.pdf - -0.25 0.00 0.25 0.50 b The mean...

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Unformatted text preview: -0.25 0.00 0.25 0.50 b) The mean values are: 3.886, 3.8611, 3.76 (barley, b+l, lupins) From the ANOVA the estimate of σ can be obtained Source DF SS MS F P C4 2 0.235 0.118 0.72 0.489 Error 76 12.364 0.163 Total 78 12.599 S = 0.4033 R-Sq = 1.87% R-Sq(adj) = 0.00% The minimum sample size could be used to calculate the standard error of a sample mean σˆ X = MS E = b 0.163 = 0.081 25 The graph would not show any differences between the diets. 13-24. μ = 55 , τ1 = -5, τ2 = 5, τ3 = -5, τ4 = 5. n(100) Φ2 = = n, a −1 = 3 4(25) Various choices for n yield: n 4 5 Φ2 4 5 a (n − 1) = 4(n − 1) Φ 2 2.24 a(n-1) 12 16 Power=1-β 0.80 0.90 Therefore, n = 5 is needed. 13-25 μ = 188 , τ1 = -13, τ2 = 2, τ3 = -28, τ4 = 12, τ5 = 27. n(1830) Φ2 = = n, a −1 = 4 a (n − 1) = 5(n − 1) 5(100) Various choices for n yield: n 2 3 Φ2 7.32 10.98 Φ 2.7 3.13 a(n-1) 5 10 Power=1-β 0.55 0.95 Therefore, n = 3 is needed. Section 13-3 13-26 a) Analysis of Variance for UNIFORMITY Source DF SS MS WAFERPOS 3 16.220 5.407 Error 8 5.217 0.652 Total 11 21.437 F 8.29 P 0.008 Reject H0, and conclude that there are significant differences among wafer positions. MSTreatments − MS E 5.407 − 0.652 = = 1.585 n 3 2 c) σˆ = MS E = 0.652 b) σˆ τ2 = d) Greater variability at wafer position 1. There is some slight curvature in the normal probability plot. 13-19 Residuals Versus WAFERPOS Residuals Versus the Fitted Values (response is UNIFORMI) (response is UNIFORMI) 1 Residual Residual 1 0 -1 0 -1 1 2 3 4 1 2 3 WAFERPOS 4 Fitted Value Normal Probability Plot of the Residuals (response is uniformi) 2 Normal Score 1 0 -1 -2 -1 0 1 Residual 13-27 a) Analysis of Variance for OUTPUT Source DF SS MS LOOM 4 0.3416 0.0854 Error 20 0.2960 0.0148 Total 24 0.6376 F 5.77 P 0.003 Reject H0, there are significant differences among the looms. MSTreatments − MS E 0.0854 − 0.0148 = = 0.01412 n 5 2 c) σˆ = MS E = 0.0148 b) σˆ τ2 = d) Residuals are acceptable Residuals Versus the Fitted Values Residuals Versus LOOM (response is OUTPUT) 0.2 0.2 0.1 0.1 Residual Residual (response is OUTPUT) 0.0 0.0 -0.1 -0.1 -0.2 -0.2 1 2 3 4 3.8 5 3.9 4.0 Fitted Value LOOM 13-20 4.1 Normal Probability Plot of the Residuals (response is OUTPUT) 2 Normal Score 1 0 -1 -2 -0.2 -0.1 0.0 0.1 0.2 Residual 13-28. a) Yes, the different batches of raw material significantly affect mean yield at α = 0.01 because pvalue is small. Source Batch Error Total DF 5 24 29 SS 56358 58830 115188 MS 11272 2451 F 4.60 P 0.004 b) Variability between batches MS Treatments − MS E 11272 − 2451 = = 1764.2 n 5 2 c) Variability within batches σˆ = MSE = 2451 σˆτ2 = d) The normal probability plot and the residual plots show that the model assumptions are reasonable. 13-21 Residual Plots for Yield Normal Probability Plot of the Residuals Residuals Versus the Fitted Values 99 100 50 Residual Percent 90 50 0 -50 10 1 -100 -50 0 Residual 50 -100 100 100 6 50 4 13-29. -75 -50 -25 0 25 Residual 50 75 1600 0 -50 2 0 1550 Fitted Value Residuals Versus the Order of the Data 8 Residual Frequency Histogram of the Residuals 1500 -100 100 2 4 6 a) Analysis of Variance for BRIGHTNENESS Source DF SS MS F CHEMICAL 3 54.0 18.0 0.75 Error 16 384.0 24.0 Total 19 438.0 8 10 12 14 16 18 20 22 24 26 28 30 Observation Order P 0.538 Do not reject H0, there is no significant difference among the chemical types. 18.0 − 24.0 = −1 . 2 5 b) σˆ τ2 = c) σˆ 2 = 24.0 set equal to 0 d) Variability is smaller in chemical 4. There is some curvature in the normal probability plot. Residuals Versus CHEMICAL Residuals Versus the Fitted Values (response is BRIGHTNE) (response is BRIGHTNE) 10 5 5 Residual Residual 10 0 0 -5 -5 1 2 3 78 4 79 80 Fitted Value CHEMICAL 13-22 81 82 Normal Probability Plot of the Residuals (response is BRIGHTNE) 2 Normal Score 1 0 -1 -2 -5 0 5 10 Residual 13-30 a) 2 2 σˆ total = σˆ position + σˆ 2 = 2.237 b) 2 σˆ position = 0.709 2 σˆ total c) It could be reduced to 0.6522. This is a reduction of approximately 71%. 13-31. a) Instead of testing the hypothesis that the individual treatment effects are zero, we are testing whether there is variability in protein content between all diets. H 0 : σ τ2 = 0 H 1 : σ τ2 ≠ 0 b) The statistical model is ⎧i = 1,2,..., a y = μ + τ i + ε ij ⎨ ⎩ j = 1,2,..., n ε i ~ N (0, σ 2 ) and τ i ~ N (0, σ τ2 ) c) The last TWO observations were omitted from two diets to generate equal sample sizes with n = 25. ANOVA: Protein versus DietType Analysis of Variance for Protein Source DietType Error Total DF 2 72 74 S = 0.405122 SS 0.2689 11.8169 12.0858 MS 0.1345 0.1641 R-Sq = 2.23% F 0.82 P 0.445 R-Sq(adj) = 0.00% σ 2 = MS E = 0.1641 MS tr − MS E 0.1345 − 0.1641 σ τ2 = = = −0.001184 n 25 13-23 ...
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