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Chapter 11: Q11.9-13SE (page 784)

In a two-way layout with \(K\) observations in each cell (\(K \ge 2\)), construct a test of the null hypothesis that all the main effects for factor \(A\) and factor\(B\), and also all the interactions, are 0.

Short Answer

Expert verified

Reject \({H_0}\) if \(U_ * ^2 = \frac{{IJ(K - 1)\left( {S_A^2 + S_B^2 + {S_{{\rm{Int\;}}}}} \right)}}{{(IJ - 1)S_{{\rm{R{\o}id\;}}}^2}} \ge F_{IJ - 1,IJ(K - 1)}^{ - 1}\left( {1 - {\alpha _0}} \right)\)

Step by step solution

01

Definition for probability and statistics

  • Probability is concerned with forecasting the possibility of future events, whereas statistics is concerned with analysing the frequency of previous events.
  • Probability is largely a theoretical discipline of mathematics that investigates the implications of mathematical notions.
02

Determine the null hypothesis

  • Given: Two-way layout with \(K\)observations in each cell.
  • \(K = {\rm{ Number of repetitions }} = K\)
  • We are interested in testing whether the effects of both factors (\(A\)and\(B\)) are \(0\) , while the interactions are also zero.
  • \({H_0}:{\alpha _i} = {\beta _j} = {\gamma _{ij}} = 0{\rm{ for }}i = 1, \ldots ,I,j = 1, \ldots ,J{H_1}:{H_0}{\rm{ is not true}}\)
  • By the theorem for the two-way layout with \(K\)repetitions per cell, \(\frac{{S_{{\rm{Resid }}}^2}}{{{\sigma ^2}}}\)has the \({\chi ^2}\)distribution with \(IJ(K - 1)\)degrees of freedom, \(\frac{{S_A^2}}{{{\sigma ^2}}}\)has the \({\chi ^2}\)distribution with \(I - 1\)degrees of freedom, \(\frac{{S_B^2}}{{{\sigma ^2}}}\)has the \({\chi ^2}\)distribution with \(J - 1\)degrees of freedom, and \(\frac{{S_{{\rm{Int}}}^2}}{{{\sigma ^2}}}\)has the \({\chi ^2}\)distribution with \((I - 1)(J - 1)\)degrees of freedom.
  • Since\(\frac{{S_A^2}}{{{\sigma ^2}}},\frac{{S_B^2}}{{{\sigma ^2}}}\), and \(\frac{{{S_{Int}}^2}}{{{\sigma ^2}}}\)freedom are the sum of the degrees of freedom for each term).
  • \(\frac{{S_A^2 + S_B^2 + S_{{\rm{Int}}}^2}}{{{\sigma ^2}}} = \frac{{S_A^2}}{{{\sigma ^2}}} + \frac{{S_B^2}}{{{\sigma ^2}}} + \frac{{S_{{\rm{Int}}}^2}}{{{\sigma ^2}}}\~\chi _{(I - 1) + (J - 1) + (I - 1)(J - 1)}^2 = \chi _{IJ - 1}^2\)
  • Since the ratio of two chi-square distributed random variables is an\({\rm{F}}\)distributed random variable with the same degrees of freedom, we can use the test statistic \(U_*^2\)to test the hypotheses.
  • \(U_*^2\)Has the \({\rm{F}}\)distribution with \((IJ - 1)\)and \((IJ(K - 1))\)degrees of freedom if the null hypothesis \({H_0}\) is true.
  • \(\begin{array}{c}U_ * ^2 = \frac{{\left. {S_A^2 + S_B^2 + S_{{\rm{Int}}}^2} \right)(IJ - 1){\sigma ^2}}}{{\frac{{S_{{\rm{Resid}}}^2}}{{IJ(K - 1){\sigma ^2}}}}}\\ = \frac{{IJ(K - 1)\left( {S_A^2 + S_B^2 + {S_{Int}}} \right)}}{{(IJ - 1)S_{{\rm{Resid\;}}}^2}}\end{array}\)
  • We can then use the test statistic \(U_*^2\)to test the null hypothesis, where we will reject the null hypothesis \({H_0}\)when \(U_*^2\)is larger than or equal to the critical value\(F_{IJ - 1,IJ(K - 1)}^{ - 1}\left( {1 - {\alpha _0}} \right)\).
  • \({\rm{\;Reject\;}}{H_ - }0{\rm{\;if\;}}U_ * ^2 = \frac{{IJ(K - 1)\left( {S_A^2 + S_B^2 + {S_{{\rm{Int\;}}}}} \right)}}{{(IJ - 1)S_{{\rm{Resid\;}}}^2}} \ge F_{IJ - 1,IJ(K - 1)}^{ - 1}\left( {1 - {\alpha _0}} \right)\)

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Most popular questions from this chapter

Question:For the conditions of Exercise \({\bf{12}}\), and the data in Table \({\bf{11}}{\bf{.2}}\), determine the value of \({{\bf{R}}^{\bf{2}}}\), as defined by Eq.\({\bf{(11}}{\bf{.5}}{\bf{.26)}}\).

Consider again the conditions of Exercise 12, For what value of the temperature \({\bf{x}}\) can the durability of a specimen of the alloy be predicted with the smallest M.S.E.?

Prove that \({{\bf{\sigma }}^{{\bf{'2}}}}\), defined in Eq.\({\bf{(11}}{\bf{.5}}{\bf{.8)}}\)is an unbiased estimator of \({{\bf{\sigma }}^{\bf{2}}}\). You may assume that \({{\bf{S}}^{\bf{2}}}\)has a \({{\bf{X}}^{\bf{2}}}\) distribution with \({\bf{n - p}}\) degrees of freedom.

Question: Consider a problem of simple linear regression as described in Sec.\({\bf{11}}{\bf{.2}}\) and let \({{\bf{R}}^{\bf{2}}}\), be defined by Eq.\({\bf{(11}}{\bf{.5}}{\bf{.26)}}\) of this section. Show that

\({{\bf{R}}^{\bf{2}}}{\bf{ = }}\frac{{{{\left( {\mathop {\sum {\left( {{{\bf{x}}_{\bf{i}}}{\bf{ - \bar x}}} \right)\left( {{{\bf{y}}_{\bf{i}}}{\bf{ - \bar y}}} \right)} }\limits_{{\bf{i = 1}}}^{\bf{n}} } \right)}^{\bf{2}}}}}{{\left( {\mathop {\sum {{{\left( {{{\bf{x}}_{\bf{i}}}{\bf{ - \bar x}}} \right)}^{\bf{2}}}} }\limits_{{\bf{i = 1}}}^{\bf{n}} } \right)\left( {\mathop {\sum {{{\left( {{{\bf{y}}_{\bf{i}}}{\bf{ - \bar y}}} \right)}^{\bf{2}}}} }\limits_{{\bf{i = 1}}}^{\bf{n}} } \right)}}\)

Consider again the conditions of Exercise 12, and let\(\theta = {c_1}{\beta _1} - {\beta _0}\), where \({c_1}\) is a constant. Determine an unbiased estimator \(\hat \theta \) of \(\theta \). For what value of \({c_1}\) will the M.S.E. of \(\hat \theta \) be smallest?
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