Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+)
Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+)
1st Edition
ISBN: 9780137548552
Author: Michael Akritas
Publisher: PEARSON+
Question
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Chapter 1.8, Problem 17E

a.

To determine

Plot the interaction plot with pH being the trace factor.

a.

Expert Solution
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Explanation of Solution

The interaction plot with pH being the trace factor is given as follows:

Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+), Chapter 1.8, Problem 17E

b.

To determine

Check whether there is interaction between the factors pH and temperature.

b.

Expert Solution
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Answer to Problem 17E

There is interaction between the factors pH and temperature.

Explanation of Solution

From the interaction plot obtained in Part (a), it is clear that the lines of pH I and pH II are not parallel to each other. Therefore, there is interaction between the factors pH and temperature.

c.

To determine

Calculate the main pH effects and the main temperature effects.

c.

Expert Solution
Check Mark

Answer to Problem 17E

The main pH effects are αI=0.125 and αII=0.125.

The temperature effects are βA=6.375,βB=0.375,βC=2.625, and βD=4.125.

Explanation of Solution

The overall mean is computed as follows:

μ¯..=108+103+101+100+111+104+100+988=103.125

The main pH effects are calculated as follows:

αI=μ¯I.μ¯..=108+103+101+1004103.125=0.125αII=μ¯II.μ..=111+104+100+984103.125=0.125

Therefore, the main pH effects are αI=0.125 and αII=0.125.

The main temperature effects are calculated as follows:

βA=μ¯A.μ¯..=108+1112103.125=6.375βB=μ¯B.μ..=103+1042103.125=0.375βC=μ¯C.μ..=101+1002103.125=2.625βD=μ¯D.μ..=100+982103.125=4.125

Therefore, the main Pygmalion effects are βA=6.375,βB=0.375,βC=2.625, and βD=4.125.

d.

To determine

Calculate the interaction effects.

d.

Expert Solution
Check Mark

Answer to Problem 17E

The interaction effects are γIA=1.375,γIB=0.375,γIC=0.625, γID=1.125,γIIA=1.375,γIIB=0.375,γIIC=0.625,and γIID=1.125.

Explanation of Solution

The interaction effects are calculated as follows:

γIA=μIA(μ¯..+αI+βA)=108(103.1250.125+6.375)=1.375γIB=μIB(μ¯..+αI+βB)=103(103.1250.125+0.375)=0.375γIC=μIC(μ¯..+αI+βC)=101(103.1250.1252.625)=0.625γID=μID(μ¯..+αI+βD)=100(103.1250.1254.125)=1.125γIIA=μIIA(μ¯..+αII+βA)=111(103.125+0.125+6.375)=1.375γIIB=μIIB(μ¯..+αII+βB)=104(103.125+0.125+0.375)=0.375γIIC=μIIC(μ¯..+αII+βC)=100(103.125+0.1252.625)=0.625γIID=μIID(μ¯..+αII+βD)=98(103.125+0.1254.125)=1.125

Therefore, the interaction effects are γIA=1.375,γIB=0.375,γIC=0.625, γID=1.125,γIIA=1.375,γIIB=0.375,γIIC=0.625,and γIID=1.125.

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Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+)

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