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## Probability (Part 2)

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**5B**Chapter Probability (Part 2) Contingency Tables Counting Rules McGraw-Hill/Irwin © 2008 The McGraw-Hill Companies, Inc. All rights reserved.**Variable 1**Col 1 Col 2 Col 3 Row 1 Row 2 Row 3 Row 4 Variable 2 Contingency Tables • What is a Contingency Table? • A contingency table is a cross-tabulation of frequencies into rows and columns. Cell • A contingency table is like a frequency distribution for two variables.**Contingency Tables**• Example: Salary Gains and MBA Tuition • Consider the following cross-tabulation table for n = 67 top-tier MBA programs:**The frequencies indicate that MBA graduates of high-tuition**schools do tend to have large salary gains. Contingency Tables • Example: Salary Gains and MBA Tuition • Are large salary gains more likely to accrue to graduates of high-tuition MBA programs? • Also, most of the top-tier schools charge high tuition. • More precise interpretations of this data can be made using the concepts of probability.**For example, find the marginal probability of a medium**salary gain (P(S2)). 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. 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).**Contingency Tables**• Marginal Probabilities • Find the marginal probability of a low tuition P(T1). .2388 16/67 = P(T1) = • There is a 24% chance that a top-tier school’s MBA tuition is under $40.000.**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(T1S3)).**Contingency Tables**• Joint Probabilities • So, using the cross-tabulation table, P(T1S3) = 1/67 = .0149 • There is less than a 2% chance that a top-tier school has both low tuition and large salary gains.**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.**Contingency Tables**• Conditional Probabilities • Find the probability that the salary gains are small (S1) given that the MBA tuition is large (T3). .1563 5/32 = P(S1|T3) = • What does this mean?**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). • What do you conclude about events S3 and T1?**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:**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 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.**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.**Contingency Tables**• Example: Payment Method and Purchase Quantity • Here is the contingency table of frequencies:**Calculate the marginal probability that a customer will use**cash to make the payment. Contingency Tables • Example: Payment Method and Purchase Quantity • 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?**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 = .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.**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.**Once coded, tabulate the frequency in each cell of the**contingency table using MINITAB’s Stat | Tables | Cross Tabulation 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.**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 occurn1 x n2 x … x nm ways.**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 12RT4855: Lime-A-Way cleaner – 16 ounceLL3319: Rust-Oleum primer – gray 15 ounce**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.**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.**Counting Rules**• Factorials • The number of ways that n items can be arranged in a particular order is nfactorial. • 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.**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.**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**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 • In other words, how many ways can the r items be arranged, treating each arrangement as different (i.e., XYZ is different from ZYX)?**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:**ABC, ACB, BAC, BCA, CAB, CBA**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: • Each of these can be arranged in 6 distinct ways: ABCABDABEACD ACE ADEBCDBCEBDE CDE • Since there are 10 groups of 3 customers and 6 arrange-ments 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**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:**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