The article “Anodic Fenton Treatment of Treflan MTF” describes a two-factor experiment designed to study the sorption of the herbicide trifluralin.

Bergen 2021-01-15 Answered

The article “Anodic Fenton Treatment of Treflan MTF” describes a two-factor experiment designed to study the sorption of the herbicide trifluralin. The factors are the initial trifluralin concentration and the \(\displaystyle{F}{e}^{{{2}}}\ :\ {H}_{{{2}}}\ {O}_{{{2}}}\) delivery ratio. There were three replications for each treatment. The results presented in the following table are consistent with the means and standard deviations reported in the article. \(\begin{array}{|c|c|}\hline \text{Initial Concentration (M)} & \text{Delivery Ratio} & \text{Sorption (%)} \\ \hline 15 & 1:0 & 10.90 \quad 8.47 \quad 12.43 \\ \hline 15 & 1:1 & 3.33 \quad 2.40 \quad 2.67 \\ \hline 15 & 1:5 & 0.79 \quad 0.76 \quad 0.84 \\ \hline 15 & 1:10 & 0.54 \quad 0.69 \quad 0.57 \\ \hline 40 & 1:0 & 6.84 \quad 7.68 \quad 6.79 \\ \hline 40 & 1:1 & 1.72 \quad 1.55 \quad 1.82 \\ \hline 40 & 1:5 & 0.68 \quad 0.83 \quad 0.89 \\ \hline 40 & 1:10 & 0.58 \quad 1.13 \quad 1.28 \\ \hline 100 & 1:0 & 6.61 \quad 6.66 \quad 7.43 \\ \hline 100 & 1:1 & 1.25 \quad 1.46 \quad 1.49 \\ \hline 100 & 1:5 & 1.17 \quad 1.27 \quad 1.16 \\ \hline 100 & 1:10 & 0.93 \quad 0.67 \quad 0.80\\ \hline \end{array}\) a) Estimate all main effects and interactions. b) Construct an ANOVA table. You may give ranges for the P-values. c) Is the additive model plausible? Provide the value of the test statistic, its null distribution, and the P-value.

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Bentley Leach
Answered 2021-01-16 Author has 109 answers

Step 1 Given: I=3
K=3 a) The mean is the sum of all values divided by the number of data values: X1.=3.6992
X=2.9744 Step 2 The main effects are the difference of the mean of the values in the category decreased by the overall mean. Main effects concentration
X1.  X=3.6992  2.9744=0.7248
X2.  X=2.6492  2.9744= 0.3252
X3.  X=2.575  2.9744= 0.3994
Main effects delivery ratio
X.1  X=8.2011  2.9744=5.2267
X.2  X=1.9656  2.9744= 1.0088
X.3  X=0.9322  2.9744= 2.0422

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