# 11.4: Test of Independence

• • OpenStax
• OpenStax
$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

Tests of independence involve using a contingency table of observed (data) values.

The test statistic for a test of independence is similar to that of a goodness-of-fit test:

$\sum_{(i \cdot j)} \frac{(O-E)^{2}}{E}$

where:

• $$O =$$ observed values
• $$E =$$ expected values
• $$i =$$ the number of rows in the table
• $$j =$$ the number of columns in the table

There are $$i \cdot j$$ terms of the form $$\frac{(O-E)^{2}}{E}$$.

The expected value for each cell needs to be at least five in order for you to use this test.

A test of independence determines whether two factors are independent or not. You first encountered the term independence in Probability Topics. As a review, consider the following example.

##### Example $$\PageIndex{1}$$

Suppose $$A =$$ a speeding violation in the last year and $$B =$$ a cell phone user while driving. If $$A$$ and $$B$$ are independent then $$P(A \text{ AND } B) = P(A)P(B)$$. $$A \text{ AND } B$$ is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phone while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not.

Let $$y =$$ expected number of drivers who used a cell phone while driving and received speeding violations.

If $$A$$ and $$B$$ are independent, then $$P(A \text{ AND } B) = P(A)P(B)$$. By substitution,

$\frac{y}{755} = \left(\frac{70}{755}\right)\left(\frac{305}{755}\right) \nonumber$

Solve for $$y$$:

$y = \frac{(70)(305)}{755} = 28.3 \nonumber$

About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations.

In a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of two factors, the null hypothesis states that the factors are independent and the alternative hypothesis states that they are not independent (dependent). If we do a test of independence using the example, then the null hypothesis is:

$$H_{0}$$: Being a cell phone user while driving and receiving a speeding violation are independent events.

If the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation.

The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit.

The number of degrees of freedom for the test of independence is:

$df = (\text{number of columns} - 1)(\text{number of rows} - 1) \nonumber$

The following formula calculates the expected number ($$E$$):

$E = \frac{\text{(row total)(column total)}}{\text{total number surveyed}} \nonumber$

##### Exercise $$\PageIndex{1}$$

A sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. Ninety-seven were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll?

About 16 students are expected to be music students and on the honor roll.

##### Example $$\PageIndex{2}$$

In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. In Table $$\PageIndex{1}$$ is a sample of the adult volunteers and the number of hours they volunteer per week.

Table $$\PageIndex{1}$$: Number of Hours Worked Per Week by Volunteer Type (Observed). The table contains observed (O) values (data).
Type of Volunteer 1–3 Hours 4–6 Hours 7–9 Hours Row Total
Community College Students 111 96 48 255
Four-Year College Students 96 133 61 290
Nonstudents 91 150 53 294
Column Total 298 379 162 839

Is the number of hours volunteered independent of the type of volunteer? Use the $$p$$-value method with $$\alpha=0.02$$.

Determine the hypothesis:

The observed table and the question at the end of the problem, "Is the number of hours volunteered independent of the type of volunteer?" tell you this is a test of independence. The two factors are number of hours volunteered and type of volunteer. This test is always right-tailed.

• $$H_{0}$$: The number of hours volunteered is independent of the type of volunteer.
• $$H_{a}$$: The number of hours volunteered is dependent on the type of volunteer.

Calculate the evidence:

The expected results are in Table $$\PageIndex{2}$$.

Table $$\PageIndex{2}$$: Number of Hours Worked Per Week by Volunteer Type (Expected). The table contains expected($$E$$) values (data).
Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours
Community College Students 90.57 115.19 49.24
Four-Year College Students 103.00 131.00 56.00
Nonstudents 104.42 132.81 56.77

For example, the calculation for the expected frequency for the top left cell is

$E = \frac{(\text{row total})(\text{column total})}{\text{total number surveyed}} = \frac{(255)(298)}{839} = 90.57 \nonumber$

Next, calculate the test statistic. For each cell, calculate $$\frac{(O-E)^{2}}{E}$$.

For example, the calculation for the top left cell is

$\frac{(O-E)^{2}}{E}=\frac{(111-90.57)^{2}}{90.57}=\frac{20.43^{2}}{90.57}=\frac{417.3849}{90.57}=4.61 \nonumber$

Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours
Community College Students 4.61 3.20 0.03
Four-Year College Students 0.48 0.03 0.45
Nonstudents 1.72 2.22 0.25

Add up all the cells in the previous table to find the test statistic, $$\chi^{2} = 12.99$$.

Next work to find the $$p$$-value. All Chi-Square tests are right-tailed, so use Excel formula $$=\text{CHISQ.DIST.RT}(x,df)$$.

The test statistic, $$\chi^{2} = 12.99$$, is the $$x$$ and the degrees of freedom are

$df = (3 \text{ columns} – 1)(3 \text{ rows} – 1) = (2)(2) = 4 \nonumber$

Enter $$=\text{CHISQ.DIST.RT}(12.99,4)=0.0113$$ Figure $$\PageIndex{1}$$.

Make a Decision:

Compare $$\alpha$$ and the $$p\text{-value}$$: $$\alpha=0.02$$ is given. $$p\text{-value} = 0.0113$$. $$\alpha > p\text{-value}$$.

Since $$\alpha > p\text{-value}$$, reject $$H_{0}$$. This means that the factors are not independent.

Conclusion: At a 2% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another.

##### Exercise $$\PageIndex{2}$$

The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. Table $$\PageIndex{3}$$ shows the results:

Table $$\PageIndex{3}$$
Industry Sector 2000 2010 2020 Total
Nonagriculture wage and salary 13,243 13,044 15,018 41,305
Goods-producing, excluding agriculture 2,457 1,771 1,950 6,178
Services-providing 10,786 11,273 13,068 35,127
Agriculture, forestry, fishing, and hunting 240 214 201 655
Nonagriculture self-employed and unpaid family worker 931 894 972 2,797
Secondary wage and salary jobs in agriculture and private household industries 14 11 11 36
Secondary jobs as a self-employed or unpaid family worker 196 144 152 492
Total 27,867 27,351 31,372 86,590

We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom.

• $$H_{0}$$: The number of jobs is independent of the year.
• $$H_{a}$$: The number of jobs is dependent on the year.
$$df = 12$$ Figure $$\PageIndex{2}$$.
##### Example $$\PageIndex{3}$$

De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. Table shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events. Use the critical value method with $$\alpha=0.10$$.

Need to Succeed in School vs. Anxiety Level
Need to Succeed in School High
Anxiety
Med-high
Anxiety
Medium
Anxiety
Med-low
Anxiety
Low
Anxiety
Row Total
High Need 35 42 53 15 10 155
Medium Need 18 48 63 33 31 193
Low Need 4 5 11 15 17 52
Column Total 57 95 127 63 58 400

Solution

Determine the hypothesis:

$$H_{0}$$: The anxiety level is independent of the need to succeed in school.

$$H_{a}$$: The anxiety level is dependent on the need to succeed in school.

This test is always right-tailed.

Calculate the evidence:

Use the Excel formula $$=\text{CHISQ.INV.RT}(\alpha,df)$$ to find the critical value.

$$\alpha=0.10$$ as given in the problem statement. The degrees of freedom are

$df = (5 \text{ columns} – 1)(3 \text{ rows} – 1) = (4)(2) = 8 \nonumber$

So enter into Excel the formula $$=\text{CHISQ.INV.RT}(0.10,8)=13.3616$$

Next work on calculating the test statistic. First, find the expected values for each cell. The formula for each cell will be $$E = \frac{(\text{row total})(\text{column total})}{\text{total number surveyed}}$$.

Need to Succeed in School High
Anxiety
Med-high
Anxiety
Medium
Anxiety
Med-low
Anxiety
Low
Anxiety
High Need 22.09 36.81 49.21 24.41 22.48
Medium Need 27.50 45.84 61.28 30.40 27.99
Low Need 7.41 12.35 16.51 8.19 7.54

Next, calculate $$\frac{(O-E)^{2}}{E}$$ for each cell.

Need to Succeed in School High
Anxiety
Med-high
Anxiety
Medium
Anxiety
Med-low
Anxiety
Low
Anxiety
High Need 7.54 0.73 0.29 3.63 6.93
Medium Need 3.28 0.10 0.05 0.22 0.32
Low Need 1.57 4.37 1.84 5.66 11.87

Next add up all the cells from the last table to find $$\sum \frac{(O-E)^{2}}{E}$$. This gives the test statistic as $$\chi^{2} = 48.4$$

Make a Decision:

Since this is a right-tailed test, everything larger than the critical value will be in the rejection region. The critical value is $$13.3616$$, so everything larger than $$13.3616$$ is the rejection region. The test statistic is $$\chi^{2} = 48.4$$, which is larger than the critical value, $$13.3616$$, so it is in the rejection region. So we will Reject the Null Hypothesis.

Determine the conclusion:

At a 10% level of significance, from the data, there is sufficient evidence to conclude that the anxiety level and the need to succeed in school are dependent on one another.

##### Exercise $$\PageIndex{3}$$

Refer back to the information in Table 11.4.3. How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020?