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BCOR 1020 Business Statistics

BCOR 1020 Business Statistics. Lecture 8 – February 12, 2007. Overview. Chapter 5 – Probability Contingency Tables Tree Diagrams Counting Rules. Variable 1. Col 1 Col 2 Col 3. Row 1 Row 2 Row 3 Row 4. Variable 2. Chapter 5 – Contingency Tables. What is a Contingency Table?

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BCOR 1020 Business Statistics

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  1. BCOR 1020Business Statistics Lecture 8 – February 12, 2007

  2. Overview • Chapter 5 – Probability • Contingency Tables • Tree Diagrams • Counting Rules

  3. Variable 1 Col 1 Col 2 Col 3 Row 1 Row 2 Row 3 Row 4 Variable 2 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. Cell

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

  5. 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.

  6. 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(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

  7. Chapter 5 – Contingency Tables Marginal 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. .2388 16/67 = P(T1) =

  8. 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

  9. 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. .0149 1/67 = P(T1S3) =

  10. 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): .1563 5/32 = P(S1|T3) =

  11. 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

  12. 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

  13. 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:

  14. 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

  15. 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.

  16. 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.

  17. 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.

  18. 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.

  19. Chapter 5 – Tree Diagrams • 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

  20. The tree diagram shows all events along with their marginal, conditional and joint 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. • For example, consider the probability of a low expense Bond… P(B and L) = (.5238)(.4773) = .2500 Consider the tree on the next slide…

  21. Chapter 5 – Tree Diagrams Tree Diagram for Fund Type and Expense Ratios:

  22. 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

  23. 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

  24. Chapter 5 – 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 • 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.

  25. 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…

  26. 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…

  27. 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!

  28. 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

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