) 8.517 (B) 7.947 (C) -6.170 (D) 1.823 To classify this individual, we note that since C0(4,5) > C1(4,5), we classify this individual into Group 0, that is, we predict that this individual will not buy. 3. What is this person's posterior probability P1 of membership in the 'Buy' group? P1 = exp(C1)/[exp(C0)+exp(C1)] = exp(7.947)/[exp(7.947)+exp(8.517)] = 1/[1+exp(8.517− 7.947)] = 1/[1 + exp(0.570)] = ? (A) .747 (B) .570 (C) .361 (D) .639 4. What is this person's posterior probability of membership in the "Not Buy" group ? (A) .747 (B) .570 (C) .
Notation: For groups k = 0, 1, let C Fk (d, p) denote the value of the classification function
for Group k, given Durability = d and Performance = p.
What are the values of the two classification
a Durability rating of d = 4 and a Performance rating of p = 5?
1. For Group 0: CF0(4, 5) = ? C0(d, p) = −6.170 + 1.823d + 1.479p; C(4, 5) = −6.170 + 1.823(4) + 1.479(5) =?
(A) 8.517 (B) 7.947 (C) -6.170 (D) 1.823
2. For Group 1: CF1(4, 5) = ? C1(d, p) = −25.619 + 5.309d + 2.466p; C(4, 5) = −25.619 + 5.309(4) + 2.466(5) = ?
(A) 8.517 (B) 7.947 (C) -6.170 (D) 1.823
To classify this individual, we note that since C0(4,5) > C1(4,5), we classify this individual into Group 0, that is, we predict that this individual will not buy.
3. What is this person's posterior
P1 = exp(C1)/[exp(C0)+exp(C1)] = exp(7.947)/[exp(7.947)+exp(8.517)] = 1/[1+exp(8.517− 7.947)] = 1/[1 + exp(0.570)] = ?
(A) .747 (B) .570 (C) .361 (D) .639
4. What is this person's posterior probability of membership in the "Not Buy" group ? (A) .747 (B) .570 (C) .361 (D) .639
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