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5B-1

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Probability (Part 2) Contingency Tables Contingency Tables Counting Rules Counting Rules Chapter 5B5B McGraw-Hill/Irwin© 2008 The McGraw-Hill Companies, Inc. All rights reserved.

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5B-3 A contingency table is a cross- tabulation of frequencies into rows and columns.A contingency table is a cross- tabulation of frequencies into rows and columns. Row 1 Row 2 Row 3 Row 4 Variable 2 Variable 1 Col 1 Col 2 Col 3 Cell A contingency table is like a frequency distribution for two variables.A contingency table is like a frequency distribution for two variables. Contingency Tables What is a Contingency Table?

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5B-4 Consider the following cross-tabulation table for n = 67 top-tier MBA programs:Consider the following cross-tabulation table for n = 67 top-tier MBA programs: Contingency Tables Example: Salary Gains and MBA Tuition

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5B-5 Are large salary gains more likely to accrue to graduates of high-tuition MBA programs?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.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.Also, most of the top-tier schools charge high tuition. More precise interpretations of this data can be made using the concepts of probability.More precise interpretations of this data can be made using the concepts of probability. Contingency Tables Example: Salary Gains and MBA Tuition

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5B-6 The marginal probability of a single event is found by dividing a row or column total by the total sample size.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(S 2 )).For example, find the marginal probability of a medium salary gain (P(S 2 )). P(S 2 ) = 33/67 =.4925 Conclude that about 49% of salary gains at the top-tier schools were between $50,000 and $100,000 (medium gain).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

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5B-7 Find the marginal probability of a low tuition P(T 1 ).Find the marginal probability of a low tuition P(T 1 ). P(T 1 ) = 16/67 =.2388 There is a 24% chance that a top-tier school’s MBA tuition is under $ There is a 24% chance that a top-tier school’s MBA tuition is under $ Contingency Tables Marginal Probabilities

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5B-8 A joint probability represents the intersection of two events in a cross-tabulation table.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(T 1 S 3 )).Consider the joint event that the school has low tuition and large salary gains (denoted as P(T 1 S 3 )). Contingency Tables Joint Probabilities

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5B-9 So, using the cross-tabulation table,So, using the cross-tabulation table, P(T 1 S 3 ) = 1/67 =.0149 There is less than a 2% chance that a top-tier school has both low tuition and large salary gains.There is less than a 2% chance that a top-tier school has both low tuition and large salary gains. Contingency Tables Joint Probabilities

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5B-10 Found by restricting ourselves to a single row or column (the condition).Found by restricting ourselves to a single row or column (the condition). For example, knowing that a school’s MBA tuition is high (T 3 ), we would restrict ourselves to the third row of the table.For example, knowing that a school’s MBA tuition is high (T 3 ), we would restrict ourselves to the third row of the table. Contingency Tables Conditional Probabilities

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5B-11 Find the probability that the salary gains are small (S 1 ) given that the MBA tuition is large (T 3 ).Find the probability that the salary gains are small (S 1 ) given that the MBA tuition is large (T 3 ). P(S 1 | T 3 ) = 5/32 =.1563 What does this mean?What does this mean? Contingency Tables Conditional Probabilities

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5B-12 To check for independent events in a contingency table, compare the conditional to the marginal probabilities.To check for independent events in a contingency table, compare the conditional to the marginal probabilities. For example, if large salary gains (S 3 ) were independent of low tuition (T 1 ), then P(S 3 | T 1 ) = P(S 3 ).For example, if large salary gains (S 3 ) were independent of low tuition (T 1 ), then P(S 3 | T 1 ) = P(S 3 ). ConditionalMarginal P(S 3 | T 1 )= 1/16 =.0625P(S 3 ) = 17/67 =.2537 What do you conclude about events S 3 and T 1 ?What do you conclude about events S 3 and T 1 ? Contingency Tables Independence

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5B-13 Calculate the relative frequencies below for each cell of the cross-tabulation table to facilitate probability calculations.Calculate the relative frequencies below for each cell of the cross-tabulation table to facilitate probability calculations. Contingency Tables Relative Frequencies Symbolic notation for relative frequencies:Symbolic notation for relative frequencies:

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5B-14 Here are the resulting probabilities (relative frequencies). For example,Here are the resulting probabilities (relative frequencies). For example, Contingency Tables Relative Frequencies P(T 1 and S 1 ) = 5/67 P(T 2 and S 2 ) = 11/67 P(T 3 and S 3 ) = 15/67 P(S 1 ) = 17/67 P(T 2 ) = 19/67

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5B-15 Contingency Tables Relative Frequencies The nine joint probabilities sum to since these are all the possible intersections.The 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.Summing the across a row or down a column gives marginal probabilities for the respective row or column.

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5B-16 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.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?Why would this information be useful to the store manager? The manager collected a random sample of 368 customer transactions.The manager collected a random sample of 368 customer transactions. Contingency Tables Example: Payment Method and Purchase Quantity

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5B-17 Here is the contingency table of frequencies:Here is the contingency table of frequencies: Contingency Tables Example: Payment Method and Purchase Quantity

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5B-18 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 Contingency Tables Example: Payment Method and Purchase Quantity Now, is this probability the same if we condition on number of items purchased?Now, is this probability the same if we condition on number of items purchased?

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5B-19 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?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.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 Example: Payment Method and Purchase Quantity

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5B-20 Contingency tables require careful organization and are created from raw data.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.Consider the data of salary gain and tuition for n = 67 top-tier MBA schools. Contingency Tables How Do We Get a Contingency Table?

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5B-21 The data should be coded so that the values can be placed into the contingency table.The data should be coded so that the values can be placed into the contingency table. Contingency Tables How Do We Get a Contingency Table? Once coded, tabulate the frequency in each cell of the contingency table using MINITAB’s Stat | Tables | Cross Tabulation

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5B-22 If event A can occur in n 1 ways and event B can occur in n 2 ways, then events A and B can occur in n 1 x n 2 ways.If event A can occur in n 1 ways and event B can occur in n 2 ways, then events A and B can occur in n 1 x n 2 ways. In general, m events can occur n 1 x n 2 x … x n m ways.In general, m events can occur n 1 x n 2 x … x n m ways. Counting Rules Fundamental Rule of Counting

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5B-23 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)?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 ounceFor example, AF1078: hex-head 6 cm bolts – box of 12 RT4855: Lime-A-Way cleaner – 16 ounce LL3319: Rust-Oleum primer – gray 15 ounce Counting Rules Example: Stock-Keeping Labels

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5B-24 View the problem as filling six empty boxes: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 6 th.There are 26 ways to fill either the 1st or 2nd box and 10 ways to fill the 3rd through 6 th. Therefore, there are 26 x 26 x 10 x 10 x 10 x 10 = 6,760,000 unique inventory labels.Therefore, there are 26 x 26 x 10 x 10 x 10 x 10 = 6,760,000 unique inventory labels. Counting Rules Example: Stock-Keeping Labels

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5B-25 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).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.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.However, the number of each type of shirt to be stocked depends on prior demand. Counting Rules Example: Shirt Inventory

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5B-26 The number of ways that n items can be arranged in a particular order is n factorial.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 factorial is the product of all integers from 1 to n. Factorials are useful for counting the possible arrangements of any n items.Factorials are useful for counting the possible arrangements of any n items. n! = n(n–1)(n–2)...1 There are n ways to choose the first, n-1 ways to choose the second, and so on.There are n ways to choose the first, n-1 ways to choose the second, and so on. Counting Rules Factorials

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5B-27 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.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

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5B-28 A home appliance service truck must make 3 stops (A, B, C).A home appliance service truck must make 3 stops (A, B, C). In how many ways could the three stops be arranged?In how many ways could the three stops be arranged? 3! = 3 x 2 x 1 = 6 List all the possible arrangements:List all the possible arrangements: {ABC, ACB, BAC, BCA, CAB, CBA} How many ways can you arrange 9 baseball players in batting order rotation?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 Factorials

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5B-29 A permutation is an arrangement in a particular order of r randomly sampled items from a group of n items and is denoted by n P rA permutation is an arrangement in a particular order of r randomly sampled items from a group of n items and is denoted by n P 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)?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 Permutations

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5B-30 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.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 order is important so each possible arrangement of the three service calls is different. The number of possible permutations is:The number of possible permutations is: Counting Rules Example: Appliance Service Cans

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5B-31 The 60 permutations with r = 3 out of the n = 5 calls can be enumerated.The 60 permutations with r = 3 out of the n = 5 calls can be enumerated. ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE There are 10 distinct groups of 3 customers:There are 10 distinct groups of 3 customers: Each of these can be arranged in 6 distinct ways: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 arrange-ments per group, there are 10 x 6 = 60 permutations.Since there are 10 groups of 3 customers and 6 arrange-ments per group, there are 10 x 6 = 60 permutations. Counting Rules Example: Appliance Service Cans

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5B-32 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 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 n C rA combination is denoted n C r Counting Rules Combinations

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5B-33 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.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.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.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:The number of possible combinations is: Counting Rules Example: Appliance Service Calls Revisited

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5B 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 Counting Rules Example: Appliance Service Calls Revisited

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