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Simplify E(MSTr) for the random effects model when\({J_1} = {J_2} = \ldots = {J_I} = J.\)

Short Answer

Expert verified

\(E(MSTr) = {\sigma ^2} + J \cdot \sigma _A^2\)

Step by step solution

01

Finding the value of E(MSTr) for value\({J_1} = {J_2} =  \ldots  = J\)\(E(MSTr) = {\sigma ^2} + \frac{1}{{I - 1}}\left( {n - \frac{{\sum\nolimits_{i = 1}^I {J_i^2} }}{n}} \right)\sigma _A^2,\)

Hence, the expected value when \({J_1} = {J_2} = \ldots = J\) is

\(E(MSTr) = {\sigma ^2} + \frac{1}{{I - 1}}\left( {n - \frac{{\sum\nolimits_{i = 1}^I {J_i^2} }}{n}} \right)\sigma _A^2,\)

\( = {\sigma ^2} + \frac{1}{{I - 1}}\left( {n - \frac{{\sum\nolimits_{i = 1}^I {{J^2}} }}{n}} \right)\sigma _A^2\)

\( = {\sigma ^2} + \frac{1}{{I - 1}}\left( {n - \frac{{I \cdot {J^2}}}{n}} \right)\sigma _A^2\)

\( = {\sigma ^2} + \frac{{n - J}}{{I - 1}}\sigma _A^2\)

\( = {\sigma ^2} + J \cdot \sigma _A^2,\)

02

E(MSTr) value for n

\(n = \sum\limits_{i = 1}^I {{J_i}} = \sum\limits_{i = 1}^I {{J^2} = I \cdot J} \)

\(E(MSTr) = {\sigma ^2} + J \cdot \sigma _A^2\)

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Most popular questions from this chapter

When sample sizes are equal\(\left( {{J_i} = J} \right)\), the parameters\({\alpha _1},{\alpha _2},...{\alpha _I}\,\)of the alternative parameterization are restricted by\(\Sigma {\alpha _i} = 0\). For unequal sample sizes, the most natural restriction is\(\Sigma {J_i}{\alpha _i} = 0\). Use this to show that

\(E\left( {MSTr} \right) = {\alpha ^2} + \frac{1}{{I - 1}}\Sigma {J_i}\alpha _i^2\)

What is\(E\left( {MSTr} \right)\)when\({H_0}\)is true? (This expectation is correct if\(\Sigma {J_i}{\alpha _i} = 0\,\)is replaced by the restriction\(\Sigma {\alpha _i} = 0\)(or any other single linear restriction on the ai 鈥檚 used to reduce the model to I independent parameters), but\(\Sigma {J_i}{\alpha _i} = 0\)simplifies the algebra and yields natural estimates for the model parameters (in particular,\({\mathop {\,\,\,\,\alpha }\limits^{\,\,\,\,\,\^} _i} = {\mathop X\limits^\_ _i}\, - \,\mathop X\limits^\_ ..\,\,\,\,{H_0}\)).)

Reconsider Example 10.8 involving an investigation of the effects of different heat treatments on the yield point of steel ingots.

a. If J = 8 and\(\sigma = 1\), what is\(\beta \)for a level .05 F test when\({\mu _1} = {\mu _2},\,\,\,{\mu _3} = {\mu _1} - 1\,\), and\({\mu _4} = {\mu _1} + 1\)?

b. For the alternative of part (a), what value of J is necessary to obtain\(\beta = .05\)?

c. If there are I = 5 heat treatments, J = 10, and\(\sigma = 1\), what is\(\beta \)for the level .05 F test when four of the\({\mu _i}\)鈥檚 are equal and the fifth differs by 1 from the other four?

Four types of mortars-ordinary cement mortar(OCM), polymer impregnated mortar(PIM), resin mortar(RM), and polymer cement mortar(PCM)- were subjected to a compression test to measure strength (MPa). Three strength observations for each mortar type are given in the article 鈥淧olymer Mortar Composite Matrices For Maintance- Free Highly Durable Ferrocement鈥漚nd are reproduced here. Construct an ANOVA table. Using a \(.05\)significance level , determine whether the data suggests that the true mean strength is not the same for all the four mortar types. If you determine that the true mean strengths are not all equal, use Turkey鈥檚 method to identify the significant differences.

\(\begin{aligned}{*{20}{c}}{OCM}&{32.15}&{35.53}&{34.20}\\{PIM}&{126.32}&{126.80}&{134.79}\\{RM}&{117.91}&{115.02}&{114.58}\\{PCM}&{29.09}&{30.87}&{29.80}\end{aligned}\)

Suppose that \({X_{ij}}\) is a binomial variable with parameters n and \({P_i}\) (so approximately normal when \(n{p_i} \ge 10\)and \(n{q_i} \ge 10\)).Then since \({\mu _i} = n{p_i}\), \(V({X_{ij}}) = \sigma _i^2 = n{p_i}(1 - {p_i}) = {\mu _i}(1 - {\mu _i}/n)\).How should the \({X_{ij}}\)鈥檚 be transformed so as to stabilize the variance?(Hint: \(g({\mu _i}) = {\mu _i}(1 - {\mu _i}/n)).\)

Refer to Exercise 38. What is b for the test when true average DNA content is identical for three of the diets and falls below this common value by 1 standard deviation (s) for the other two diets?

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