Math Core

In Unit 2 you described association in a two-way table by comparing conditional distributions. But a sample can show differences just by chance. The chi-square tests for homogeneity and independence use the same $\chi^2$ statistic you met in the goodness-of-fit test to decide whether the pattern in a two-way table is strong enough to be real.

Two questions, one table

Two-way tables arise from two different study designs, and each design asks its own question.

Definition

Homogeneity versus independence

A chi-square test for homogeneity compares the distribution of one categorical variable across two or more populations or treatments. The data come from separate random samples (or from groups in a randomized experiment).
A chi-square test for independence checks whether two categorical variables are associated in one population. The data come from a single random sample, with each individual classified on both variables.

The quickest way to tell them apart is to ask: how many samples were taken? Several samples (or treatment groups), one variable measured: homogeneity. One sample, two variables measured: independence.

The hypotheses follow the design.

Test	$H_0$	$H_a$
Homogeneity	The distribution of the variable is the same for every population.	The distribution is not the same for all populations.
Independence	There is no association between the two variables (they are independent).	There is an association between the two variables.

The mechanics that follow are identical for both tests.

Expected counts in a two-way table

A school district randomly samples students from three high schools and asks how they usually get to school.

	Bus	Car	Walk/Bike	Total
School A	$42$	$38$	$20$	$100$
School B	$30$	$58$	$32$	$120$
School C	$48$	$44$	$28$	$120$
Total	$120$	$140$	$80$	$340$

If $H_0$ is true and the commute distribution is the same at all three schools, then each school should match the overall distribution. Overall, $\dfrac{120}{340}$ of students ride the bus, so School A, with $100$ students, should have about $100 \times \dfrac{120}{340} \approx 35.29$ bus riders. That reasoning gives a general formula.

Expected counts and degrees of freedom

For each cell of a two-way table,

\text{expected count} = \frac{(\text{row total})(\text{column total})}{\text{grand total}}.

The statistic is $\chi^2 = \sum \dfrac{(O - E)^2}{E}$ over every cell, and

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

Expected counts don't have to be whole numbers; keep a few decimal places. For the commute data, df $= (3-1)(3-1) = 4$ .

Conditions

The conditions match the goodness-of-fit test, adjusted for the design:

Random: separate random samples from each population, or groups formed by random assignment (homogeneity); a single random sample (independence).
10%: each sample is at most $10\%$ of its population when sampling without replacement.
Large counts: all expected counts are at least $5$ .

Worked example: Test for homogeneity: commuting at three schools

Use the commute table. Do the data give convincing evidence at $\alpha = 0.05$ that the distribution of commute method differs among the three schools?

State. $H_0$ : the distribution of commute method is the same at all three schools. $H_a$ : the distribution is not the same at all three schools.

Plan. Chi-square test for homogeneity. Random: separate random samples from each school. $10\%$ : assume each sample is less than $10\%$ of its school's enrollment (a high school with at least $1{,}200$ students). Large counts: the expected counts are

	Bus	Car	Walk/Bike
School A	$35.29$	$41.18$	$23.53$
School B	$42.35$	$49.41$	$28.24$
School C	$42.35$	$49.41$	$28.24$

All are at least $5$ .

Do. The nine contributions are

	Bus	Car	Walk/Bike
School A	$1.274$	$0.245$	$0.529$
School B	$3.603$	$1.493$	$0.502$
School C	$0.753$	$0.593$	$0.002$

Summing, $\chi^2 \approx 8.994$ with df $= 4$ . In the df $= 4$ row of the table, $8.994$ is between $7.779$ ( $0.10$ ) and $9.488$ ( $0.05$ ). Technology gives $P \approx 0.061$ .

Conclude. Because $0.061 > 0.05$ , fail to reject $H_0$ . There is not convincing evidence that the distribution of commute method differs among the three schools.

Common mistake

Use counts, never row or column percents, in the table you analyze. Also, don't mix up the two tests in your conclusion. A homogeneity conclusion talks about whether distributions differ among populations; an independence conclusion talks about whether two variables are associated. And neither test shows causation unless the data came from a randomized experiment.

Worked example: Test for independence: age and texting

A random sample of $180$ adults in a city was asked their age group and whether they prefer to reach friends by texting or calling.

	Text	Call	Total
18–29	$36$	$24$	$60$
30–49	$29$	$41$	$70$
50+	$15$	$35$	$50$
Total	$80$	$100$	$180$

Is there convincing evidence at $\alpha = 0.01$ of an association between age group and communication preference in this city? Which cells contribute most?

State. $H_0$ : there is no association between age group and preference. $H_a$ : there is an association.

Plan. Chi-square test for independence. One random sample of adults; $180$ is less than $10\%$ of the city's adults. Expected counts: $\dfrac{60 \cdot 80}{180} \approx 26.67$ , $33.33$ ; $31.11$ , $38.89$ ; $22.22$ , $27.78$ . All are at least $5$ .

Do.

	Text	Call
18–29	$3.267$	$2.613$
30–49	$0.143$	$0.115$
50+	$2.347$	$1.878$

$\chi^2 \approx 10.363$ with df $= (3-1)(2-1) = 2$ . In the table, $10.363$ is between $9.210$ ( $0.01$ ) and $10.597$ ( $0.005$ ). Technology: $P \approx 0.0056$ .

Conclude. Because $0.0056 < 0.01$ , reject $H_0$ . There is convincing evidence of an association between age group and communication preference among adults in this city.

The largest contributions come from the youngest group, who texted more than expected ( $36$ versus $26.67$ ), and the oldest group, who texted less than expected ( $15$ versus $22.22$ ).

A 2-by-2 table and the two-proportion z test

When the table has just two rows and two columns, a chi-square test for homogeneity asks the same question as a two-sided two-proportion $z$ test from Unit 6, and the two tests always agree: $\chi^2 = z^2$ , and the P-values are equal.

Worked example: Two ways to test the same 2-by-2 table

In a randomized experiment, $100$ students received a reminder text about a scholarship deadline and $100$ did not. Of those reminded, $45$ applied early; of those not reminded, $30$ applied early. Compare the chi-square statistic with the two-proportion $z$ statistic.

The table is

	Applied early	Did not	Total
Reminder	$45$	$55$	$100$
No reminder	$30$	$70$	$100$
Total	$75$	$125$	$200$

Expected counts: $\dfrac{100 \cdot 75}{200} = 37.5$ and $62.5$ in each row. So

\chi^2 = \frac{7.5^2}{37.5} + \frac{7.5^2}{62.5} + \frac{7.5^2}{37.5} + \frac{7.5^2}{62.5} = 1.5 + 0.9 + 1.5 + 0.9 = 4.8,

with df $= 1$ and $P \approx 0.028$ .

For the $z$ test, the pooled proportion is $\hat{p}_C = \dfrac{75}{200} = 0.375$ , and

z = \frac{0.45 - 0.30}{\sqrt{0.375(0.625)\left(\dfrac{1}{100} + \dfrac{1}{100}\right)}} \approx \frac{0.15}{0.06847} \approx 2.191.

Indeed $2.191^2 \approx 4.80$ , and the two-sided P-value is also $0.028$ . Because the reminders were randomly assigned, this result supports a cause-and-effect conclusion.

Tip

When a test is significant, compare the observed and expected counts in the cells with the largest contributions. That turns "there is an association" into a specific, useful statement about how the groups differ.

Practice

Problems 1, 4, 5 and 8 use this table. A random sample of $60$ ninth graders and a separate random sample of $100$ twelfth graders at a large school were asked what setting they prefer while studying.

	Silence	Music	Background noise	Total
9th grade	$30$	$20$	$10$	$60$
12th grade	$25$	$35$	$40$	$100$
Total	$55$	$55$	$50$	$160$

Practice 1

What is the expected count of ninth graders who prefer silence, assuming the distribution of study setting is the same for both grades?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

A researcher builds a two-way table with $4$ rows and $3$ columns. How many degrees of freedom does the chi-square test have?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

A polling group selects one random sample of $1{,}000$ adults and records each person's region of the country and their preferred type of news source. Which test is appropriate for determining whether region and news source are related?

Practice 4

For the study-setting table, calculate the chi-square statistic. Round to two decimal places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Continue the study-setting problem ( $\chi^2 \approx 13.38$ ). Find the P-value using technology. Round to four decimal places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Which pair of hypotheses is correct for a chi-square test for independence between pet ownership (dog, cat, none) and housing type (house, apartment) among adults in a city?

Practice 7

A two-way table with $3$ rows and $4$ columns gives $\chi^2 = 11.2$ in a chi-square test for independence. Which conclusion is correct at $\alpha = 0.05$ ?

Practice 8

The study-setting test is significant. Which statement best describes the cell with the largest contribution to $\chi^2$ ?