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When factors A and B are fixed but factor C is random and the restricted model is used (see the footnote on page 438; there is a technical complication with the unrestricted model here), and EsMSEd \( = {\sigma ^2}\)

  1. Based on these expected mean squares, what F ratios \({H_0}:\sigma _{ABC}^2 = 0;{H_0}:\sigma _C^2 = 0\)\({H_{0:}}\gamma _B^{AB} = 0\;for all\;i.\;i\;and\;{H_{{a_i}}}{\alpha _1} = L = {\alpha _0} = 0?\)
  2. In an experiment to assess the effects of age, type of soil, and day of production on compressive strength of cement/soil mixtures, two ages (A), four types of soil (B), and three days (C, assumed random) were used, with L 5 2 observations made for each combination of factor levels. The resulting sums of squares were SSA 5 14,318.24, SSB 5 9656.40,

SSC 52270.22, SSAB 53408.93, SSAC 51442.58, SSBC 53096.21, SSABC 52832.72, and SSE 5 8655.60. Obtain the ANOVA table and carry out all tests using level .01. 33. Because of potential variability in aging due to different

Short Answer

Expert verified

It appears that there is no interaction or main effects present.

Step by step solution

01

Step 1:Calculating \(E(MSABC)\)

The given values are \({\rm{\;(where\;}}E(MSABC)\)is modified here,perhaps there is mistake in the book)

02

Step 2:Find F ratios

a)The classic F ratios should not work here because the C is random factor.

\(\frac{{E(MSABC)}}{{E(MSE)}} = \frac{{{\sigma ^2} + L \cdot \sigma _{ABC}^2}}{{{\sigma ^2}}} = 1 + \frac{{\sigma _{ABC}^2}}{{{\sigma ^2}}}\)

Which is equal to 1 when \(\sigma _{ABC}^2 = 0\)and is larger than 1 when \(\sigma _{ABC}^2 > 0\)which indicates that the following F ratio is appropriate

\({F_{ABC}} = \frac{{MSABC}}{{MSE}}\)

For testing

\({H_0}:\sigma _{ABC}^2 = 0{\rm{\;versus\;}}{H_a}:\sigma _{ABC}^2 > 0.{\rm{\;}}\)

03

Step 3:For testing when \(\sigma _C^2 = 0\)

Similarly from

\(\frac{{E(MSC)}}{{E(MSE)}} = \frac{{{\sigma ^2} + I \cdot J \cdot L \cdot \sigma _C^2}}{{{\sigma ^2}}} = 1 + \frac{{I \cdot J \cdot L \cdot \sigma _C^2}}{{{\sigma ^2}}}\)

which is equal to 1 when \(\sigma _C^2 = 0\)and is larger than 1 when \(\sigma _{ABC}^2 > 0\) which indicates F ratio is appropriate

\({F_C} = \frac{{MSC}}{{MSE}}\)

For testing

\({H_0}:\sigma _C^2 = 0{\rm{\;versus\;}}{H_a}:\sigma _C^2 > 0.{\rm{\;}}\)

04

For testing when \(\gamma _{ij}^{AB} = 0\)

From

which is equal to 1 when \(\gamma _{ij}^{AB} = 0,{\rm{\;for all\;}}i,j,{\rm{\;which indicates that the following\;}}\)indicates F ratio is appropriate.

\({F_{AB}} = \frac{{MSAB}}{{MSABC}}\)

For testing

\({H_0}:\gamma _{ij}^{AB} = 0{\rm{\;for all\;}}i,j,{\rm{\;versus\;}}{H_a}:{\rm{\;at least one\;}}{\gamma _ - }i{j^{AB}} \ne 0.{\rm{\;}}\)

05

For testing when \({\alpha _1} = {\alpha _2} =  \ldots  = {\alpha _I} = 0\)

From

which is equal to 1 \({\rm{\;when\;}}{\alpha _1} = {\alpha _2} = \ldots = {\alpha _I} = 0\)which indicates F ratio is appropriate.

\({F_A} = \frac{{MSA}}{{MSAC}}\)

For testing

\({H_0}:{\rm{\;all\;}}{\alpha _i} = 0{\rm{\;versus\;}}{H_a}:{\rm{\;at least one\;}}{\alpha _ - }i \ne 0.\)

06

Step 6:Total sum of squares SST

b)The given sum of squares are given,all expect the total sum of squares SST

\(\begin{aligned}{*{20}{c}}{SSA = 14,318.24;}\\{SSB = 9656.40;}\\{SSC = 2270.22;}\\{SSAB = 3408.93;}\\{SSAC = 1442.58;}\\{SSBC = 3096.21;}\\{SSABC = 2832.72}\\{SSE = 8655.60.}\end{aligned}\)

07

Table with corresponding values

The following table with corresponding values

08

Fundamental Identity

\(SST = SSA + SSB + SSC + SSAB + SSAC + SSBC + SSABC + SSE\)

By the fundamental identity,value of the missing values SST is

\(\begin{aligned}{*{20}{c}}{SST = SSA + SSB + SSC + SSAB + SSAC + SSBC + SSABC + SSE}\\{ = 14,318.24 + 9656.4 + 2270.22 + 3408.93 + }\\{ + 1442.58 + 3096.21 + 2832.72 + 8655.60}\\{ = 45,680.9.}\end{aligned}\)

The degrees of freedom are

\(\begin{aligned}{*{20}{c}}{d{f_T} = IJKL - 1 = 2 \cdot 4 \cdot 3 \cdot 2 - 1 = 47}\\{d{f_E} = IJK(L - 1) = 2 \cdot 4 \cdot 3 \cdot (2 - 1) = 24;}\\{d{f_A} = I - 1 = 2 - 1 = 1;}\\{d{f_B} = J - 1 = 4 - 1 = 3;}\\{d{f_C} = K - 1 = 3 - 1 = 2;}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{d{f_{AB}} = (I - 1)(J - 1) = (2 - 1) \cdot (4 - 1) = 3}\\{d{f_{AC}} = (I - 1)(K - 1) = (2 - 1) \cdot (3 - 1) = 2}\\{d{f_{BC}} = (J - 1)(K - 1) = (4 - 1) \cdot (3 - 1) = 6}\\{d{f_{ABC}} = (I - 1)(J - 1)(K - 1) = (2 - 1) \cdot (4 - 1) \cdot (3 - 1) = 6.}\end{aligned}\)

09

Step 9:Means squares

The mean squares are

\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{d{f_A}}} \cdot SSA = \frac{1}{1} \cdot 14,318.24 = 14,318.24}\\{MSB = \frac{1}{{d{f_B}}} \cdot SSB = \frac{1}{3} \cdot 9656.4 = 3218.80}\\{MSC = \frac{1}{{d{f_C}}} \cdot SSC = \frac{1}{2} \cdot 2270.22 = 1135.11}\\{MSAB = \frac{1}{{d{f_1}}} \cdot SSAB = \frac{1}{3} \cdot 3408.93 = 1136.31}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{MSAC = \frac{1}{{d{f_{AC}}}} \cdot SSAC = \frac{1}{2} \cdot 1442.58 = 721.29}\\{MSBC = \frac{1}{{d{f_{BC}}}} \cdot SSBC = \frac{1}{6} \cdot 3096.21 = 516.04}\\{MSABC = \frac{1}{{d{f_{ABC}}}} \cdot SSABC = \frac{1}{6} \cdot 2832.72 = 472.12}\\{MSE = \frac{1}{{d{f_E}}} \cdot SSE = \frac{1}{{24}} \cdot 8655.60 = 360.65}\end{aligned}\)

The corresponding f values a\({\rm{\$ }} \setminus {\rm{\;textbf\{ as proved in\;}}(a)\} {\rm{\$ , by symmetry,\;}}\) are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSAC}} = \frac{{14,318.24}}{{721.29}} = 19.85}\\{{f_B} = \frac{{MSB}}{{MSBC}} = \frac{{3218.80}}{{516.04}} = 6.24}\\{{f_C} = \frac{{MSC}}{{MSE}} = \frac{{1135.11}}{{360.65}} = 3.15}\\{{f_{AB}} = \frac{{MSAB}}{{MSABC}} = \frac{{1136.31}}{{472.12}} = 2.41}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{f_{AC}} = \frac{{MSAC}}{{MSABC}} = \frac{{721.29}}{{472.12}} = 2.00}\\{{f_{BC}} = \frac{{MSBC}}{{MSE}} = \frac{{516.04}}{{360.65}} = 1.43}\\{{f_{ABC}} = \frac{{MSABC}}{{MSE}} = \frac{{13.53}}{{360.65}} = 1.31}\end{aligned}\)

10

Calculating P values

The P value can be computed using a software.

\({P_A} = P\left( {F > {f_A}} \right) = P(F > 19.85) = 0.047\)

P-value is the area under the \({F_{I - 1,(I - 1)(J - 1)}}{\rm{\;curve to the right of the test statistic value\;}}{f_A}{\rm{.\;}}\)

\({P_B} = P\left( {F > {f_B}} \right) = P(F > 6.24) = 0.034\)

P-value is the area under the \({F_{J - 1,(J - 1)(K - 1)}}{\rm{\;curve to the right of the test statistic value\;}}{f_B}{\rm{.\;}}\)

\({P_C} = P\left( {F > {f_C}} \right) = P(F > 3.15) = 0.061\)

P-value is the area under the \({F_{K - 1.IJK(L - 1)}}{\rm{\;curve to the right of the test statistic value\;}}{f_C}{\rm{.\;}}\)

\({P_{AB}} = P\left( {F > {f_{AB}}} \right) = P(F > 2.41) = 0.165\)

P-value is the area under the\({F_{(I - 1)(J - 1),(I - 1)(J - 1)(K - 1)}}{\rm{\;curve to the right of the test statistic value\;}}{f_{AB}}{\rm{.\;}}\) \({P_{AC}} = P\left( {F > {f_{AC}}} \right) = P(F > 2.00) = 0.216\)

P-value is the area under the \(F(I - 1)(K - 1),(I - 1)(J - 1)(K - 1){\rm{\;curve to the right of the test statistic value\;}}{f_{AC}}\)

\({P_{BC}} = P\left( {F > {f_{BC}}} \right) = P(F > 1.43) = 0.244\)

P-value is the area under the \({F_{(J - 1)(K - 1),IJK(L - 1)}}{\rm{\;curve to the right of the test statistic value\;}}{f_{BC}}{\rm{.\;}}\)

\({P_{ABC}} = P\left( {F > {f_{ABC}}} \right) = P(F > 1.31) = 0.291\)

P-value is the area under the

\({F_{(I - 1)(J - 1)(K - 1),IJK(L - 1)}}{\rm{\;curve to the right of the test statistic value\;}}{f_{ABC}}\)

11

Step 11:ANOVA table values

To complete the ANOVA table,the left values are the critical values.

\(\begin{aligned}{*{20}{c}}{{F_{0.01,1,2}} = 98.50}\\{{F_{0.01,3,6}} = 9.78}\\{{F_{0.01,2,24}} = 5.61}\\{{F_{0.01,3,6}} = 9.78}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,6}} = 10.92}\\{{F_{0.01,6,24}} = 3.67}\\{{F_{0.01,6,24}} = 3.67}\end{aligned}\)

12

Step 12:ANOVA table

Finally,ANOVA table becomes

13

Step 13:Relevant Hypotheses

The relevant Hypotheses are

\(\begin{aligned}{*{20}{c}}{{H_{0A}}:{\alpha _1} = {\alpha _2} = \ldots = {\alpha _I} = 0{\rm{\;versus\;}}{H_{aA}}:{\rm{\;at least one\;}}{\alpha _ - }i \ne 0;}\\{{H_{0B}}:{\beta _1} = {\beta _2} = \ldots = {\beta _J} = 0{\rm{\;versus\;}}{H_{aB}}:{\rm{\;at least one\;}}{\beta _ - }j \ne 0;}\\{{H_{0C}}:\sigma _C^2 = 0{\rm{\;versus\;}}{H_{aC}}:\sigma _{ABC}^2 > 0,}\end{aligned}\)

For the interactions

\({H_{0AB}}:\gamma _{ij}^{AB} = 0{\rm{\;for all\;}}i,j{\rm{\;versus\;}}{H_{aAB}}:{\rm{\;at least one\;}}{\gamma _ - }i{j^{AB}} \ne 0{\rm{;\;}}\)

\({H_{0AC}}:\gamma _{ij}^{AC} = 0{\rm{\;for all\;}}i,j{\rm{\;versus\;}}{H_{aAC}}:{\rm{\;at least one\;}}{\gamma _ - }i{j^{AC}} \ne 0{\rm{;\;}}\)

\({H_{0BC}}:\gamma _{ij}^{BC} = 0{\rm{\;for all\;}}i,j{\rm{\;versus\;}}{H_{aBC}}:{\rm{\;at least one\;}}{\gamma _ - }i{j^{BC}} \ne 0{\rm{,\;}}\)

And for the interactions ABC

\({H_{0ABC}}:\sigma _{ABC}^2 = 0\quad {\rm{\;versus\;}}{H_{aABC}}:\sigma _{ABC}^2 > 0\)

14

Step 14:Find the test statistics value

\({\rm{\;When testing hypotheses\;}}{H_{0A}}{\rm{\;versus\;}}{H_{aA}}{\rm{,\;}}\)the test statistics value is

\({f_A} = \frac{{MSA}}{{MSAC}},\)

P-value is the area under the \({F_{I - 1,(I - 1)(J - 1)}}{\rm{\;curve to the right of the test statistic value\;}}{f_A}{\rm{.\;}}\)

\({\rm{\;When testing hypotheses\;}}{H_{0B}}{\rm{\;versus\;}}{H_{aB}}\), the test statistics value is

\({f_B} = \frac{{MSB}}{{MSBC}},\)

P-value is the area under the \({F_{J - 1,(J - 1)(K - 1)}}{\rm{\;curve to the right of the test statistic value\;}}{f_B}{\rm{.\;}}\)

the test statistics value is

\({f_C} = \frac{{MSC}}{{MSE}}\)

P-value is the area under the \({F_{K - 1.IJK(L - 1)}}{\rm{\;curve to the right of the test statistic value\;}}{f_C}{\rm{.\;}}\)

15

Step 15:Find the significance level

For factor A

\(\begin{aligned}{*{20}{c}}{{F_{0.01,1,2}} = 98.5 > 19.85 = {f_A}}\\{{P_A} = 0.047 > 0.01 = \alpha ,}\end{aligned}\)

\({\rm{\;do not reject null hypothesis\;}}{H_{0A}}\)

\({\rm{\;at given significance level\;}}\alpha {\rm{. Factor (main effect)\;}}\)A is not statiscally significant.

For factor B

\(\begin{aligned}{*{20}{c}}{{F_{0.01,3,6}} = 9.78 > 6.24 = {f_B}}\\{{P_B} = 0.034 > 0.01 = \alpha }\end{aligned}\)

\({\rm{\;do not reject null hypothesis\;}}{H_{0B}}\)

\({\rm{\;at given significance level\;}}\alpha {\rm{. Factor (main effect)\;}}\)B is not statiscally significant.

For factor C

\(\begin{aligned}{*{20}{r}}{{F_{0.01,2,24}} = 5.61 > 3.15 = {f_C}}\\{{P_C} = 0.061 > 0.01 = \alpha }\end{aligned}\)

\({\rm{\;do not reject null hypothesis\;}}{H_{0C}}\)

\({\rm{\;at given significance level\;}}\alpha {\rm{. Factor (main effect)\;}}\)C is not statiscally significant.

16

Step 16:Find the test statistic value

When testing hypotheses \({H_{0AB}}{\rm{\;versus\;}}{H_{aAB}}\), the test statistic value is \({f_{AB}} = \frac{{MSAB}}{{MSABC}}\)and the p-value is the area under the \({F_{(I - 1)(J - 1),(I - 1)(J - 1)(K - 1)}}\)curve to the right of the test statistic value \({f_{AB}}\).

When testing hypotheses \({H_{0AC}}{\rm{\;versus\;}}{H_{aAC}}\), the test statistic value is \({f_{AC}} = \frac{{MSAC}}{{MSABC}},\) and

the p-value is the area under the \({F_{(I - 1)(J - 1),(I - 1)(J - 1)(K - 1)}}\)curve to the right of the test statistic value\({f_{AC}}\).

When testing hypotheses\({H_{0BC}}{\rm{\;versus\;}}{H_{aBC}}{\rm{,\;}}\) the test statistic value is \({f_{BC}} = \frac{{MSBC}}{{MSE}}\)and

the p-value is the area under the \({F_{(I - 1)(J - 1),(I - 1)(J - 1)(K - 1)}}\)curve to the right of the test statistic value \({f_{BC}}\)

When testing hypotheses \({H_{0ABC}}{\rm{\;versus\;}}{H_{aABC}}\), the test statistic value is \({f_{ABC}} = \frac{{MSABC}}{{MSE}}\)and

the p-value is the area under the \({F_{(I - 1)(J - 1),(I - 1)(J - 1)(K - 1)}}\)curve to the right of the test statistic value \({f_{ABC}}.\)

17

There is no interaction or main effects present

For interaction AB

\(\begin{aligned}{*{20}{c}}{{F_{0.01,3,6}} = 9.78 > 2.41 = {f_{AB}};}\\{{P_{AB}} = 0.165 > 0.01 = \alpha ,}\end{aligned}\)

\({\rm{\;do not reject null hypothesis\;}}{H_{0AB}}\)

\({\rm{\;at given significance level}}{\rm{. Interaction\;}}AB{\rm{\;is not statistically significant}}{\rm{.\;}}\)

For interaction AC

\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,5}} = 10.92 > 2.00 = {f_{AC}}}\\{{P_{AB}} = 0.216 > 0.01 = \alpha }\end{aligned}\)

\({\rm{\;do not reject null hypothesis\;}}{H_{0AC}}\)

\({\rm{\;at given significance level}}{\rm{. Interaction\;}}AC{\rm{\;is not statistically significant}}{\rm{.\;}}\)

For interaction BC

\(\begin{aligned}{*{20}{c}}{{F_{0.01,6,24}} = 3.67 > 1.43 = {f_{BC}}}\\{{P_{BC}} = 0.244 > 0.01 = \alpha }\end{aligned}\)

\({\rm{\;do not reject null hypothesis\;}}{H_{0BC}}\)

\({\rm{\;at given significance level}}{\rm{. Interaction\;}}BC{\rm{\;is not statistically significant}}{\rm{.\;}}\)

For interaction ABC

\(\begin{aligned}{*{20}{c}}{{F_{0.01,6,24}} = 3.67 > 1.31 = {f_{ABC}}}\\{{P_{ABC}} = 0.291 > 0.01 = \alpha }\end{aligned}\)

\({\rm{\;do not reject null hypothesis\;}}{H_{0ABC}}\)

\({\rm{\;at given significance level}}{\rm{. Interaction\;}}ABC{\rm{, the three way interaction, is not statistically significant}}{\rm{.\;}}\)

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Most popular questions from this chapter

Four different coatings are being considered for corrosion protection of metal pipe. The pipe will be buried in three different types of soil. To investigate whether the amount of corrosion depends either on the coating or on the type of soil, 12 pieces of pipe are selected. Each piece is coated with one of the four coatings and buried in one of the three types of soil for a fixed time, after which the amount of corrosion (depth of maximum pits, in .0001 in.) is determined. The data appears in the table.

a. Assuming the validity of the additive model, carry out the ANOVA analysis using an ANOVA table to see whether the amount of corrosion depends on either the type of coating used or the type of soil. Use=.05.

b. Compute\(\hat \mu ,{\hat \alpha _1},{\hat \alpha _2},{\hat \alpha _3},{\hat \alpha _4},{\hat \beta _1},{\hat \beta _2},\;and\;{\hat \beta _3}\)

Show how a \(10 \times (1 - \alpha ){\rm{\% }} + Cl\) for \({\alpha _i} - {\alpha _i}\)can be obtained. Then compute a 95% interval for \({\alpha _2} - {\alpha _3}\)using the data from Exercise 19. (Hint: With \({\alpha _2} - {\alpha _3}\)the result of Exercise 24(a) indicates how to obtain \(\hat \theta \). Then compute \(V(\hat \theta )\)and\(\sigma _{\dot \theta }^{}\), and obtain an estimate of \(\sigma _{\dot \theta }^{}\)by using \(\sqrt {{\rm{MSE}}} \) to estimate a (which identifies the appropriate number of df).)

When both factors are random in a two-way ANOVA experiment with K replications per combination of factor levels, the expected mean squares are \(E({\rm{MSE}}) = {\sigma ^2},E({\rm{MSA}}) = {\sigma ^2} + K\sigma _G^2 + JK\sigma _A^2,E({\rm{MSB}}) = {\sigma ^2} + K\sigma _G^2 + IK\sigma _B^2,{\rm{\;and\;}}E({\rm{MSAB}}) = {\sigma ^2} + K\sigma _{{G^2}}^2\) a. What F ratio is appropriate for testing \({H_{O{C^ * }}}\sigma _G^2 = 0\) versus \({H_{a{B^ * }}}\sigma _B^2 > 0\) ?

b. Answer part (a) for testing \({H_{0{\rm{A}}}}:\sigma _A^2 = 0\)versus \({H_{aA}}:\sigma _A^2 > 0{\rm{\;and\;}}{H_{0B}}:\sigma _B^2 = 0{\rm{\;versus\;}}H_{aB}^ * \sigma _B^2 > 0.\)

Use the fact that \(E\left( {{X_{ij}}} \right) = \mu \pm {\alpha _i} + {\beta _j}\)with \(\Sigma {\alpha _i} = \Sigma {\beta _j} = 0\) to show that\(E\left( {{{\bar X}_{i \times }} - {{\bar X}_{..}}} \right) = {\alpha _i}\). , so that \({\hat \alpha _i} = {\bar X_{i \times }} - \bar X\)is an unbiased estimator for\({\alpha _i}\).

Suppose that in the experiment described in Exercise 6 the five houses had actually been selected at random from among those of a certain age and size, so that factor B is random rather than fixed. Test \({H_0}:\sigma _B^2 = 0\) versus \({H_a}:\sigma _B^2 > 0\)using a level .01 test.

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