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Introductory Business Statistics

3.4 Contingency Tables and Probability Trees

Introductory Business Statistics3.4 Contingency Tables and Probability Trees
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  1. Preface
  2. 1 Sampling and Data
    1. Introduction
    2. 1.1 Definitions of Statistics, Probability, and Key Terms
    3. 1.2 Data, Sampling, and Variation in Data and Sampling
    4. 1.3 Levels of Measurement
    5. 1.4 Experimental Design and Ethics
    6. Key Terms
    7. Chapter Review
    8. Homework
    9. References
    10. Solutions
  3. 2 Descriptive Statistics
    1. Introduction
    2. 2.1 Display Data
    3. 2.2 Measures of the Location of the Data
    4. 2.3 Measures of the Center of the Data
    5. 2.4 Sigma Notation and Calculating the Arithmetic Mean
    6. 2.5 Geometric Mean
    7. 2.6 Skewness and the Mean, Median, and Mode
    8. 2.7 Measures of the Spread of the Data
    9. Key Terms
    10. Chapter Review
    11. Formula Review
    12. Practice
    13. Homework
    14. Bringing It Together: Homework
    15. References
    16. Solutions
  4. 3 Probability Topics
    1. Introduction
    2. 3.1 Terminology
    3. 3.2 Independent and Mutually Exclusive Events
    4. 3.3 Two Basic Rules of Probability
    5. 3.4 Contingency Tables and Probability Trees
    6. 3.5 Venn Diagrams
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Bringing It Together: Practice
    12. Homework
    13. Bringing It Together: Homework
    14. References
    15. Solutions
  5. 4 Discrete Random Variables
    1. Introduction
    2. 4.1 Hypergeometric Distribution
    3. 4.2 Binomial Distribution
    4. 4.3 Geometric Distribution
    5. 4.4 Poisson Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  6. 5 Continuous Random Variables
    1. Introduction
    2. 5.1 Properties of Continuous Probability Density Functions
    3. 5.2 The Uniform Distribution
    4. 5.3 The Exponential Distribution
    5. Key Terms
    6. Chapter Review
    7. Formula Review
    8. Practice
    9. Homework
    10. References
    11. Solutions
  7. 6 The Normal Distribution
    1. Introduction
    2. 6.1 The Standard Normal Distribution
    3. 6.2 Using the Normal Distribution
    4. 6.3 Estimating the Binomial with the Normal Distribution
    5. Key Terms
    6. Chapter Review
    7. Formula Review
    8. Practice
    9. Homework
    10. References
    11. Solutions
  8. 7 The Central Limit Theorem
    1. Introduction
    2. 7.1 The Central Limit Theorem for Sample Means
    3. 7.2 Using the Central Limit Theorem
    4. 7.3 The Central Limit Theorem for Proportions
    5. 7.4 Finite Population Correction Factor
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  9. 8 Confidence Intervals
    1. Introduction
    2. 8.1 A Confidence Interval for a Population Standard Deviation, Known or Large Sample Size
    3. 8.2 A Confidence Interval for a Population Standard Deviation Unknown, Small Sample Case
    4. 8.3 A Confidence Interval for A Population Proportion
    5. 8.4 Calculating the Sample Size n: Continuous and Binary Random Variables
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  10. 9 Hypothesis Testing with One Sample
    1. Introduction
    2. 9.1 Null and Alternative Hypotheses
    3. 9.2 Outcomes and the Type I and Type II Errors
    4. 9.3 Distribution Needed for Hypothesis Testing
    5. 9.4 Full Hypothesis Test Examples
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  11. 10 Hypothesis Testing with Two Samples
    1. Introduction
    2. 10.1 Comparing Two Independent Population Means
    3. 10.2 Cohen's Standards for Small, Medium, and Large Effect Sizes
    4. 10.3 Test for Differences in Means: Assuming Equal Population Variances
    5. 10.4 Comparing Two Independent Population Proportions
    6. 10.5 Two Population Means with Known Standard Deviations
    7. 10.6 Matched or Paired Samples
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. Bringing It Together: Homework
    14. References
    15. Solutions
  12. 11 The Chi-Square Distribution
    1. Introduction
    2. 11.1 Facts About the Chi-Square Distribution
    3. 11.2 Test of a Single Variance
    4. 11.3 Goodness-of-Fit Test
    5. 11.4 Test of Independence
    6. 11.5 Test for Homogeneity
    7. 11.6 Comparison of the Chi-Square Tests
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. Bringing It Together: Homework
    14. References
    15. Solutions
  13. 12 F Distribution and One-Way ANOVA
    1. Introduction
    2. 12.1 Test of Two Variances
    3. 12.2 One-Way ANOVA
    4. 12.3 The F Distribution and the F-Ratio
    5. 12.4 Facts About the F Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  14. 13 Linear Regression and Correlation
    1. Introduction
    2. 13.1 The Correlation Coefficient r
    3. 13.2 Testing the Significance of the Correlation Coefficient
    4. 13.3 Linear Equations
    5. 13.4 The Regression Equation
    6. 13.5 Interpretation of Regression Coefficients: Elasticity and Logarithmic Transformation
    7. 13.6 Predicting with a Regression Equation
    8. 13.7 How to Use Microsoft Excel® for Regression Analysis
    9. Key Terms
    10. Chapter Review
    11. Practice
    12. Solutions
  15. A | Statistical Tables
  16. B | Mathematical Phrases, Symbols, and Formulas
  17. Index

Contingency Tables

A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.

Example 3.20

Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:

Speeding violation in the last year No speeding violation in the last year Total
Uses cell phone while driving 25 280 305
Does not use cell phone while driving 45 405 450
Total 70 685 755
Table 3.2

The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.

Calculate the following probabilities using the table.

a. Find P(Driver is a cell phone user).

Solution 3.20

a. number of cell phone users total number in study  =  305 755 number of cell phone users total number in study  =  305 755

b. Find P(Driver had no violation in the last year).

Solution 3.20

b. number that had no violation total number in study  =  685 755 number that had no violation total number in study  =  685 755

c. Find P(Driver had no violation in the last year was a cell phone user).

Solution 3.20

c. 280 755 280 755

d. Find P(Driver is a cell phone user driver had no violation in the last year).

Solution 3.20

d. ( 305 755  +  685 755 )   280 755  =  710 755 ( 305 755  +  685 755 )   280 755  =  710 755

e. Find P(Driver is a cell phone user || driver had a violation in the last year).

Solution 3.20

e. 25 70 25 70 (The sample space is reduced to the number of drivers who had a violation.)

f. Find P(Driver had no violation last year || driver was not a cell phone user)

Solution 3.20

f. 405 450 405 450 (The sample space is reduced to the number of drivers who were not cell phone users.)

Try It 3.20

Table 3.3 shows the number of athletes who stretch before exercising and how many had injuries within the past year.

Injury in last year No injury in last year Total
Stretches 55 295 350
Does not stretch 231 219 450
Total 286 514 800
Table 3.3
  1. What is P(athlete stretches before exercising)?
  2. What is P(athlete stretches before exercising||no injury in the last year)?

Example 3.21

Table 3.4 shows a random sample of 100 hikers and the areas of hiking they prefer.

Sex The coastline Near lakes and streams On mountain peaks Total
Female 18 16 ___ 45
Male ___ ___ 14 55
Total ___ 41 ___ ___
Table 3.4 Hiking Area Preference

a. Complete the table.

Solution 3.21

a.

Sex The coastline Near lakes and streams On mountain peaks Total
Female 18 16 11 45
Male 16 25 14 55
Total 34 41 25 100
Table 3.5 Hiking Area Preference

b. Are the events "being female" and "preferring the coastline" independent events?

Let F = being female and let C = preferring the coastline.

  1. Find P(FC)P(FC).
  2. Find P(F)P(C)

Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.

Solution 3.21

b.

  1. P(FC)= 18 100 P(FC)= 18 100 = 0.18
  2. P(F)P(C) = ( 45 100 )( 34 100 ) ( 45 100 )( 34 100 ) = (0.45)(0.34) = 0.153

P(F C)P(FC)P(F)P(C), so the events F and C are not independent.

c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams.

  1. What word tells you this is a conditional?
  2. Fill in the blanks and calculate the probability: P(___||___) = ___.
  3. Is the sample space for this problem all 100 hikers? If not, what is it?
Solution 3.21

c.

  1. The word 'given' tells you that this is a conditional.
  2. P(M||L) = 25 41 25 41
  3. No, the sample space for this problem is the 41 hikers who prefer lakes and streams.



d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks.

  1. Find P(F).
  2. Find P(P).
  3. Find P(FP)P(FP).
  4. Find P(FP)P(FP).
Solution 3.21

d.

  1. P(F) = 45 100 45 100
  2. P(P) = 25 100 25 100
  3. P(FP)P(FP) = 11 100 11 100
  4. P(FP)P(FP) = 45 100 45 100 + 25 100 25 100 - 11 100 11 100 = 59 100 59 100
Try It 3.21

Table 3.6 shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path.

Gender Lake path Hilly path Wooded path Total
Female 45 38 27 110
Male 26 52 12 90
Total 71 90 39 200
Table 3.6
  1. Out of the males, what is the probability that the cyclist prefers a hilly path?
  2. Are the events “being male” and “preferring the hilly path” independent events?

Example 3.22

Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is 1 5 1 5 and the probability he is not caught is 4 5 4 5 . If he goes out the second door, the probability he gets caught by Alissa is 1 4 1 4 and the probability he is not caught is 3 4 3 4 . The probability that Alissa catches Muddy coming out of the third door is 1 2 1 2 and the probability she does not catch Muddy is 1 2 1 2 . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is 1 3 1 3 .

Caught or not Door one Door two Door three Total
Caught 1 15 1 15 1 12 1 12 1 6 1 6 ____
Not caught 4 15 4 15 3 12 3 12 1 6 1 6 ____
Total ____ ____ ____ 1
Table 3.7 Door Choice
  • The first entry 1 15 = ( 1 5 ) ( 1 3 ) 1 15 = ( 1 5 )( 1 3 ) is P(Door OneCaught)P(Door OneCaught)
  • The entry 4 15 = ( 4 5 )( 1 3 ) 4 15 =( 4 5 )( 1 3 ) is P(Door OneNot Caught)P(Door OneNot Caught)

Verify the remaining entries.

a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.

Solution 3.22

a.

Caught or not Door one Door two Door three Total
Caught 1 15 1 15 1 12 1 12 1 6 1 6 19601960
Not caught 4 15 4 15 3 12 3 12 1 6 1 6 41604160
Total 515515 412412 2626 1
Table 3.8 Door Choice

b. What is the probability that Alissa does not catch Muddy?

Solution 3.22

b. 41604160

c. What is the probability that Muddy chooses Door One Door Two given that Muddy is caught by Alissa?

Solution 3.22

c. 919919

Example 3.23

Table 3.9 contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.

Year Robbery Burglary Rape Vehicle Total
2008 145.7 732.1 29.7 314.7
2009 133.1 717.7 29.1 259.2
2010 119.3 701 27.7 239.1
2011 113.7 702.2 26.8 229.6
Total
Table 3.9 United States Crime Index Rates Per 100,000 Inhabitants 2008–2011

TOTAL each column and each row. Total data = 4,520.7

  1. Find P(2009Robbery)P(2009Robbery).
  2. Find P(2010Burglary)P(2010Burglary).
  3. Find P(2010Burglary)P(2010Burglary).
  4. Find P(2011||Rape).
  5. Find P(Vehicle||2008).
Solution 3.23

a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575

Try It 3.23

Table 3.10 relates the weights and heights of a group of individuals participating in an observational study.

Weight/height Tall Medium Short Totals
Obese 18 28 14
Normal 20 51 28
Underweight 12 25 9
Totals
Table 3.10
  1. Find the total for each row and column
  2. Find the probability that a randomly chosen individual from this group is Tall.
  3. Find the probability that a randomly chosen individual from this group is Obese and Tall.
  4. Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese.
  5. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
  6. Find the probability a randomly chosen individual from this group is Tall and Underweight.
  7. Are the events Obese and Tall independent?

Tree Diagrams

Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams can be used to visualize and solve conditional probabilities.

Tree Diagrams

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.

Example 3.24

In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.

This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8B and 3R. The second branch has a set of two lines (8B and 3R) for each line of the first branch. Multiply along each line to find 64BB, 24BR, 24RB, and 9RR.
Figure 3.2 Total = 64 + 24 + 24 + 9 = 121

The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as:

R1R1; R1R2; R1R3; R2R1; R2R2; R2R3; R3R1; R3R2; R3R3

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.

a. List the 24 BR outcomes: B1R1, B1R2, B1R3, ...

Solution 3.24

a. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3

b. Using the tree diagram, calculate P(RR).

Solution 3.24

b. P(RR) = ( 3 11 )( 3 11 ) ( 3 11 )( 3 11 ) = 9 121 9 121

c. Using the tree diagram, calculate P(RBBR)P(RBBR).

Solution 3.24

c. P(RBBR)P(RBBR) = ( 3 11 )( 8 11 ) ( 3 11 )( 8 11 ) + ( 8 11 )( 3 11 ) ( 8 11 )( 3 11 ) = 48 121 48 121

d. Using the tree diagram, calculate P(Ron 1st drawBon 2nd draw)P(Ron 1st drawBon 2nd draw).

Solution 3.24

d. P(Ron 1st drawBon 2nd draw)P(Ron 1st drawBon 2nd draw) = ( 3 11 )( 8 11 ) ( 3 11 )( 8 11 ) = 24 121 24 121

e. Using the tree diagram, calculate P(R on 2nd draw||B on 1st draw).

Solution 3.24

e. P(R on 2nd draw||B on 1st draw) = P(R on 2nd||B on 1st) = 24 88 24 88 = 3 11 3 11

This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. 24 88 24 88 = 3 11 3 11 .

f. Using the tree diagram, calculate P(BB).

Solution 3.24

f. P(BB) =  64 121 64 121

g. Using the tree diagram, calculate P(B on the 2nd draw||R on the first draw).

Solution 3.24

g. P(B on 2nd draw||R on 1st draw) =  8 11 8 11

There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then 24 33 24 33 .

Try It 3.24

In a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF).

This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 12F and 40N. The second branch has a set of two lines (12F and 40N) for each line of the first branch. Multiply along each line to find 144FF, 480FN, 480NF, and 1,600NN.
Figure 3.3

Example 3.25

An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. "Without replacement" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, ( 3 11 )( 2 10 )= 6 110 ( 3 11 )( 2 10 )= 6 110 .

This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8/11 and R 3/11. The second branch has a set of 2 lines for each first branch line. Below B 8/11 are B 7/10 and R 3/10. Below R 3/11 are B 8/10 and R 2/10. Multiply along each line to find BB 56/110, BR 24/110, RB 24/110, and RR 6/110.
Figure 3.4 Total = 56+24+24+6 110 = 110 110 =1 56+24+24+6 110 = 110 110 =1

NOTE

If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.


Calculate the following probabilities using the tree diagram.

a. P(RR) = ________

Solution 3.25

a. P(RR) = ( 3 11 )( 2 10 )= 6 110 ( 3 11 )( 2 10 )= 6 110

b. Fill in the blanks:

P(RBBR)P(RBBR) = ( 3 11 )( 8 10 ) + (___)(___) =  48 110 ( 3 11 )( 8 10 ) + (___)(___) =  48 110

Solution 3.25

b. P(RBBR)P(RBBR) = ( 3 11 )( 8 10 ) ( 3 11 )( 8 10 ) + ( 8 11 )( 3 10 ) ( 8 11 )( 3 10 ) = 48 110 48 110

c. P(R on 2nd||B on 1st) =

Solution 3.25

c. P(R on 2nd||B on 1st) = 3 10 3 10

d. Fill in the blanks.

P(Ron 1stBon 2nd)P(Ron 1stBon 2nd) = (___)(___) = 24 100 24 100

Solution 3.25

d. P(Ron 1stBon 2nd)P(Ron 1stBon 2nd) = ( 3 11 )( 8 10 ) ( 3 11 )( 8 10 ) = 24 100 24 100

e. Find P(BB).

Solution 3.25

e. P(BB) = ( 8 11 )( 7 10 ) ( 8 11 )( 7 10 )

f. Find P(B on 2nd||R on 1st).

Solution 3.25

f. Using the tree diagram, P(B on 2nd||R on 1st) = P(R||B) = 8 10 8 10 .

If we are using probabilities, we can label the tree in the following general way.

This is a tree diagram for a two-step experiment. The first branch shows first outcome: P(B) and P(R). The second branch has a set of 2 lines for each line of the first branch: the probability of B given B = P(BB), the probability of R given B = P(RB), the probability of B given R = P(BR), and the probability of R given R = P(RR).
  • P(R||R) here means P(R on 2nd||R on 1st)
  • P(B||R) here means P(B on 2nd||R on 1st)
  • P(R||B) here means P(R on 2nd||B on 1st)
  • P(B||B) here means P(B on 2nd||B on 1st)

Try It 3.25

In a standard deck, there are 52 cards. Twelve cards are face cards (F) and 40 cards are not face cards (N). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.

This is a tree diagram with branches showing frequencies of each draw. The first branch shows 2 lines: F 12/52 and N 40/52. The second branch has a set of 2 lines (F 11/52 and N 40/51) for each line of the first branch. Multiply along each line to find FF 121/2652, FN 480/2652, NF 480/2652, and NN 1560/2652.
Figure 3.5
  1. Find P(FNNF)P(FNNF).
  2. Find P(N||F).
  3. Find P(at most one face card).
    Hint: "At most one face card" means zero or one face card.
  4. Find P(at least on face card).
    Hint: "At least one face card" means one or two face cards.

Example 3.26

A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.

This is a tree diagram with branches showing probabilities of kitten choices. The first branch shows two lines: T 4/9 and B 5/9. The second branch has a set of 2 lines for each first branch line. Below T 4/9 are T 3/8 and B 5/8. Below B 5/9 are T 4/8 and B 4/8. Multiply along each line to find probabilities of possible combinations.
  1. What is the probability that both kittens are tabby?

    a. ( 1 2 )( 1 2 ) ( 1 2 )( 1 2 ) b. ( 4 9 )( 4 9 ) ( 4 9 )( 4 9 ) c. ( 4 9 )( 3 8 ) ( 4 9 )( 3 8 ) d. ( 4 9 )( 5 9 ) ( 4 9 )( 5 9 )
  2. What is the probability that one kitten of each coloring is selected?

    a. ( 4 9 )( 5 9 ) ( 4 9 )( 5 9 ) b. ( 4 9 )( 5 8 ) ( 4 9 )( 5 8 ) c. ( 4 9 )( 5 9 )+( 5 9 )( 4 9 ) ( 4 9 )( 5 9 )+( 5 9 )( 4 9 ) d. ( 4 9 )( 5 8 )+( 5 9 )( 4 8 ) ( 4 9 )( 5 8 )+( 5 9 )( 4 8 )
  3. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
  4. What is the probability of choosing two kittens of the same color?
Solution 3.26

a. c, b. d, c. 4 8 4 8 , d. 32 72 32 72

Try It 3.26

Suppose there are four red balls and three yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?

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