4141-Final-sol
2 Pages

4141-Final-sol

Course Number: BUS 4141, Fall 2009

College/University: LSU

Word Count: 319

Rating:

Document Preview

ISDS 4141 INTRODUCTION TO DATA MINING Final Exam Problem 1. (1) True ( X ); False ( ) (2) True ( ); False ( X ) (3) True ( X ); False ( ) (4) True ( X ); False ( ) (5) True ( X ); False ( ) (6) True ( ); False ( X ) (7) True ( ); False ( X ) (8) True ( X ); False ( ) (9) True ( ); False ( X ) (10) True ( X ); False ( ) (11) True ( ); False ( X )(12) True ( ); False ( X ) (13) True ( X ); False ( )(14) True ( X );...

Unformatted Document Excerpt
Coursehero >> Louisiana >> LSU >> BUS 4141

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

4141 ISDS INTRODUCTION TO DATA MINING Final Exam Problem 1. (1) True ( X ); False ( ) (2) True ( ); False ( X ) (3) True ( X ); False ( ) (4) True ( X ); False ( ) (5) True ( X ); False ( ) (6) True ( ); False ( X ) (7) True ( ); False ( X ) (8) True ( X ); False ( ) (9) True ( ); False ( X ) (10) True ( X ); False ( ) (11) True ( ); False ( X )(12) True ( ); False ( X ) (13) True ( X ); False ( )(14) True ( X ); False ( ) (15) True ( ); False ( X ) Problem 2. (a) P[Peanut Butter, Bread] = 10/100 = 0.1 (b) P[Bread | Peanut Butter] = 10/40 = 0.25 (c) P[Peanut Butter, Bread ] / (P[Peanut Butter ] P[Bread ]) = 0.1/(0.2*0.4) = 1.25 (d) (10*50 10*30)^2 / (20 * 80 * 60 * 40) * 100 =1.0417 Problem 3. (a) y-bar = b0 + b1 * x-bar x-bar = (y-bar b0) / b1 = 600 (b) E[Y-hat] = -395 + 1.825 * 700 = $882.5 (c) 1- SSE/(n-k-1) / SSTO/(n-1) = 0.7926 (d) Alternatives H0: 1 = 0 H1: 1 0 Decision rules If |t*| t8, 0.025 ( = 2.306 ), conclude Ho Test statistic t* = 5.94969 Conclusion Conclude H1. Problem 4. (a) = exp(bx)/[1+ exp(bx)] where bx = -9.6 + 0.1*X1 + 0.8*X2 (b) bx = -9.6 + 0.1*50 + 0.8*5 = -0.6 = exp(bx) / [1+ exp(bx)] = 35.43% (c) -9.6 + 0.1*X1 + 0.8*X2 = 0 or 1 X1 + 8 X2 = 96 Problem 5. (a) Minimax rule Bayes rule Reject a2 with E[L] = $110 Accept a1 E[L] ith = $44 ISDS 4141 Dr. Chun ISDS 4141 2/2 (b) If x2 (Good), then a1 (Accept). E[L] = $60 Problem 6. (a) P[Yes | Child] = 57/109 = 52.29% (b) P[Yes, No] = 57/109 * 52/109 * 2 = 49.89% (c) P[Yes | Adult Female] => P[Adult | Yes] P[Female | Yes] P[Yes] = 654/711 * 344/711 * 711/2201 = 0.14376 P[No | Adult Female] => P[Adult | No] P[Female | No] P[No] = 1438/1490 * 126/1490 * 1490/2201 = 0.05524 Since P[Yes | Adult Female] > P[No | Adult Female], classify it as Yes (d) He(Y) = 0.6291 H(Y) = He(Y) / ln(2) = 0.9077 He(Y|X) = He(X, Y) - He(X) = 0.8218 0.1971 = 0.6247 H(Y|X) = He(Y|X) / ln(2) = 0.9012 H(Y) H(Y|X) = 0.9077 0.9012 = 0.0065 (d) Classification Yes No Yes 57 654 True class No 52 1438 Total 109 2092 Error rate = (52 + 654)/1438 = 32.076% Problem 7. (a) r = 24 / sqrt(32 * 72) = 0.5 (b) d(A, B) = 4 + 6 + 0.1 = 10.1 (c) d(A, B) = sqrt(16 + 36 + 0.01) = 7.2118 (d) d(A, B) = SQRT(1 + 1 + 1) = 1.732 Problem 8. Multicorrelinearity problem! Expensive to maintain the model Hard to analyze and understand Total 711 1490 2201 Problem 9. Nave Bayesian classifiers assume that the effect of an attribute value on a given class is independent of the values of the other attributes. This assumption is called class conditional independence.
MOST POPULAR MATERIALS FROM LSU