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ANOVA and Regression Analysis for Factor Experiments: Stat 875 HW10 - Prof. Xiaomi Hu, Assignments of Statistics

Wichita State University (WSU)Statistics
Prof. Xiaomi Hu

The solutions to problem 10 of statistics 875 homework. It includes the anova table for a model with percentage of hardwood concentration (phc), pressure (p), and cooking time (ct) as factors. Conclusions about the significance of interactions and main effects are drawn at a 0.05 significance level. Additionally, the document presents the results of a regression analysis with factors time (t) and culture medium (c), including the estimated factor effects and ss, and the parameter table from sas output.

Typology: Assignments

Pre 2010

Uploaded on 08/18/2009

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Stat 875 HW10
1. p204 5.18
(a) Present the ANOVA table for model
yijkl =µ+αi+βj+γk+ (αβ )ij + (αγ)ik + (βγ)jk + (αβγ)ij k +²
Let PHC be the percentage of hardwood concentration, P be the
pressure, and CT be the cooking time.
Sum of
Source DF Squares Mean Square F Value Pr > F
Model 17 59.72888889 3.51346405 9.61 <.0001
Error 18 6.58000000 0.36555556
C. Total 35 66.30888889
----------------------------------------------------
PHC 2 7.76388889 3.88194444 10.62 0.0009
P 2 19.37388889 9.68694444 26.50 <.0001
CT 1 20.25000000 20.25000000 55.40 <.0001
PHC*P 4 6.09111111 1.52277778 4.17 0.0146
PHC*CT 2 2.08166667 1.04083333 2.85 0.0843
P*CT 2 2.19500000 1.09750000 3.00 0.0750
PHC*P*CT 4 1.97333333 0.49333333 1.35 0.2903
(b) Based on the ANOVA table draw your conclusions at level 0.05.
(Conclusions only)
The interaction of PHC, PC and CT is not significant
The interaction of PHC and CT is not significant
The interaction of P and CT is not significant.
1
pf2

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Stat 875 HW

  1. p204 5.

(a) Present the ANOVA table for model

yijkl = μ + αi + βj + γk + (αβ)ij + (αγ)ik + (βγ)jk + (αβγ)ijk + ≤

Let PHC be the percentage of hardwood concentration, P be the pressure, and CT be the cooking time.

Sum of Source DF Squares Mean Square F Value Pr > F

Model 17 59.72888889 3.51346405 9.61 <. Error 18 6.58000000 0. C. Total 35 66.

PHC 2 7.76388889 3.88194444 10.62 0. P 2 19.37388889 9.68694444 26.50 <. CT 1 20.25000000 20.25000000 55.40 <. PHCP 4 6.09111111 1.52277778 4.17 0. PHCCT 2 2.08166667 1.04083333 2.85 0. PCT 2 2.19500000 1.09750000 3.00 0. PHCP*CT 4 1.97333333 0.49333333 1.35 0.

(b) Based on the ANOVA table draw your conclusions at level 0.05. (Conclusions only)

The interaction of PHC, PC and CT is not significant The interaction of PHC and CT is not significant The interaction of P and CT is not significant.

  1. p265 6.

The signs of the estimates depend on how ±1 were assigned to the levels. But they should agree in (a) and (b).

(a) Let T be the time factor and C be the culture medium factor. Find estimated factor effects and SS by fill out the table below.

ab a b (1) Est. Effect SS T 1 1 -1 -1 4.9583 590. C 1 -1 1 -1 -0.6250 9. T*C 1 -1 -1 1 -1.9583 92. Mean: 32 37.1667 26 23.

(b) Run SAS proc reg; and present the part of SAS output on param- eter table

Parameter Standard Variable DF Estimate Error t Value Pr > |t|

Intercept 1 29.62500 0.46135 64.21 <. x1 1 4.95833 0.46135 10.75 <. x2 1 -0.62500 0.46135 -1.35 0. x12 1 -1.95833 0.46135 -4.24 0.