1 Contingency Tables Tree Diagrams Bayes’ Theorem Counting Rules5 Probability (Part 2) Chapter Contingency Tables Tree Diagrams Bayes’ Theorem Counting Rules McGraw-Hill/Irwin Copyright © 2009 by The McGraw-Hill Companies, Inc. All rights reserved.
2 Contingency Tables What is a Contingency Table? Variable 1A contingency table is a cross-tabulation of frequencies into rows and columns. Variable 1 Col 1 Col 2 Col 3 Row 1 Row 2 Row 3 Row 4 Variable 2 Cell A contingency table is like a frequency distribution for two variables.
3 Contingency Tables Example: Salary Gains and MBA TuitionConsider the following cross-tabulation table for n = 67 top-tier MBA programs: (Table 5.4)
4 Contingency Tables Example: Salary Gains and MBA TuitionAre 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.
5 Contingency Tables Marginal ProbabilitiesThe 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).
6 Contingency Tables Marginal ProbabilitiesFind 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 $
7 Contingency Tables Joint ProbabilitiesA 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)).
8 Contingency Tables Joint ProbabilitiesSo, 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.
9 Contingency Tables Conditional ProbabilitiesFound 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.
10 Contingency Tables Conditional ProbabilitiesFind 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?
11 Contingency Tables IndependenceTo 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?
12 Contingency Tables Relative FrequenciesCalculate the relative frequencies below for each cell of the cross-tabulation table to facilitate probability calculations. Symbolic notation for relative frequencies:
13 Contingency Tables Relative FrequenciesHere 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 P(T2) = 19/67 5B-13
14 Contingency Tables Relative FrequenciesThe nine joint probabilities sum to 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. 5B-14
15 Contingency Tables Example: Payment Method and Purchase QuantityA 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.
16 Contingency Tables Example: Payment Method and Purchase QuantityHere is the contingency table of frequencies:
17 Contingency Tables Example: Payment Method and Purchase QuantityCalculate 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?
18 Contingency Tables Example: Payment Method and Purchase QuantityP(C | 1-5) = 30/88 = .3409 P(C | 6-10) = 46/135 = .3407 P(C | 10-20) = 31/89 = .3483 P(C | 20+) = 19/56 = .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.
19 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.
20 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
21 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.
22 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 5B-22
23 Tree Diagrams What is a Tree? = .2500The 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
24 Tree Diagrams Tree Diagram for Fund Type and Expense RatiosFigure 5.11
25 Bayes’ Theorem Thomas Bayes ( ) 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:
26 Bayes’ Theorem Bayes’ formula begins as:In some situations P(A) is not given. Therefore, the most useful and common form of Bayes’ Theorem is:
27 Bayes’ Theorem How Bayes’ Theorem WorksConsider 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 4% of time 99% of time 5B-27
28 Bayes’ Theorem How Bayes’ Theorem WorksSuppose 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.
29 Bayes’ Theorem How Bayes’ Theorem WorksOf the 580 women who test positive, 576 will actually be pregnant. So, the desired probability is: P(Pregnant│Positive Test) = 576/580 = .9931
30 Bayes’ Theorem How Bayes’ Theorem WorksNow use Bayes’s Theorem to formally derive the result P(Pregnant | Positive) = .9931: First define A = positive test B = pregnant A' = negative test B' = not pregnant From the contingency table, we know that: And the compliment of each event is: P(A' | B) = .04 P(A' | B') = .99 P(B') = .40 P(A | B) = .96 P(A | B') = .01 P(B) = .60
31 Bayes’ Theorem How Bayes’ Theorem Works P(A | B)P(B) P(B | A) =P(A | B)P(B) + P(A | B')P(B') = (.96)(.60) (.96)(.60) + (.01)(.40) = .576 = .576 .580 = .9931 So, there is a 99.31% chance that a woman is pregnant, given that the test is positive.
32 After positive test resultBayes’ 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 Posterior After positive test result P(B) = .60 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.
33 Bayes’ Theorem How Bayes’ Theorem WorksA tree diagram helps visualize the situation.
34 Bayes’ Theorem How Bayes’ Theorem WorksThe 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).
35 Bayes’ Theorem General Form of Bayes’ TheoremA generalization of Bayes’s Theorem allows event B to be polytomous (B1, B2, … Bn) rather than dichotomous (B and B').
36 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
37 Bayes’s Theorem Example: Hospital Trauma CentersApplying the general form of Bayes’ Theorem, find P(B1 | A). 0.
38 Bayes’ Theorem Example: Hospital Trauma CentersConclude that the probability that the malpractice suit was filed in hospital 1 is or 13.89%. All the posterior probabilities for each hospital can be calculated and then compared: (Table 5.19)
39 Malpractice Suit Filed No Malpractice Suit FiledBayes’ Theorem Example: Hospital Trauma Centers Intuitively, imagine there were 10,000 patients and calculate the frequencies: Hospital Malpractice Suit Filed No Malpractice Suit Filed Total 1 5 4,995 5,000 2 15 2,985 3,000 3 16 1,984 2,000 36 9,964 10,000 = 5,000 x .001 = 5, = 10,000x.5 = 3,000 x .005 = 3, = 10,000x.3 = 2,000 x .008 = 1, = 10,000x.2
40 Malpractice Suit Filed No Malpractice Suit FiledBayes’ 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 P(A)=36/10000=.0036 P(A')=.9964 1.0000
41 Bayes’ Theorem Example: Hospital Trauma CentersConsider the following visual description of the problem:
42 Bayes’ Theorem Example: Hospital Trauma CentersThe initial sample space consists of 3 mutually exclusive and collectively exhaustive events (hospitals B1, B2, B3).
43 Bayes’ Theorem Example: Hospital Trauma CentersAs indicated by their relative areas, B1 is 50% of the sample space, B2 is 30% and B3 is 20%. 30% 50% 20%
44 Bayes’ Theorem Example: Hospital Trauma CentersBut, 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) P(B1 | A) P(B3 | A)
45 Counting Rules Fundamental Rule of CountingIf 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.
46 Counting Rules Example: Stock-Keeping LabelsHow 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
47 Counting Rules Example: Stock-Keeping LabelsView 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.
48 Counting Rules Example: Shirt InventoryL.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.
49 Counting Rules Factorials n! = n(n–1)(n–2)...1The 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.
50 Counting Rules FactorialsAs 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.
51 Counting Rules FactorialsA 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
52 Counting Rules PermutationsA permutation is an arrangement in a particular order of r randomly sampled items from a group of n items and is denoted by nPr In other words, how many ways can the r items be arranged, treating each arrangement as different (i.e., XYZ is different from ZYX)?
53 Counting Rules Example: Appliance Service Cansn = 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:
54 Counting Rules Example: Appliance Service CansThe 60 permutations with r = 3 out of the n = 5 calls can be enumerated. There are 10 distinct groups of 3 customers: Each of these can be arranged in 6 distinct ways: ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABC, ACB, BAC, BCA, CAB, CBA Since there are 10 groups of 3 customers and 6 arrange-ments per group, there are 10 x 6 = 60 permutations.
55 Counting Rules CombinationsA 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
56 Counting Rules Example: Appliance Service Calls Revisitedn = 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:
57 Counting Rules Example: Appliance Service Calls Revisited10 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
58 Applied Statistics in Business & EconomicsEnd of Chapter 5B 5B-58