BCOR 1020Business Statistics

Lecture 8 – February 12, 2007

Overview

• Chapter 5 – Probability– Contingency Tables– Tree Diagrams– Counting Rules

Chapter 5 – Contingency Tables

What is a Contingency Table?• A contingency table is a cross-tabulation of

frequencies into rows and columns.– It is like a frequency distribution for two variables.

Row 1Row 1

Row 2Row 2

Row 3Row 3

Row 4Row 4Var

iabl

e 2

Var

iabl

e 2

Variable 1Variable 1Col 1 Col 2 Col 3Col 1 Col 2 Col 3

CellCell

Chapter 5 – Contingency Tables

Example: Salary Gains and MBA Tuition• Consider the following cross-tabulation table for

n = 67 top-tier MBA programs:

Chapter 5 – 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.

Chapter 5 – Contingency TablesMarginal 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)).

• Conclude that about 49% of salary gains at the top-tier schools were between $50,000 and $100,000 (medium gain).

P(S2) =

33/67 = .4925

Chapter 5 – Contingency TablesMarginal Probabilities:• Find the marginal probability of a low tuition

P(T1).

• There is a 24% chance that a top-tier school’s MBA tuition is under $40.000.

P(T1) =

16/67 =.2388

Clickers

Consider the overhead of the cross-tabulation of salary gains and MBA tuitions. Find the marginal probability of a large salary gain (P(S3)).

A = 17/67

B = 17/33

C = 19/67

D = 32/67

Chapter 5 – 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)).

• There is less than a 2% chance that a top-tier school has both low tuition and large salary gains.

P(T1 S3) = 1/67 = .0149

Chapter 5 – 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.

• To find the probability that the salary gains are small (S1) given that the MBA tuition is large (T3):

P(S1 | T3) = 5/32 = .1563

Clickers

Consider the overhead of the cross-tabulation of salary gains and MBA tuitions. Find the probability that the salary gains are large (S3) given that the MBA tuition is large (T3).P(S3 | T3) = ?

A = 5/15

B = 15/32

C = 12/32

D = 12/15

Chapter 5 – 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? (Clickers)

A = Dependent or B = Independent

Conditional Marginal

P(S3 | T1)= 1/16 = .0625 P(S3) = 17/67 = .2537

Chapter 5 – 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:

Chapter 5 – 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

Chapter 5 – 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.

Chapter 5 – 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.

Chapter 5 – 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 the appropriate menus in our statistical analysis software.

Chapter 5 – 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.

Chapter 5 – Tree Diagrams

• 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

Chapter 5 – Tree Diagrams

• 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.

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

• For example, consider the probability of a low expense Bond…

= (.5238)(.4773) = .2500 P(B and L)

Consider the tree on the next slide…

Chapter 5 – Tree DiagramsTree Diagram for Fund Type and Expense Ratios:Tree Diagram for Fund Type and Expense Ratios:

Chapter 5 – 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.– For example, consider the number of different

possibilities for license plates if each plate consists of three letters followed by a three-digit number. How many possibilities are there?

26 x 26 x 26 x 10 x 10 x 10 = 17,576,000

Chapter 5 – Counting Rules

Sampling with or without replacement:• Sampling with replacement occurs when an object is

selected and then replaced before the next object is selected. (i.e. the object can be selected again).– For example, our license plate example.

• Sampling without replacement occurs when an object is selected and then not replaced (i.e. the object cannot be selected again).– For example, consider the number of different possibilities for

license plates if each plate consists of three letters followed by a three-digit number and no letters or numbers can be repeated…

26 x 25 x 24 x 10 x 9 x 8 = 11,232,000

Chapter 5 – Counting RulesFactorials:• 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• By definition, 0! = 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.

Chapter 5 – Counting Rules

Permutations and Combinations:• A permutation is an arrangement in a particular

order of r randomly sampled items from a group of n items (i.e., XYZ is not the same as ZYX).– If r items are randomly selected (with replacement)

from n items, then the number of permutations is…

nr

– If r items are randomly selected (without replacement) from n items, then the number of permutations, denoted by nPr is…

!

( )!rn

nP

n r

Chapter 5 – Counting Rules

Permutations and 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).– If r items are randomly selected (without replacement)

from n items, then the number of combinations can be determined by dividing out the number of distinct orderings of the r items (r!) from the number of permutations.

– The number of combinations, denoted by nCr is…!

!( )!rn

nC

r n r

Chapter 5 – Counting Rules

Example – Lottery Odds:• Consider the Colorado Lottery drawing…

• There are 42 balls, numbered 1 – 42. (n = 42)• 6 balls are selected at random. (r = 6)• Order is unimportant. (combinations, not

permutations)

• How many different combinations are possible?

)!(!

!

rnr

nCrn

!36!6

!42 786,245,5

The probability that a single ticket will have the winning combination of numbers is 1 in 5,245,786!

ClickersConsider a standard deck of playing cards – which consists of 52 cards. If five cards are drawn at random and order is of no importance, how many distinct 5-card poker hands are possible?

A = 2,598,960B = 3,168,367C = 311,875,200D = 380,204,032