3.4: Contingency Tables

15

Contingency Tables

jkesler

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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.

 

1. Find P(Driver is a cell phone user).
   
2. Find P(driver had no violation in the last year).
3. Find P(Driver had no violation in the last year AND was a cell phone user).
4. Find P(Driver is a cell phone user OR driver had no violation in the last year).
5. Find P(Driver is a cell phone user GIVEN driver had a violation in the last year).
6. Find P(Driver had no violation last year GIVEN driver was not a cell phone user)

Solutions:

1. $\frac{\text{number of cell phone users}}{\text{total number in study}}= \frac{305}{755} = 0.404$
   
2. $\frac{\text{number that had no violation}}{\text{total number in study}}= \frac{685}{755} = 0.9073$
3. $\frac{280}{755} = 0.3709$
4. $\left(\frac{305}{755}+\frac{685}{755}\right)-\frac{280}{755} = \frac{710}{755}=0.9404$
5. $\frac{25}{70}=0.3571$
6. $\frac{405}{450}=0.9$

Try It 3.20

Table 3.3 below 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 below 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 3.4Hiking Area Preference

1. Complete the table.
2. Are the events “being female” and “preferring the coastline” independent events?
   Let F = being female and let C = preferring the coastline.
   1. Find P(F AND C).
   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.

3. 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?
4. 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(F AND P).
   4. Find P(F OR P).

Try It 3.21

Table 3.6 below 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 $\frac{1}{5}$ and the probability he is not caught is $\frac{4}{5}$ . If he goes out the second door, the probability he gets caught by Alissa is $\frac{1}{4}$ and the probability he is not caught is $\frac{3}{4}$. The probability that Alissa catches Muddy coming out of the third door is $\frac{1}{2}$  and the probability she does not catch Muddy is $\frac{1}{2}$ . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is $\frac{1}{3}$.

Caught or Not	Door One	Door Two	Door Three	Total
Caught	$\frac{1}{15}$	$\frac{1}{12}$	$\frac{1}{6}$	____
Not Caught	$\frac{4}{15}$	$\frac{3}{12}=\frac{1}{4}$	$\frac{1}{6}$	____
Total	____	____	____	1

Table 3.7Door Choice

• The first entry $\frac{1}{15}=\left( \frac{1}{5} \right) \left(  \frac{1}{3}\right)$ is $P(\text{Door One AND Caught})$
• The entry $\frac{4}{15}=\left( \frac{4}{5} \right) \left(  \frac{1}{3}\right)$ is $P(\text{Door One AND Not Caught})$

Verify the remaining entries.

 

1. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.
2. What is the probability that Alissa does not catch Muddy?
3. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

Solution

1. Caught or Not	Door One	Door Two	Door Three	Total
   Caught	$\frac{1}{15}$	$\frac{1}{12}$	$\frac{1}{6}$	$\frac{19}{60}$
   Not Caught	$\frac{4}{15}$	$\frac{3}{12}=\frac{1}{4}$	$\frac{1}{6}$	$\frac{41}{60}$
   Total	$\frac{5}{15}=\frac{1}{3}$	$\frac{4}{12}=\frac{1}{3}$	$\frac{2}{6}=\frac{1}{3}$	1

2. $\frac{41}{60}=0.6833$
3. $\frac{9}{19}=0.4737$

Example 3.23

Table 3.9 below 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 3.9United States Crime Index Rates Per 100,000 Inhabitants 2008–2011

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

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

Solution

1. 0.0294
2. 0.1551
3. 0.7165
4. 0.2365
5. 0.2575

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 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

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?