mont4e_sm_ch13_sec03

# mont4e_sm_ch13_sec03 - Therefore n = 3 is needed Section...

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Unformatted text preview: 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) = MS E = 0.652 b) ^ 2 = d) Greater variability at wafer position 1. There is some slight curvature in the normal probability plot. 13-19 Residuals Versus WAFERPOS (response is UNIFORMI) Residuals Versus the Fitted Values (response is UNIFORMI) 1 1 Residual 0 Residual 1 2 3 4 0 -1 -1 1 2 3 4 WAFERPOS Fitted Value Normal Probability Plot of the Residuals (response is uniformi) 2 1 Normal Score 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) = MS E = 0.0148 b) ^ 2 = d) Residuals are acceptable Residuals Versus LOOM (response is OUTPUT) 0.2 Residuals Versus the Fitted Values (response is OUTPUT) 0.2 0.1 0.1 Residual Residual 0.0 0.0 -0.1 -0.1 -0.2 1 2 3 4 5 -0.2 3.8 3.9 4.0 4.1 LOOM Fitted Value 13-20 Normal Probability Plot of the Residuals (response is OUTPUT) 2 1 Normal Score 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 = 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 MS Treatments - MS E 11272 - 2451 = = 1764.2 n 5 ^2 c) Variability within batches = MSE = 2451 ^ 2 = d) The normal probability plot and the residual plots show that the model assumptions are reasonable. b) Variability between batches 13-21 Residual Plots for Yield Normal Probability Plot of the Residuals 99 90 50 10 1 -100 -50 0 Residual 50 100 Residual Percent 100 50 0 -50 -100 Residuals Versus the Fitted Values 1500 1550 Fitted Value 1600 Histogram of the Residuals 8 Frequency 6 4 2 0 -75 -50 -25 0 25 Residual 50 75 100 Residuals Versus the Order of the Data 100 Residual 50 0 -50 -100 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 Observation Order 13-29. 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 P 0.538 Do not reject H0, there is no significant difference among the chemical types. b) ^ 2 = 18.0 - 24.0 = -1 . 2 5 set equal to 0 c) ^ 2 = 24.0 Residuals Versus the Fitted Values (response is BRIGHTNE) d) Variability is smaller in chemical 4. There is some curvature in the normal probability plot. Residuals Versus CHEMICAL (response is BRIGHTNE) 10 10 Residual Residual 0 -5 5 5 0 -5 1 2 3 4 78 79 80 81 82 CHEMICAL Fitted Value 13-22 Normal Probability Plot of the Residuals (response is BRIGHTNE) 2 1 Normal Score 0 -1 -2 -5 0 5 10 Residual 13-30 a) b) ^2 ^2 ^ total = position + 2 = 2.237 ^2 position = 0.709 ^2 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 : 2 = 0 H 1 : 2 0 b) The statistical model is i = 1,2,..., a y = + i + ij j = 1,2,..., n i ~ N (0, 2 ) and i ~ N (0, 2 ) 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 SS 0.2689 11.8169 12.0858 MS 0.1345 0.1641 F 0.82 P 0.445 S = 0.405122 R-Sq = 2.23% R-Sq(adj) = 0.00% 2 = MS E = 0.1641 MS tr - MS E 0.1345 - 0.1641 2 = = = -0.001184 n 25 13-23 ...
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