9.3. The eigenvalues and eigen vectors of the correlation matrix p in Exercise 9.1 are 1 = 1.96, e = (.625,.593.507) λα .68 es = [-219,- 491,1843] = .36, es = 1.749,-.638, -177) (a) Assuming an m = 1 factor model, calculate the loading matrix L and matrix of specific variances y using the principal component solution method. Compare the results with those in Exercise 9.1. (b) What proportion of the total population variance is explained by the first common factor? 9.4. Given p and in Exercise 9.1 and an m = 1 factor model, calculate the reduced correlation matrix P = p - and the principal factor solution for the loading matrix L Is the result consistent with the information in Exercise 9.1? Should it be?
9.1. Show that the covariance matrix 1.0.63 .45 P = .63 1.0 35 = .45 35 1.0 for the p = 3 standardized random variables 2, 22, and Z can be generated by the m = 1 factor model Z= 9F + 1 Z2 = .7F, + E2 Zz = 5F, + 3 where Var (Ft) = 1, Cov (€, Fi) = 0, and .1900 V = Cov(E) 0 .51 0 LO 0 .75 That is, write p in the form p LL' + Y =
9.3. The eigenvalues and eigen vectors of the correlation matrix p in Exercise 9.1 are 1 = 1.96, e = (.625,.593.507) λα
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9.3. The eigenvalues and eigen vectors of the correlation matrix p in Exercise 9.1 are 1 = 1.96, e = (.625,.593.507) λα
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