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5. Chapter. Probability (Part 2). Contingency Tables Tree Diagrams Bayes’s Theorem (optional) Counting Rules (optional). 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?.

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Probability (Part 2)

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Probability part 2

5

Chapter

Probability (Part 2)

Contingency Tables

Tree Diagrams

Bayes’s Theorem (optional)

Counting Rules (optional)


Probability part 2

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.


Probability part 2

Contingency Tables

  • Example: Salary Gains and MBA Tuition

  • Consider the following cross-tabulation table for n = 67 top-tier MBA programs:


Probability part 2

  • 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.


Probability part 2

  • 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).


Probability part 2

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.


Probability part 2

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)).


Probability part 2

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.


Probability part 2

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.


Probability part 2

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(T1|S3) =

  • What does this mean?


Probability part 2

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?


Probability part 2

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:


Probability part 2

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


Probability part 2

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.


Probability part 2

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.


Probability part 2

Contingency Tables

  • Example: Payment Method and Purchase Quantity

  • Here is the contingency table of frequencies:


Probability part 2

  • 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?


Probability part 2

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.


Probability part 2

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.


Probability part 2

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.


Probability part 2

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.


Probability part 2

Tree Diagrams

  • What is a Tree?

  • To label the tree, first calculate conditional probabilities by dividing each cell frequency by its column total.

= .5238

  • For example,

= 11/21

P(L | B)

  • Here is the table of conditional probabilities


Probability part 2

  • The tree diagram shows all events along with their marginal, conditional and joint probabilities.

Tree Diagrams

  • What is a Tree?

  • 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


Probability part 2

Tree Diagrams

  • Tree Diagram for Fund Type and Expense Ratios


Probability part 2

Bayes’s 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:


Probability part 2

Bayes’s Theorem

  • Bayes’s formula begins as:

  • In some situations P(A) is not given. Therefore, the most useful and common form of Bayes’s Theorem is:


Probability part 2

False Negative

False Positive

96% of time

4% of time

1% of time

99% of time

Bayes’s Theorem

  • How Bayes’s 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”?


Probability part 2

Bayes’s Theorem

  • How Bayes’s 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.


Probability part 2

Bayes’s Theorem

  • How Bayes’s 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


Probability part 2

  • From the contingency table, we know that:

Bayes’s Theorem

  • How Bayes’s Theorem Works

  • Now use Bayes’s Theorem to formally derive the result P(Pregnant | Positive) = .9931:

  • First defineA = positive test B = pregnantA' = negative test B' = not pregnant

  • And the compliment of each event is:

P(A | B) = .96P(A | B') = .01P(B) = .60

P(A' | B) = .04P(A' | B') = .99 P(B') = .40


Probability part 2

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

=

=

=

.9931

.576 + .04

.580

Bayes’s Theorem

  • How Bayes’s Theorem Works

  • So, there is a 99.31% chance that a woman is pregnant, given that the test is positive.


Probability part 2

Bayes’s Theorem

  • How Bayes’s 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.

  • Bayes’s Theorem is useful when a direct calculation of a conditional probability is not permitted due to lack of information.


Probability part 2

Bayes’s Theorem

  • How Bayes’s Theorem Works

  • A tree diagram helps visualize the situation.


Probability part 2

Bayes’s Theorem

  • How Bayes’s 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 whosenumerator is P(B A).


Probability part 2

Bayes’s Theorem

  • General Form of Bayes’s Theorem

  • A generalization of Bayes’s Theorem allows event B to be polytomous (B1, B2, … Bn) rather than dichotomous (B and B').


Probability part 2

Bayes’s Theorem

  • Example: Hospital Trauma Centers

  • 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 filedBi= patient was treated at trauma center i


Probability part 2

0.

Bayes’s Theorem

  • Example: Hospital Trauma Centers

  • Applying the general form of Bayes’ Theorem, find P(B1 | A).


Probability part 2

Bayes’s 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:


Probability part 2

= 10,000x.2

= 10,000x.3

= 10,000x.5

= 5,000 - 5

= 2,000 x .008

= 3,000 x .005

= 5,000 x .001

= 3,000 - 15

= 1,984 - 16

Bayes’s Theorem

  • Example: Hospital Trauma Centers

  • Intuitively, imagine there were 10,000 patients and calculate the frequencies:


Probability part 2

Bayes’s Theorem

  • Example: Hospital Trauma Centers

  • Now, use these frequencies to find the probabilities needed for Bayes’ Theorem.

  • For example,


Probability part 2

Bayes’s Theorem

  • Example: Hospital Trauma Centers

  • Consider the following visual description of the problem:


Probability part 2

Bayes’s Theorem

  • Example: Hospital Trauma Centers

  • The initial sample space consists of 3 mutually exclusive and collectively exhaustive events (hospitals B1, B2, B3).


Probability part 2

Bayes’s Theorem

  • Example: Hospital Trauma Centers

  • As indicated by their relative areas, B1is 50% of the sample space, B2 is 30% and B3 is 20%.

30%

50%

20%


Probability part 2

P(B2 | A)

P(B3 | A)

P(B1 | A)

Bayes’s 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.


Probability part 2

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.


Probability part 2

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


Probability part 2

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.


Probability part 2

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.


Probability part 2

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.


Probability part 2

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.


Probability part 2

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


Probability part 2

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)?


Probability part 2

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:


Probability part 2

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.


Probability part 2

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


Probability part 2

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:


Probability part 2

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


Probability part 2

Applied Statistics in Business and Economics

End of Chapter 5


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