Question: question 5.20 {r} library (knitr) concentration = factor (rep (c (2, 4, 8) , each = 2)) dat2 = c(196. 6, 196.0, 198.5, 197.2, 197.5,



question 5.20 {r} library (knitr) concentration = factor (rep (c (2, 4, 8) , each = 2)) dat2 = c(196. 6, 196.0, 198.5, 197.2, 197.5, 196.6, 197. 7, 196.0, 196.0, 196.9, 195.6, 196.2, 199 . 8, 199.4, 198. 4, 197.6, 197 .4, 198.1, 198 . 4, 198.6, 197.5, 198.1, 197.6, 198.4, 199. 6, 200.4, 198. 7, 198.0, 197.0, 197.8, 200 . 6, 200.9, 199. 6, 199.0, 198.5, 199.8) paper = as . data . frame (matrix (dat2, now = 6)) paper = cbind (concentration, paper) colnames (paper) [2:7] = c('3.0 hours - pressure 400 '3.0 hours - pressure 500', '3.0 hours - pressure 650' , '3.5 hours - pressure 400', '3.5 hours - pressure 500' , '3.5 hours - pressure 650') kable (paper) {r} paper = pivot_longer( paper , cols = -c(' concentration' ), values_to = 'strength'5.20 The percentage of hardwood concentration in raw pulp, the vat pressure. and the cooking time of the pulp are being investigated for their effects on the strength of paper. Three levels of hardwood concentration. three levels of pressure, and two cooking times are selected. A factorial experiment with two replicates is conducted, and the following data are obtained: Cooking Time 3.0 Hours Percentage of Hardwood \"9+ Concentration 400 500 650 2 196.6 197.7 199.8 196.0 196.0 199.4 4 198.5 196.0 198.4 197.2 196.9 197-6 8 197.5 195.6 197.4 196.6 196.2 198.1 Cooking Time 4.0 Hours Percentage of Hardwood .__._P"e Concentration 400 500 650 2 198.4 199.6 200.6 1 98.6 200.4 200.9 4 197.5 198.7 199.6 198.1 198.0 199.0 8 197.6 197.0 198.5 198.4 197.8 199.8 (a) Analyze the data and draw conclusions. Use ti = 0.05. (a) Analyze the data and draw conclusions. Use a = 0.05. (I1) Prepare appropriate residual plots and comment on the model's adequacy. (e) Under what set of conditions would you operate this process? Why
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