A2sol - Question 1 Problem 2 a support(cfw_E = 8 10 = 0.8 support(cfw_B D = 2 10 = 0.2 support(cfw_B D E = 2 10 = 0.2 b confidence(BD E =

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Question 1: Problem 2 a) support({E}) = 8 / 10 = 0.8 support({B, D}) = 2 / 10 = 0.2 support({B, D, E}) = 2 / 10 = 0.2 b) confidence(BD → E) = support({B, D, E}) / support({B, D}) = 0.2 / 0.2 = 1.0 confidence(E → BD) = support({B, D, E}) / support({E}) = 0.2 / 0.8 = 0.25 No, confidence is not a symmetric measure. As seen in the above example, the numerators are the same, however, the denominators are different values. c) support({E}) = 4 / 5 = 0.8 support({B, D}) = 5 / 5 = 1.0 support({B, D, E}) = 4 / 5 = 0.8 d) confidence(BD → E) = support({B, D, E}) / support({B, D}) = 0.8 / 1.0 = 0.8 confidence(E → BD) = support({B, D, E}) / support({E}) = 0.8 / 0.8 = 1.0 e) s1 ≤ s2. There is no relationship between c1 and c2. Question 2: Problem 6 a) Total number of items in the data set {Beer, Bread, Butter, Cookies, Diaper, Milk}, d = 6 Therefore, total possible association rules = 3 d – 2 d+1 + 1 = 3 6 – 2 7 + 1 = 729 – 128 + 1 = 602 b) 4 c) 6 C 3 = 20 d) Support: {Bread, Butter} = 0.5
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/17/2011 for the course CIS 4930 taught by Professor Staff during the Fall '08 term at University of Florida.

Page1 / 2

A2sol - Question 1 Problem 2 a support(cfw_E = 8 10 = 0.8 support(cfw_B D = 2 10 = 0.2 support(cfw_B D E = 2 10 = 0.2 b confidence(BD E =

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online