11.3 - Chi-Square Test of Independence

11.3 - Chi-Square Test of Independence

The chi-square (\(\chi^2\)) test of independence is used to test for a relationship between two categorical variables. Recall that if two categorical variables are independent, then \(P(A) = P(A \mid B)\). The chi-square test of independence uses this fact to compute expected values for the cells in a two-way contingency table under the assumption that the two variables are independent (i.e., the null hypothesis is true).

Even if two variables are independent in the population, samples will vary due to random sampling variation. The chi-square test is used to determine if there is evidence that the two variables are not independent in the population using the same hypothesis testing logic that we used with one mean, one proportion, etc.

Again, we will be using the five step hypothesis testing procedure:

1. Check assumptions and write hypotheses.

The assumptions are that the sample is randomly drawn from the population and that all expected values are at least 5 (we will see what expected values are later).

Our hypotheses are:

     \(H_0:\) There is not a relationship between the two variables in the population (they are independent)

     \(H_a:\) There is a relationship between the two variables in the population (they are dependent)

Note: When you're writing the hypotheses for a given scenario, use the names of the variables, not the generic "two variables."

2. Calculate an appropriate test statistic
Chi-Square Test Statistic
\(\chi^2=\sum \dfrac{(Observed-Expected)^2}{Expected}\)
Expected Cell Value
\(E=\dfrac{row\;total \; \times \; column\;total}{n}\)
3. Determine the p-value

The p-value can be found using Minitab Express. Look up the area to the right of your chi-square test statistic on a chi-square distribution with the correct degrees of freedom. Chi-square tests are always right-tailed tests. 

Degrees of Freedom: Chi-Square Test of Independence
\(df=(number\;of\;rows-1)(number\;of\;columns-1)\)
4. Make a decision

If \(p \leq \alpha\) reject the null hypothesis.

If \(p>\alpha\) fail to reject the null hypothesis.

5. State a "real world" conclusion

Write a conclusion in terms of the original research question.


11.3.1 - Example: Gender and Online Learning

11.3.1 - Example: Gender and Online Learning

A sample of 314 Penn State students was asked if they have ever taken an online course. Their genders were also recorded. The contingency table below was constructed. Use a chi-square test of independence to determine if there is a relationship between gender and whether or not someone has taken an online course.

  Have you taken an online course?
  Yes No
Men 43 63
Women 95 113
1. Check assumptions and write hypotheses

\(H_0:\) There is not a relationship between gender and whether or not someone has taken an online course (they are independent)

\(H_a:\) There is a relationship between gender and whether or not someone has taken an online course (they are dependent)

Looking ahead to our calculations of the expected values, we can see that all expected values are at least 5. This means that the sampling distribution can be approximated using the \(\chi^2\) distribution. 

2. Calculate the test statistic

In order to compute the chi-square test statistic we must know the observed and expected values for each cell. We are given the observed values in the table above. We must compute the expected values. The table below includes the row and column totals.

  Have you taken an online course?  
  Yes No  
Men 43 63 106
Women 95 113 208
  138 176 314
\(E=\dfrac{row\;total \times column\;total}{n}\)
\(E_{Men,\;Yes}=\dfrac{106\times138}{314}=46.586\)
\(E_{Men,\;No}=\dfrac{106\times176}{314}=59.414\)
\(E_{Women,\;Yes}=\dfrac{208\times138}{314}=91.414\)
\(E_{Women,\;No}=\dfrac{208 \times 176}{314}=116.586\)

Note that all expected values are at least 5, thus this assumption of the \(\chi^2\) test of independence has been met. 

Observed and expected counts are often presented together in a contingency table. In the table below, expected values are presented in parentheses.

  Have you taken an online course?  
  Yes No  
Men 43 (46.586) 63 (59.414) 106
Women 95 (91.414) 113 (116.586) 208
  138 176 314

\(\chi^2=\sum \dfrac{(O-E)^2}{E} \)

\(\chi^2=\dfrac{(43-46.586)^2}{46.586}+\dfrac{(63-59.414)^2}{59.414}+\dfrac{(95-91.414)^2}{91.414}+\dfrac{(113-116.586)^2}{116.586}=0.276+0.216+0.141+0.110=0.743\)

The chi-square test statistic is 0.743

\(df=(number\;of\;rows-1)(number\;of\;columns-1)=(2-1)(2-1)=1\)

3. Determine the p-value

We can determine the p-value by constructing a chi-square distribution plot with 1 degree of freedom and finding the area to the right of 0.743.

Distribution Plot - Chi-Square, DF=1

\(p = 0.388702\)

4. Make a decision

\(p>\alpha\), therefore we fail to reject the null hypothesis.

5. State a "real world" conclusion

There is not evidence that gender and whether or not an individual has completed an online course are related.

Note that we cannot say for sure that these two categorical variables are independent, we can only say that we do not have evidence that they are dependent.


11.3.2 - Minitab Express: Test of Independence

11.3.2 - Minitab Express: Test of Independence

If you have a data file with the responses for individual cases then you have "raw data" and can follow the directions below. If you have a table filled with data, then you have "summarized data." There is an example of conducting a chi-square test of independence using summarized data on a later page. After data entry the procedure is the same for both data entry methods.

MinitabExpress  – Chi-square Test Using Raw Data

Research question: Is there a relationship between where a student sits in class and whether they have ever cheated?

Null hypothesis: Seat location and cheating are not related in the population. 
Alternative hypothesis: Seat location and cheating are related in the population.

To perform a chi-square test of independence in Minitab Express using raw data:

  1. Open Minitab data set
  2. On a PC: Select STATISTICS > Cross Tabulation and Chi-Square
    On a Mac: Select Statistics > Tables > Cross Tabulation and Chi-Square
  3. Double-click on the variable Seating to insert it into the Rows box
  4. Double-click on the variable Ever Cheat to insert it into the Columns box
  5. Click the Display tab and check the boxes Chi-square test for association and Expected cell counts
  6. Click OK

This should result in the following output:

Cell contents grouped by Back, Front, Middle, All; First row: count, Next row: expected count

Rows: Seating | Columns: Ever Cheat
  No Yes All
Back 24 8 32
  24.21 7.79  
Front 38 8 46
  34.81 11.19  
Middle 109 39 148
  111.98 36.02  
All 1714 55 226
Chi-Square Test
  Chi-Square DF P-Value
Pearson 1.54 2 0.4633
Likelihood Ratio 1.63 2 0.4435
Video Walkthrough

Select your operating system below to see a step-by-step guide for this example.

All expected values are at least 5 so we can use the Pearson chi-square test statistic. Our results are \(\chi^2 (2) = 1.54\). \(p = 0.4633\). Because our \(p\) value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not evidence of a relationship in the population between seat location and whether a student has cheated.


11.3.2.1 - Video Example: Dog & Cat Ownership (Raw Data)

11.3.2.1 - Video Example: Dog & Cat Ownership (Raw Data)

This example uses the dataset:

1. Check any necessary assumptions and write null and alternative hypotheses.

 \(H_0:\) There is not a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students
\(H_a:\) There is a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students

Cell contents grouped by No, Yes, Missing; First row: count, Next row: expected count

Rows: Dog | Columns: Cat
  No Yes All
BaNock 183 69 252
  176.02 75.98  
Yes 183 89 272
  189.98 82.02  
Missing 1 0  
All 366 158 524

Assumption: All expected counts are at least 5. The expected counts here are 176.02, 75.98, 189.98, and 82.02, so this assumption has been met.

2. Calculate an appropriate test statistic.
Chi-Square Test
  Chi-Square DF P-Value
Pearson 1.77 1 0.1833
Likelihood Ratio 1.77 1 0.1828

Since the assumption was met in step 1, we can use the Pearson chi-square test statistic.

\(Pearson\;\chi^2 = 1.77\)

3. Determine a p value associated with the test statistic.

\(p = 0.1833\)

4. Decide between the null and alternative hypotheses.

Our p value is greater than the standard 0.05 alpha level, so we fail to reject the null hypothesis.

5. State a "real world" conclusion.

There is not evidence of a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students.


11.3.2.2 - Video Example: Coffee and Tea (Summarized Data)

11.3.2.2 - Video Example: Coffee and Tea (Summarized Data)

11.3.3 - Relative Risk

11.3.3 - Relative Risk

A chi-square test of independence will give you information concerning whether or not a relationship between two categorical variables in the population is likely. As was the case with the single sample and two sample hypothesis tests that you learned earlier this semester, with a large sample size statistical power is high and the probability of rejecting the null hypothesis is high, even if the relationship is relatively weak. In addition to examining statistical significance by looking at the p value, we can also examine practical significance by computing the relative risk.

In Lesson 2 you learned that risk is often used to describe the probability of an event occurring. Risk can also be used to compare the probabilities in two different groups. First, we'll review risk, then you'll be introduced to the concept of relative risk.

The risk of an outcome can be expressed as a fraction or as the percent of a group that experiences the outcome.

Risk
\(Risk=\dfrac{n\;with\;outcome}{total\;n}\)

Examples of Risk

Asthma

60 out of 1000 teens have asthma. The risk is \(\frac{60}{1000}=.06\). This means that 6% of all teens experience asthma.

Flu

45 out of 100 children get the flu each year. The risk is \(\frac{45}{100}=.45\) or 45%

Relative Risk
Relative risk compares the risk of a particular outcome in two different groups.
Relative Risk
\(Relative\ Risk=\dfrac{Risk\ in\ Group\ 1}{Risk\ in\ Group\ 2}\)

Thus, relative risk gives the risk for group 1 as a multiple of the risk for group 2.

Examples of Relative Risk

Asthma

Suppose, 7% of teenage girls have asthma compared to 5% of teenage boys with asthma. Using the females as group 1, what is the relative risk for girls compared to boys?

  •  \(Relative\;risk=\dfrac{7 \%}{5 \%}=1.4\) 

The risk of asthma is 1.4 times as great for teenage girls as it is for teenage boys.

Flu

Suppose that the risk of a child getting the flu this year is .45 and the risk of an adult getting the flu this year is .10. What is the relative risk of children compared to adults?

  • \(Relative\;risk=\dfrac{.45}{.10}=4.5\)

Children are 4.5 times more likely than adults to get the flu this year.

Caution

Watch out for relative risk statistics where no baseline information is given about the actual risk. For instance, it doesn't mean much to say that beer drinkers have twice the risk of stomach cancer as non-drinkers unless we know the actual risks. The risk of stomach cancer might actually be very low, even for beer drinkers. For example, 2 in a million is twice the size of 1 in a million but is would still be a very low risk. This is known as the baseline with which other risks are compared.


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