(Part 2)
5
Chapter
Probability
Contingency Tables Tree Diagrams Bayes’ Theorem Counting Rules
McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc. All rights reserved.
Contingency Tables What is a Contingency Table? Variable 1 Col 1 Col 2 Col 3
Variable 2
• A contingency table is a crosstabulation of frequencies into rows and columns.
Row 1 Row 2
Cell
Row 3 Row 4
• A contingency table is like a frequency distribution for two variables. 5B-2
Contingency Tables Example: Salary Gains and MBA Tuition • Consider the following cross-tabulation table for n = 67 top-tier MBA programs: (Table 5.4)
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Contingency Tables Example: Salary Gains and MBA Tuition • Are large salary gains more likely to accrue to graduates of high-tuition MBA programs? • The frequencies indicate that MBA graduates of high-tuition schools do tend to have large salary gains. • Also, most of the top-tier schools charge high tuition. • More precise interpretations of this data can be made using the concepts of probability. 5B-4
Contingency Tables Marginal Probabilities
• The marginal probability of a single event is found by dividing a row or column total by the total sample size. • For example, find the marginal probability of a medium salary gain (P(S2)). P(S2) = 33/67 = .4925 • Conclude that about 49% of salary gains at the top-tier schools were between $50,000 and $100,000 (medium gain).
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Contingency Tables Marginal Probabilities • Find the marginal probability of a low tuition P(T1).
P(T1) = 16/67 = .2388 • There is a 24% chance that a top-tier school’s MBA tuition is under $40.000. 5B-6
Contingency Tables Joint Probabilities • A joint probability represents the intersection of two events in a cross-tabulation table. • Consider the joint event that the school has low tuition and large salary gains (denoted as P(T1 ∩ S3)).
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Contingency Tables Joint Probabilities • So, using the cross-tabulation table, P(T1 ∩ S3) = 1/67 = .0149 • There is less than a 2% chance that a top-tier school has both low tuition and large salary gains.
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Contingency Tables Conditional Probabilities • Found by restricting ourselves to a single row or column (the condition). • For example, knowing that a school’s MBA tuition is high (T3), we would restrict ourselves to the third row of the table.
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Contingency Tables Conditional Probabilities • Find the probability that the salary gains are small (S1) given that the MBA tuition is large (T3). P(T1 | S3) = 5/32 = .1563 • What does this mean?
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Contingency Tables Independence • To check for independent events in a contingency table, compare the conditional to the marginal probabilities. • For example, if large salary gains (S3) were independent of low tuition (T1), then P(S3 | T1) = P(S3). Conditional
Marginal
P(S3 | T1)= 1/16 = .0625
P(S3) = 17/67 = .2537
• What do you conclude about events S3 and T1? 5B-11
Contingency Tables Relative Frequencies • Calculate the relative frequencies below for each cell of the cross-tabulation table to facilitate probability calculations.
• Symbolic notation for relative frequencies: 5B-12
Contingency Tables Relative Frequencies
• Here are the resulting probabilities (relative frequencies). For example,
P(T1 and S1) = 5/67 P(T2 and S2) = 11/67 P(T3 and S3) = 15/67 P(S1) = 17/67
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P(T2) = 19/67
Contingency Tables Relative Frequencies
• The nine joint probabilities sum to 1.0000 since these are all the possible intersections. • Summing the across a row or down a column gives marginal probabilities for the respective row or column.
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Contingency Tables Example: Payment Method and Purchase Quantity • A small grocery store would like to know if the number of items purchased by a customer is independent of the type of payment method the customer chooses to use. • Why would this information be useful to the store manager? • The manager collected a random sample of 368 customer transactions. 5B-15
Contingency Tables Example: Payment Method and Purchase Quantity Here is the contingency table of frequencies:
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Contingency Tables Example: Payment Method and Purchase Quantity • Calculate the marginal probability that a customer will use cash to make the payment. • Let C be the event cash. P(C) = 126/368 = .3424 • Now, is this probability the same if we condition on number of items purchased?
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Contingency Tables Example: Payment Method and Purchase Quantity P(C | 1-5) = 30/88 = .3409 P(C | 6-10) = 46/135 = .3407 P(C | 10-20) = 31/89 P(C | 20+) = 19/56
= .3483 = .3393
• P(C) = .3424, so what do you conclude about independence? • Based on this, the manager might decide to offer a cash-only lane that is not restricted to the number of items purchased. 5B-18
Contingency Tables How Do We Get a Contingency Table? • Contingency tables require careful organization and are created from raw data. • Consider the data of salary gain and tuition for n = 67 top-tier MBA schools.
5B-19
Contingency Tables How Do We Get a Contingency Table? • The data should be coded so that the values can be placed into the contingency table. Once coded, tabulate the frequency in each cell of the contingency table using MINITAB’s : Stat | Tables | Cross Tabulation 5B-20
Tree Diagrams What is a Tree? • A tree diagram or decision tree helps you visualize all possible outcomes. • Start with a contingency table. • For example, this table gives expense ratios by fund type for 21 bond funds and 23 stock funds.
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Tree Diagrams What is a Tree? • To label the tree, first calculate conditional probabilities by dividing each cell frequency by its column total. • For example, P(L | B) = 11/21 = .5238 • Here is the table of conditional probabilities
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Tree Diagrams What is a Tree? • The tree diagram shows all events along with their marginal, conditional and joint probabilities. • To calculate joint probabilities, use P(A ∩ B) = P(A | B)P(B) = P(B | A)P(A) • The joint probability of each terminal event on the tree can be obtained by multiplying the probabilities along its branch. • For example, P(B ∩ L) = P(L | B)P(B) = (.5238)(.4773) = .2500 5B-23
Tree Diagrams Tree Diagram for Fund Type and Expense Ratios Figure 5.11
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Bayes Bayes’’ Theorem • Thomas Bayes (1702-1761) provided a method (called Bayes’s Theorem) of revising probabilities to reflect new probabilities. • The prior (marginal) probability of an event B is revised after event A has been considered to yield a posterior (conditional) probability. • Bayes’s formula is:
5B-25
P( A | B) P( B) P( B | A) = P( A)
Bayes Bayes’’ Theorem • Bayes’ formula begins as: P( A | B) P( B) P( B | A) = P( A)
• In some situations P(A) is not given. Therefore, the most useful and common form of Bayes’ Theorem is:
P( A | B) P( B) P( B | A) = P( A | B) P( B ) + P( A | B ') P( B ') 5B-26
Bayes Bayes’’ Theorem How Bayes’ Theorem Works
• Consider an over-the-counter pregnancy testing kit and it’s “track record” of determining pregnancies. • If a woman is actually pregnant, what is the test’s “track record”? • If a woman is not pregnant, what is the test’s “track record”? False Positive False Negative
Table 5.17 96% of time 1% of time 5B-27
4% of time 99% of time
Bayes Bayes’’ Theorem How Bayes’ Theorem Works • Suppose that 60% of the women who purchase the kit are actually pregnant. • Intuitively, if 1,000 women use this test, the results should look like this.
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Bayes Bayes’’ Theorem How Bayes’ Theorem Works • Of the 580 women who test positive, 576 will actually be pregnant.
• So, the desired probability is: P(Pregnant│Positive Test) = 576/580 = .9931 5B-29
Bayes Bayes’’ Theorem How Bayes’ Theorem Works • Now use Bayes’s Theorem to formally derive the result P(Pregnant | Positive) = .9931:
• First define A = positive test A' = negative test • From the contingency table, we know that: P(A | B) = .96 P(A | B') = .01 P(B) = .60 5B-30
B = pregnant B' = not pregnant • And the compliment of each event is: P(A' | B) = .04 P(A' | B') = .99 P(B') = .40
Bayes Bayes’’ Theorem How Bayes’ Theorem Works P(B | A) = = =
P(A | B)P(B) P(A | B)P(B) + P(A | B')P(B') (.96)(.60) (.96)(.60) + (.01)(.40) .576 .576 = = .9931 .576 + .04 .580
• So, there is a 99.31% chance that a woman is pregnant, given that the test is positive. 5B-31
Bayes Bayes’’ Theorem How Bayes’ Theorem Works • Bayes’s Theorem shows us how to revise our prior probability of pregnancy to get the posterior probability after the results of the pregnancy test are known. Prior Before the test P(B) = .60
Posterior After positive test result P(B | A) = .9931 →
• Bayes’s Theorem is useful when a direct calculation of a conditional probability is not permitted due to lack of information. 5B-32
Bayes Bayes’’ Theorem How Bayes’ Theorem Works • A tree diagram helps visualize the situation.
5B-33
Bayes Bayes’’ Theorem How Bayes’ Theorem Works The 2 branches showing a positive test (A) comprise a reduced sample space B ∩ A and B' ∩ A, so add their probabilities to obtain the denominator of the fraction whose numerator is P(B ∩ A). 5B-34
Bayes Bayes’’ Theorem General Form of Bayes’ Theorem • A generalization of Bayes’s Theorem allows event B to be polytomous (B1, B2, … Bn) rather than dichotomous (B and B'). P( A | Bi ) P( Bi ) P( Bi | A) = P( A | B1 ) P( B1 ) + P( A | B2 ) P( B2 ) + ... + P( A | Bn ) P( Bn )
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Bayes Bayes’’ Theorem Example: Hospital Trauma Centers (Table 5.18)
• Based on historical data, the percent of cases at 3 hospital trauma centers and the probability of a case resulting in a malpractice suit are as follows:
• let event A = a malpractice suit is filed Bi = patient was treated at trauma center i 5B-36
Bayes ’s Theorem Bayes’s Example: Hospital Trauma Centers • Applying the general form of Bayes’ Theorem, find P(B1 | A). P( A | B1 ) P( B1 ) P( B1 | A) = P( A | B1 ) P( B1 ) + P( A | B2 ) P( B2 ) + P( A | B3 ) P( B3 )
(0.001)(0.50) P( B1 | A) = (0.001)(0.50) + (0.005)(0.30) + (0.008)(0.20) 0.0005 0.0005 P( B1 | A) = = = 0.1389 0.0005 + 0.0015 + 0.0016 0.00036 0. 5B-37
Bayes Bayes’’ Theorem Example: Hospital Trauma Centers • Conclude that the probability that the malpractice suit was filed in hospital 1 is .1389 or 13.89%. • All the posterior probabilities for each hospital can be calculated and then compared:
(Table 5.19)
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Bayes Bayes’’ Theorem Example: Hospital Trauma Centers • Intuitively, imagine there were 10,000 patients and calculate the frequencies:
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Hospital
Malpractice Suit Filed
No Malpractice Suit Filed
Total
1
5
4,995
5,000
= 10,000x.5
2
15
2,985
3,000
= 10,000x.3
3
16
1,984
2,000
= 10,000x.2
Total
36
9,964
10,000
= 5,000 x .001
= 5,000 - 5
= 3,000 x .005
= 3,000 - 15
= 2,000 x .008
= 1,984 - 16
Bayes Bayes’’ Theorem Example: Hospital Trauma Centers • Now, use these frequencies to find the probabilities needed for Bayes’ Theorem. • For example, Hospital
Malpractice Suit Filed
No Malpractice Suit Filed
Total
1
P(B1|A)=5/36=.1389
P(B1|A')=.5012
P(B1)=.5
2
P(B2|A)=15/36=.4167
P(B2|A')=.2996
P(B2)=.3
3
P(B3|A)=16/36=4444
P(B3|A')=.1991
P(B3)=.2
Total
P(A)=36/10000=.0036
P(A')=.9964
1.0000
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Bayes Bayes’’ Theorem Example: Hospital Trauma Centers • Consider the following visual description of the problem:
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Bayes Bayes’’ Theorem Example: Hospital Trauma Centers • The initial sample space consists of 3 mutually exclusive and collectively exhaustive events (hospitals B1, B2, B3).
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Bayes Bayes’’ Theorem Example: Hospital Trauma Centers • As indicated by their relative areas, B1 is 50% of the sample space, B2 is 30% and B3 is 20%. 50%
30%
20% 5B-43
Bayes Bayes’’ Theorem Example: Hospital Trauma Centers • But, given that a malpractice case has been filed (event A), then the relevant sample space is reduced to the yellow area of event A. • The revised probabilities are the relative areas within event A. P(B2 | A)
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P(B1 | A)
P(B3 | A)
Counting Rules Fundamental Rule of Counting • If event A can occur in n1 ways and event B can occur in n2 ways, then events A and B can occur in n1 x n2 ways. • In general, m events can occur n1 x n2 x … x nm ways.
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Counting Rules Example: Stock-Keeping Labels • How many unique stock-keeping unit (SKU) labels can a hardware store create by using 2 letters (ranging from AA to ZZ) followed by four numbers (0 through 9)? • For example, AF1078: hex-head 6 cm bolts – box of 12 RT4855: Lime-A-Way cleaner – 16 ounce LL3319: Rust-Oleum primer – gray 15 ounce 5B-46
Counting Rules Example: Stock-Keeping Labels • View the problem as filling six empty boxes:
• There are 26 ways to fill either the 1st or 2nd box and 10 ways to fill the 3rd through 6th. • Therefore, there are 26 x 26 x 10 x 10 x 10 x 10 = 6,760,000 unique inventory labels. 5B-47
Counting Rules Example: Shirt Inventory • L.L. Bean men’s cotton chambray shirt comes in 6 colors (blue, stone, rust, green, plum, indigo), 5 sizes (S, M, L, XL, XXL) and two styles (short and long sleeves). • Their stock might include 6 x 5 x 2 = 60 possible shirts. • However, the number of each type of shirt to be stocked depends on prior demand. 5B-48
Counting Rules Factorials • The number of ways that n items can be arranged in a particular order is n factorial. • n factorial is the product of all integers from 1 to n. n! = n(n–1)(n–2)...1 • Factorials are useful for counting the possible arrangements of any n items. • There are n ways to choose the first, n-1 ways to choose the second, and so on. 5B-49
Counting Rules Factorials • As illustrated below, there are n ways to choose the first item, n-1 ways to choose the second, n-2 ways to choose the third and so on.
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Counting Rules Factorials
• A home appliance service truck must make 3 stops (A, B, C). • In how many ways could the three stops be arranged? 3! = 3 x 2 x 1 = 6 • List all the possible arrangements: {ABC, ACB, BAC, BCA, CAB, CBA} • How many ways can you arrange 9 baseball players in batting order rotation? 9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362,880
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Counting Rules Permutations • A permutation is an arrangement in a particular order of r randomly sampled items from a group of n items and is denoted by nPr
n! n Pr = (n − r )! • In other words, how many ways can the r items be arranged, treating each arrangement as different (i.e., XYZ is different from ZYX)? 5B-52
Counting Rules Example: Appliance Service Cans • n = 5 home appliance customers (A, B, C, D, E) need service calls, but the field technician can service only r = 3 of them before noon. • The order is important so each possible arrangement of the three service calls is different. • The number of possible permutations is:
n! 5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 120 = = = = 60 n Pr = (n − r )! (5 − 3)! 2! 2 5B-53
Counting Rules Example: Appliance Service Cans • The 60 permutations with r = 3 out of the n = 5 calls can be enumerated. • There are 10 distinct groups of 3 customers: ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE 5B-54
• Each of these can be arranged in 6 distinct ways: ABC, ACB, BAC, BCA, CAB, CBA
• Since there are 10 groups of 3 customers and 6 arrangements per group, there are 10 x 6 = 60 permutations.
Counting Rules Combinations • A combination is an arrangement of r items chosen at random from n items where the order of the selected items is not important (i.e., XYZ is the same as ZYX). • A combination is denoted nCr
n! nCr = r !(n − r )! 5B-55
Counting Rules Example: Appliance Service Calls Revisited • n = 5 home appliance customers (A, B, C, D, E) need service calls, but the field technician can service only r = 3 of them before noon. • This time order is not important. • Thus, ABC, ACB, BAC, BCA, CAB, CBA would all be considered the same event because they contain the same 3 customers. • The number of possible combinations is:
n! 5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 120 = = = = 10 nCr = r !(n − r )! 3!(5 − 3)! (3 ⋅ 2 ⋅ 1)(2 ⋅ 1) 12 5B-56
Counting Rules Example: Appliance Service Calls Revisited • 10 combinations is much smaller than the 60 permutations in the previous example. • The combinations are easily enumerated: ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE
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Applied Statistics in Business & Economics
End of Chapter 5B
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