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Chapter 11: Q26E (page 459)

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 \({H_{aA}}:\sigma _A^2 > 0{\rm{\;and\;}}{H_{0B}}:\sigma _B^2 = 0{\rm{\;versus\;}}H_{aB}^ * \sigma _B^2 > 0.\)

Short Answer

Expert verified

\({\rm{\;a}}{\rm{.\;}}\begin{aligned}{*{20}{r}}{{F_{AB}} = \frac{{MSAB}}{{MSE}};{\rm{\;b}}{\rm{.\;}}{F_A} = \frac{{MSA}}{{MSAB}}}\\{{F_B} = \frac{{MSB}}{{MSAB}}}\end{aligned}\)

Step by step solution

01

Find F ratio

The given values are

\(\begin{aligned}{*{20}{c}}{E(MSE) = {\sigma ^2};}\\{E(MSA) = {\sigma ^2} + K \cdot \sigma _G^2 + JK \cdot \sigma _A^2;}\\{E(MSB) = {\sigma ^2} + K \cdot \sigma _G^2 + IK \cdot \sigma _B^2;}\\{E(MSAB) = {\sigma ^2} + K \cdot \sigma _G^2.}\end{aligned}\)

  1. In order to obtain the F ratio, look at the following division (you might have to try more combinations)

\(\frac{{E(MSAB)}}{{E(MSE)}} = \frac{{{\sigma ^2} + K \cdot \sigma _G^2}}{{{\sigma ^2}}} = 1 + \frac{{K \cdot \sigma _G^2}}{{{\sigma ^2}}}\)

for which, when \(\sigma _G^2 = 0\)

\(\frac{{E(MSAB)}}{{E(MSE)}} = 1\)

and, when \(\sigma _G^2 > 0\)

Because \(\frac{{K \cdot \sigma _G^2}}{{{\sigma ^2}}} > 1.\)

This indicates that the appropriate F ratio is

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

(b):

Look at the following division

\(\frac{{E(MSA)}}{{E(MSAB)}} = \frac{{{\sigma ^2} + K \cdot \sigma _G^2 + JK \cdot \sigma _A^2}}{{{\sigma ^2} + K \cdot \sigma _G^2}} = 1 + \frac{{JK \cdot \sigma _A^2}}{{{\sigma ^2} + K \cdot \sigma _G^2}},\)

for which, when \(\sigma _A^2 = 0\)

\(\frac{{E(MSA)}}{{E(MSAB)}} = 1\)

and, when \(\sigma _A^2 > 0\)

\(\frac{{E(MSA)}}{{E(MSAB)}} > 1\)

because

\(\frac{{JK \cdot \sigma _A^2}}{{{\sigma ^2} + K \cdot \sigma _G^2}} > 1\)

This indicates that the appropriate F ratio is

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

For the other hypotheses, look at the following division

\(\frac{{E(MSB)}}{{E(MSAB)}} = \frac{{{\sigma ^2} + K \cdot \sigma _G^2 + JK \cdot \sigma _B^2}}{{{\sigma ^2} + K \cdot \sigma _G^2}} = 1 + \frac{{IK \cdot \sigma _B^2}}{{{\sigma ^2} + K \cdot \sigma _G^2}},\)

for which, when \(\sigma _B^2 = 0\)

\(\frac{{E(MSB)}}{{E(MSAB)}} = 1\)

and, when \(\sigma _B^2 > 0\)

\(\frac{{E(MSB)}}{{E(MSAB)}} > 1\)

because

\(\frac{{JK \cdot \sigma _B^2}}{{{\sigma ^2} + K \cdot \sigma _G^2}} > 1\)

This indicates that the appropriate F ratio is

\({F_B} = \frac{{MSB}}{{MSAB}}\)

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Most popular questions from this chapter

The accompanying data resulted from an experiment to investigate whether yield from a certain chemical process depended either on the formulation of a particular input or on mixer speed.

A statistical computer package gave \(SS(\;Form\;) = 2253.44SS(\;Speed\;) = 230.81,\quad SS(\;Form*Speed\;) = 18.58, andSSE = 71.87\;\)

a. Does there appear to be interaction between the factors?

b. Does yield appear to depend on either formulation or speed?

c. Calculate estimates of the main effects.

d. The fitted values are\({\hat x_{ijk}} = \hat \mu + {\hat \alpha _i} + {\hat \beta _j} + {\hat \gamma _{ij}}\), and the residuals are \({x_{ijk}} - {\hat x_{ij{k^*}}}\)Verify that the residuals \(are.23, - .87,.63,4.50, - 1.20, - 3.30, - 2.03,1.97.07, - 1.10, - .30,1.40,.67, - 1.23,.57, - 3.43, - .13,\;\;and 3.57.\;\)e. Construct a normal probability plot from the residuals given in part (d). Do they \({ \in _{ijk}}\;'s\;\)appear to be normally distributed?

The article 鈥淎n Assessment of the Effects of Treatment, Time, and Heat on the Removal of Erasable Pen Marks from Cotton and Cotton/Polyester Blend Fabrics (J. of Testing and Eval., 1991: 394鈥�397) reports the following sums of squares for the response variable degree of removal of marks:

\(\begin{aligned}{l}SSA = 39.171,SSB = .665,SSC = 21.508, SSAB 51.432,SSAC = 15.953,SSBC = 1.382,\\SSABC = 9.016, and SSE = 115.820\end{aligned}\)Four different laundry treatments, three different types of pen, and six different fabrics were used in the experiment, and there were three observations for each treatment-pen-fabric combination. Perform an analysis of variance using \(\alpha = .01\) for each test, and state your conclusions (assume fixed effects for all three factors).

In an experiment in investigale the effect of "cement factor" (number of sacks of cement per cub艒c yard) on flex= ural strength of the resulting concrele ('Studies of Flecural Strength of Concrete. Part 3: Effects of Variation in Testing Procedure, Proceedings, ASTM, 1957: 1127-1139 ), I=3 different factor values were used, I=5 different batches of cement were selected, and K=2 beams were cast from cach cement factod batch combination. Sums of squares include SSA=22,941.80, SSB=22,765.53, SSE=15,253.50, and SST - 64,954.7D. Construct the ANOVA table. Then, assuming a mixed model with cement factor (A) fixed and batches (B) random, test the three pairs of hypotheses of interest at level .05.

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\({\alpha _i}\).

An experiment was carried out to investigate the effect of species (factor A, with I = 4) and grade (factor B, with J = 3) on breaking strength of wood specimens. One observation was made for each species鈥攇rade combination鈥攔esulting in SSA = 442.0, SSB = 428.6, and SSE = 123.4. Assume that an additive model is appropriate.

  1. Test\({H_0}:{\alpha _1} = {\alpha _2} = {\alpha _3} = {\alpha _4} = 0\)(no differences in true average strength due to species) versus Ha: at least one\({\alpha _i} \ne 0\)using a level .05 test.
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