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where STR stands for the sum of transformed residuals, STR = Lp(e), and A= (n/m)Le;z/(n - p - 1). The function p(e) was introduced in formula (5.1) and is obtained by truncating i as described in Section 5.3. That is, e; = ei if leil s 1.5u, e; = -1.5u if ei < -1.5u, and e; = 1.5u if ei > 1.5u. The estimate u is 1.483MAD, as in the preceding section. The integer m is the number of residuals i that do not require truncation, that is, for which leil s 1.5u. The residuals in STRreduced and STR full are calculated by applying the M-regression procedure to, respectively, the reduced model and the full model. The estimation procedure for the reduced model differs slightly from the procedure described in the preceding section; the estimate of 0' is not iterated. Throughout the iterations required to obtain the vector of M-estimates of the regression coefficients in the reduced model, we keep the same estimate calculated from the full model. In calculating A, the residuals from the full model are used. An approximate p-value of the test is calculated in the same way as for the least-squares test, namely, as Prob[F:2: F M ], where F denotes a random variable having an F distribution with p - q and n - p - 1 degrees of freedom.
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Similarity to the Least-Squares Test. The formula for FM is very similar to the formula for F LS' In fact, if in the definition of and p(e), 1.5 is replaced by 00, then F M becomes exactly F LS' Note that if 1.5 is replaced by 00, then = i for all i and m = n, so that A coincides with u~s. Moreover, if 1.5 is replaced by 00, then p( e) = e z, so STR = SSR, and so F M coincides with F LS '
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The Aerobic Fitness Data. Let us test whether the two measurements of pulse rate make a significant contribution in equation (5.5); that is, let us test the hypothesis that f34 = f35 = o. We have p = 5 and q = 3. To calculate F M we first obtain the residuals i = y, - ~' x,, where ~ is the vector of M-estimates calculated in Section 5.6. The median of the absolute residuals leil is MAD = 0.9046, so u = (1.483)(0.9046) = 1.341 and 1.5u = (1.5)(1.341) = 2.012. For the 24 residuals between - 2.012 and 2.012, p(i) = i;, and for the other seven residuals, p(e) = 4.0241e,1 - 4.049. Adding all 31 values of p( i), we obtain STR full = 114.7. By truncating the four residuals that are below - 2.012 and the three residuals that are above 2.012, we obtain the truncated residuals if. The sum of their squares is 50.44. Now A = (31/24)(50.44)/25 = 2.606. The reduced model with f34 = f35 = 0 is Y = f30 + f3,X, + f3 z X z + f33X3 + e. The Huber M-estimates are the values b o, b" b z, b) that minimize the function LP(Yi - (b o + b,x i , + bZx iZ + bJx iJ . The estimation procedure followed for the reduced model differs slightly from the procedure
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followed for the full model, because the estimate 0- is not iterated. The same value 0- = 1.341, calculated from the full model, is used throughout the iterations required to obtain the M-estimates of the regression coefficients in the reduced model. So the value of k in the definition of p in (5.1) stays constant at k = 1.50- = (1.5)( 1.341) = 2.012. The vector of M-estimates in the reduced model is (~o, ~I' ~2' ~3) = (95.07, - 0.1844, - 0.08861, - 3.015), so the residuals in the reduced model are i = Yi - (95.07 - O.l844x il - 0.08861xi2 - 3.015x i ). For the 21 residuals between - 2.012 and 2.012, p(e) = e;, and for the other 10 residuals, p(e) = 4.0241ei l - 4.049. Adding all 31 values of p(e), we obtain STRrcduccd = 145.8. Now FM = (145.8 - 114.7)/[(5 - 3)(2.606)] = 5.964. The approximate p-value of the test is Prob[F 2': 5.964], where F is a random variable having an F distribution with 2 and 25 degrees of freedom. From the F table in the Appendix, we see that the p-value is between 0.001 and 0.01. We conclude that one or both of the pulse rate measurements have a significant relationship to oxygen consumption.
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