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3.2 Contingency Tables

LEARNING OBJECTIVES

  • Construct and interpret contingency tables.
  • Find probabilities using contingency tables.

A contingency table provides a way of displaying data that can facilitate calculating probabilities. The table can be used to describe the sample space of an experiment. Contingency tables allow us to break down a sample pace when two variables are involved.

Cell Phone Use Speeding violation in the last year No speeding violation in the last year Total
Cell phone user 25 280 305
Not a cell phone user 45 405 450
Total 70 685 755

When reading a contingency table:

  • The left-side column lists all of the values for one of the variables. In the table shown above, the left-side column shows the variable about whether or not someone uses a cell phone while driving.
  • The top row lists all of the values for the other variable. In the table shown above, the top row shows the variable about whether or not someone had a speeding violation in the last year.
  • In the body of the table, the cells contain the number of outcomes that fall into both of the categories corresponding to the intersecting row and column. In the table shown above, the number of [latex]25[/latex] at the intersection of the “cell phone user” row and “speeding violation in the last year” column tells us that there are [latex]25[/latex] people who have both of these characteristics.
  • The bottom row gives the totals in each column. In the table shown above, the number [latex]685[/latex] in the bottom of the “no speeding violation in the last year” tells us that there are [latex]685[/latex] people who did not have a speeding violation in the last year.
  • The right-side column gives the totals in each row. In the table shown above, the number [latex]305[/latex] in the right side of the “cell phone user” row tells us that there are [latex]305[/latex] people who use cell phones while driving.
  • The number in the bottom right corner is the size of the sample space. In the table shown above, the number in the bottom right corner is [latex]755[/latex], which tells us that there are [latex]755[/latex] people in the sample space.

EXAMPLE

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

Cell Phone Use Speeding violation in the last year No speeding violation in the last year Total
Cell phone user 25 280 305
Not a cell phone user 45 405 450
Total 70 685 755

Calculate the following probabilities:

  1. What is the probability that a randomly selected person is a cell phone user?
  2. What is the probability that a randomly selected person had no speeding violations in the last year?
  3. What is the probability that a randomly selected person had a speeding violation in the last year and does not use a cell phone?
  4. What is the probability that a randomly selected person uses a cell phone and had no speeding violations in the last year?

Solution

  1. [latex]\displaystyle{\text{Probability}=\frac{\text{number of cell phone users}}{\text{total number in study}}=\frac{305}{755}}[/latex]
  2. [latex]\displaystyle{\text{Probability}=\frac{\text{number of no violations}}{\text{total number in study}}=\frac{685}{755}}[/latex]
  3. [latex]\displaystyle{\text{Probability}=\frac{\text{number of violations and not cell phone users}}{\text{total number in study}}=\frac{45}{755}}[/latex]
  4. [latex]\displaystyle{\text{Probability}=\frac{\text{number of cell phone users and no violations}}{\text{total number in study}}=\frac{280}{755}}[/latex]

TRY IT

The table below shows the number of athletes who stretch before exercising and how many had injuries within the past year.

Stretch habits Injury in last year No injury in last year Total
Stretches 55 295 350
Does not stretch 231 219 450
Total 286 514 800
  1. What is the probability that a randomly selected athlete stretches before exercising?
  2. What is the probability that a randomly selected athlete had an injury in the last year?
  3. What is the probability that a randomly selected athlete does not stretch before exercising and had no injuries in the last year?
  4. What is the probability that a randomly selected athlete stretches before exercising and had no injuries in the last year?
Click to see Solution
  1. [latex]\displaystyle{\text{Probability}=\frac{350}{800}=0.4375}[/latex]
  2. [latex]\displaystyle{\text{Probability}=\frac{286}{800}=0.3575}[/latex]
  3. [latex]\displaystyle{\text{Probability}=\frac{219}{800}=0.27375}[/latex]
  4. [latex]\displaystyle{\text{Probability}=\frac{295}{800}=0.36875}[/latex]

 

EXAMPLE

The table below shows a random sample of [latex]100[/latex] hikers broken down by gender and the areas of hiking they prefer.

Gender The Coastline Near Lakes and Streams On Mountain Peaks Total
Female 18 16 45
Male 14 55
Total 41
  1. Fill in the missing values in the table
  2. What is the probability that a randomly selected hiker is female?
  3. What is the probability that a randomly selected hiker prefers to hike on the coast?
  4. What is the probability that a randomly selected hiker is male and prefers to hike near lakes and streams?
  5. What is the probability that a randomly selected hiker is female and prefers to hike in the mountains?

Solution

  1. Gender 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
  2. [latex]\displaystyle{\text{Probability}=\frac{45}{100}=0.45}[/latex]
  3. [latex]\displaystyle{\text{Probability}=\frac{34}{100}=0.34}[/latex]
  4. [latex]\displaystyle{\text{Probability}=\frac{25}{100}=0.25}[/latex]
  5. [latex]\displaystyle{\text{Probability}=\frac{11}{100}=0.11}[/latex]

TRY IT

The table below 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
  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 normal.
  4. Find the probability that a randomly chosen individual from this group is obese and short.
  5. Find the probability that a randomly chosen individual from this group is underweight and medium.
Click to see Solution
  1. Weight/Height Tall Medium Short Totals
    Obese 18 28 14 60
    Normal 20 51 28 99
    Underweight 12 25 9 46
    Totals 50 104 51 205
  2. [latex]\displaystyle{\text{Probability}=\frac{50}{205}}[/latex]
  3. [latex]\displaystyle{\text{Probability}=\frac{99}{205}}[/latex]
  4. [latex]\displaystyle{\text{Probability}=\frac{14}{205}}[/latex]
  5. [latex]\displaystyle{\text{Probability}=\frac{25}{205}}[/latex]

Video: “Ex: Basic Example of Finding Probability From a Table” by Mathispower4u [2:40] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.

Exercises

  1. A previous year, the weights, in pounds, of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data are compiled into the table.
    Shirt Number At most 210 211–250 251–290 More than 290 Total
    1–33 21 5 0 0 26
    34–66 6 18 7 4 35
    67–99 6 12 22 5 45
    Total 33 35 29 9 106

    For the following, suppose that one player from these two teams is selected at random.

    1. What is the probability that the player’s shirt number is in the 34-66 category?
    2. What is the probability that the player weighs at most 210 pounds?
    3. What is the probability that the player’s shirt number is in the 1-33 category and weighs between 211 and 250 pounds?
    Click to see Answer
    1. [latex]0.3302[/latex]
    2. [latex]0.3113[/latex]
    3. [latex]0.0472[/latex]

     

  2. The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected.
    Name Single Double Triple Home Run Total Hits
    Babe Ruth 1,517 506 136 714 2,873
    Jackie Robinson 1,054 273 54 137 1,518
    Ty Cobb 3,603 174 295 114 4,189
    Hank Aaron 2,294 624 98 755 3,771
    Total 8,471 1,577 583 1,720 12,351
    1. What is the probability that a hit was made by Jackie Robinson?
    2. What is the probability that the hit was a double?
    3. What is the probability that the hit was made by Ty Cobb?
    4. What is the probability that a hit was made by Hank Aaron and is a home run?
    5. What is the probability that a hit was a triple and hit by Babe Ruth?
    Click to see Answer
    1. [latex]0.1229[/latex]
    2. [latex]0.1277[/latex]
    3. [latex]0.3392[/latex]
    4. [latex]0.0611[/latex]
    5. [latex]0.011[/latex]

     

  3. The table shows a random sample of musicians and how they learned to play their instruments.
    Gender Self-taught Studied in School Private Instruction Total
    Female 12 38 22 72
    Male 19 24 15 58
    Total 31 62 37 130
    1. Find the probability a musician is female.
    2. Find the probability that a musician received private instruction.
    3. Find the probability that a musician is male and is self-taught.
    Click to see Answer
    1. [latex]0.5538[/latex]
    2. [latex]0.2846[/latex]
    3. [latex]0.1462[/latex]

     

  4. The table shows the political party affiliation of each of [latex]67[/latex] members of the US Senate in June 2012, and when they are up for reelection.
    Up for reelection: Democratic Party Republican Party Other Total
    November 2014 20 13 0
    November 2016 10 24 0
    Total
    1. Complete the table by filling in the totals.
    2. What is the probability that a randomly selected senator has an “Other” affiliation?
    3. What is the probability that a randomly selected senator is a Republican?
    4. What is the probability that a randomly selected senator is up for reelection in November 2016?
    5. What is the probability that a randomly selected senator is a Democrat and up for reelection in November 2014?
    Click to see Answer
    1. Up for reelection: Democratic Party Republican Party Other Total
      November 2014 20 13 0 33
      November 2016 10 24 0 34
      Total 30 37 0 67
    2. [latex]0[/latex]
    3. [latex]0.5522[/latex]
    4. [latex]0.5075[/latex]
    5. [latex]0.2985[/latex]

     

  5. The table below identifies a group of children by one of four hair colours and by type of hair.
    Hair Type Brown Blond Black Red Totals
    Wavy 20 15 3 43
    Straight 80 15 12
    Totals 20 215
    1. Complete the table.
    2. What is the probability that a randomly selected child will have wavy hair?
    3. What is the probability that a randomly selected child will have blond hair?
    4. What is the probability that a randomly selected child will have straight hair and brown hair?
    Click to see Answer
    1. Hair Type Brown Blond Black Red Totals
      Wavy 20 5 15 3 43
      Straight 80 15 65 12 172
      Totals 100 20 80 15 215
    2. [latex]0.2[/latex]
    3. [latex]0.093[/latex]
    4. [latex]0.3721[/latex]

     

“3.3 Contingency Tables” and “3.8 Exercises” from Introduction to Statistics by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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