BCOR 1020 Business 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 1 Row 2 Row 3 Row 4 Variable 2 Variable 1 Col 1 Col 2 Col 3 Cell

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 Tables Marginal 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(S 2 )). Conclude that about 49% of salary gains at the top-tier schools were between $50,000 and $100,000 (medium gain). P(S 2 ) =33/67 =.4925

Chapter 5 – Contingency Tables Marginal Probabilities: Find the marginal probability of a low tuition P(T 1 ). There is a 24% chance that a top-tier school’s MBA tuition is under $ P(T 1 ) = 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(S 3 )). 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(T 1  S 3 )). There is less than a 2% chance that a top-tier school has both low tuition and large salary gains. P(T 1  S 3 ) = 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 (T 3 ), we would restrict ourselves to the third row of the table. To 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

Clickers Consider the overhead of the cross-tabulation of salary gains and MBA tuitions. Find the probability that the salary gains are large (S 3 ) given that the MBA tuition is large (T 3 ). P(S 3 | T 3 ) = ? 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 (S 3 ) were independent of low tuition (T 1 ), then P(S 3 | T 1 ) = P(S 3 ). What do you conclude about events S 3 and T 1 ? (Clickers) A = DependentorB = Independent ConditionalMarginal P(S 3 | T 1 )= 1/16 =.0625P(S 3 ) = 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(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

Chapter 5 – Contingency Tables Relative Frequencies: 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.

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 Diagrams Tree Diagram for Fund Type and Expense Ratios:

Chapter 5 – Counting Rules Fundamental Rule of Counting: 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. –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 Rules Factorials: 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… nrnr –If r items are randomly selected (without replacement) from n items, then the number of permutations, denoted by n P r is…

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 n C r is…

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

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