10.4 Test for Homogeneity
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
SLO 5
Identify appropriate statistical techniques and use technology-based statistical analysis to describe, interpret, and communicate results.
Learning Objectives
By the end of this section, you will be able to:
- explain what a test for homogeneity decides, and how that differs from a goodness-of-fit test and a test of independence;
- state the null and alternative hypotheses for a test for homogeneity;
- find the degrees of freedom for a test for homogeneity from the number of columns;
- carry a test for homogeneity through to a decision, and state precisely what the conclusion does and does not claim.
The goodness-of-fit test in Section 10.2 decided whether one population fits a given distribution — one that someone handed you in advance, as percentages or as counts. That is not always the question. Often there is no published distribution to test against, and what you actually want to know is whether two groups look the same as each other: do men and women college students live in the same mix of places, do families and singles drive the same mix of cars, did voter preferences shift after an earthquake. In each case the distribution is unknown, and it is the comparison that matters. The test for homogeneity is built for exactly that.
10.4.1 The Question This Test Answers
The goodness-of-fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution.
A test for homogeneity decides whether two populations have the same distribution across a categorical variable, without requiring that distribution to be known in advance.
To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence. The expected counts come from the margins exactly as in Definition 10.3.2, and the statistic sums \(\frac{(O-E)^{2}}{E}\) over every cell.
Sections 10.3 and 10.4 compute the identical number from the identical table. What separates them is where the data came from, and therefore what the answer means. In a test of independence you sample ONE population and record two characteristics of each person — 839 volunteers, each with a type and an hour count. In a test for homogeneity you sample TWO populations separately and record one characteristic — 250 men and 300 women, each with a living arrangement. The sample sizes in a homogeneity test are chosen by the researcher rather than produced by the data, which is why the question shifts from "are these two traits related?" to "do these two groups match?"
Definition 10.4.1 — Test for Homogeneity: Are the two traits related?.
A researcher surveys 200 randomly chosen night-shift workers and 200 randomly chosen day-shift workers, asking each one which of four commute methods they use. A second researcher surveys 400 randomly chosen city residents, recording each person's commute method and whether they own a car. Which study calls for a test for homogeneity and which calls for a test of independence? Explain how you can tell.
Solution — telling the two designs apart
Step 1 — Look at how many populations were sampled in the first study. Two groups were sampled separately, night-shift and day-shift workers, and the researcher fixed both sample sizes at 200.
Step 2 — Look at how many characteristics were recorded in the first study. Just one: commute method.
Step 3 — Classify the first study. Two populations, one categorical variable, sample sizes chosen by the researcher. This is a test for homogeneity.
Step 4 — Apply the same two questions to the second study. One population was sampled — city residents — and two characteristics were recorded for each person, commute method and car ownership.
Step 5 — Classify the second study. One population, two characteristics. This is a test of independence.
Answer: The night-shift/day-shift study is a test for homogeneity; the city-residents study is a test of independence. The giveaway is how many populations were sampled and how many characteristics were recorded on each person.
10.4.2 The Hypotheses
Both hypotheses are stated in words, and both are about whole distributions rather than about any individual category:
- \(H_{0}\): The distributions of the two populations are the same.
- \(H_{a}\): The distributions of the two populations are not the same.
Notice there is no direction and no named category in either statement. A homogeneity test never asks whether one group prefers a particular response more than the other — only whether the overall patterns match.
A school district compares the grade distributions (A, B, C, D, F) of two high schools. Write the null and alternative hypotheses for a test for homogeneity, and explain why "\(H_{a}\): School A gives more A's than School B" would be wrong.
Solution — writing hypotheses about distributions
Step 1 — Write the null hypothesis about the whole distribution. \(H_{0}\): The distribution of grades at School A is the same as the distribution of grades at School B.
Step 2 — Write the alternative as its plain negation. \(H_{a}\): The distribution of grades at School A is not the same as the distribution of grades at School B.
Step 3 — Say why the proposed alternative fails. "School A gives more A's" names one category and gives a direction. The chi-square statistic sums a contribution from every cell into a single number, so the test cannot attribute a result to one column or say which way it went.
Answer: \(H_{0}\): the two grade distributions are the same; \(H_{a}\): they are not the same. The proposed alternative is wrong because it is directional and about a single category, and this test measures the whole table at once.
10.4.3 The Test Statistic and Its Degrees of Freedom
Test statistic. Use a \(\chi^{2}\) test statistic. It is computed in the same way as the test for independence — build each expected count from its row total, its column total and the sample size, then sum \(\frac{(O-E)^{2}}{E}\) across every cell.
This looks different from the independence formula, but it is the same formula wearing different clothes. A homogeneity test compares two populations, so the table always has exactly two rows, and
$$ (\text{rows} - 1)(\text{columns} - 1) = (2 - 1)(\text{columns} - 1) = \text{columns} - 1 $$The rows drop out because there are only ever two of them. Count the response categories, subtract one, and you have the degrees of freedom.
Definition 10.4.2 — Degrees of Freedom for a Test for Homogeneity: A homogeneity table always has exactly two rows — one per population — so its degrees of freedom are just columns − 1.
A test for homogeneity compares two groups across seven response categories. Find the degrees of freedom two ways: with the homogeneity formula and with the independence formula. Then state the degrees of freedom if one response category were merged into another.
Solution — one rule, two spellings
Step 1 — Use the homogeneity formula.
$$ df = \text{number of columns} - 1 = 7 - 1 = 6 $$Step 2 — Use the independence formula. The table has two rows, one per population, and seven columns.
$$ df = (\text{rows} - 1)(\text{columns} - 1) = (2 - 1)(7 - 1) = (1)(6) = 6 $$Step 3 — Confirm they agree. Both give 6, as they must, because multiplying by \((2-1) = 1\) changes nothing.
Step 4 — Merge one category. Merging two response categories leaves six columns, so \(df = 6 - 1 = 5\).
Answer: \(df = 6\) by either formula; after merging one category, \(df = 5\).
10.4.4 Requirements and Common Uses
Requirements. All expected values in the table must be greater than or equal to five.
This is the third time the condition has appeared, and it is checked the same way every time — on the expected counts, cell by cell, before computing anything else. A small observed count is not by itself a problem; a small expected count is. When a category fails, merge it into an adjacent one, which costs a column and therefore one degree of freedom.
Common uses. Comparing two populations: men versus women, before versus after, east versus west. The variable is categorical with more than two possible response values.
Two groups of 60 people each are surveyed about their preferred news source, with five categories. One category has an observed count of 3 in the first group and 4 in the second. Explain whether the test can proceed, and what you would need to check first.
Solution — observed counts are not what the condition tests
Step 1 — Recall exactly what the condition constrains. The requirement is that every expected count be at least five. The observed counts of 3 and 4 are not what is checked.
Step 2 — Build the expected counts for that category. Its column total is \(3 + 4 = 7\), each row total is 60, and the total surveyed is 120. So each expected count in that column is
$$ E = \frac{(60)(7)}{120} = 3.5 $$Step 3 — Compare with the requirement. Both expected counts in that column are 3.5, which is below five, so the condition fails.
Step 4 — State the repair. Merge that news source into an adjacent category. Four columns then remain, so the degrees of freedom drop from \(5 - 1 = 4\) to \(4 - 1 = 3\).
Answer: The test cannot proceed as it stands. The expected counts in that column are 3.5, below five, so the category must be merged, leaving \(df = 3\).
10.4.5 Comparing Two Groups
Do families and singles have the same distribution of cars? Suppose that 100 randomly selected families and 200 randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in Table 10.4.1. Test at a level of significance of 0.05.
| Sport | Sedan | Hatchback | Truck | Van/SUV | |
|---|---|---|---|---|---|
| Family | 5 | 15 | 35 | 17 | 28 |
| Single | 45 | 65 | 37 | 46 | 7 |
Solution — a homogeneity test that rejects decisively
Step 1 — State the hypotheses.
\(H_{0}\): The distribution of cars for families is the same as the distribution of cars for singles.
\(H_{a}\): The distribution of cars for families is not the same as the distribution of cars for singles.
Step 2 — Find the degrees of freedom. There are five response categories.
$$ df = \text{number of columns} - 1 = 5 - 1 = 4 $$The distribution for the test is \(\chi_{4}^{2}\).
Step 3 — Check the condition. The row totals are 100 and 200; the column totals are 50, 80, 72, 63 and 35. The smallest expected count is
$$ E = \frac{(100)(35)}{300} \approx 11.67 $$which clears five, so the test may proceed. The observed count of 5 sport cars among families is not what the condition looks at.
Step 4 — Calculate the test statistic. Summing \(\frac{(O-E)^{2}}{E}\) over all ten cells gives \(\chi^{2} \approx 62.91\).
Step 5 — Find the p-value. \(P(\chi^{2} > 62.91)\) with \(df = 4\) is about \(7 \times 10^{-13}\), effectively zero.
Step 6 — Compare and decide. With \(\alpha = 0.05\) and a p-value near zero, \(\alpha >\) p-value, so reject \(H_{0}\).
Answer: \(\chi^{2} \approx 62.91\) with \(df = 4\) and a p-value near zero. At the 5% level of significance there is sufficient evidence to conclude that families and singles do not have the same distribution of cars.
Do men and women college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected men college students and 300 randomly selected women college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in Table 10.4.2.
| Dormitory | Apartment | With Parents | Other | |
|---|---|---|---|---|
| Men | 72 | 84 | 49 | 45 |
| Women | 91 | 86 | 88 | 35 |
Solution — a full test for homogeneity
Step 1 — State the hypotheses.
\(H_{0}\): The distribution of living arrangements for men college students is the same as the distribution of living arrangements for women college students.
\(H_{a}\): The distribution of living arrangements for men college students is not the same as the distribution of living arrangements for women college students.
Step 2 — Find the degrees of freedom.
$$ df = \text{number of columns} - 1 = 4 - 1 = 3 $$Distribution for the test: \(\chi_{3}^{2}\).
Step 3 — Check the condition. The smallest expected count in the table is about 36.4, comfortably above five.
Step 4 — Calculate the test statistic. Using a calculator or computer, \(\chi^{2} = 10.1287\).
Step 5 — Write the probability statement. \(p\text{-value} = P(\chi^{2} > 10.1287) = 0.0175\).
Step 6 — Compare \(\alpha\) and the p-value. Since no \(\alpha\) is given, assume \(\alpha = 0.05\). The p-value is 0.0175, so \(\alpha >\) p-value.
Step 7 — Make a decision. Since \(\alpha >\) p-value, reject \(H_{0}\). This means that the distributions are not the same.
Answer: \(\chi^{2} = 10.1287\) with \(df = 3\) and a p-value of 0.0175. At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for men and women college students are not the same.
Using the TI-83, 83+, 84, 84+ Calculator
Press the MATRIX key and arrow over to EDIT. Press 1: [A]. Press 2 ENTER 4 ENTER to size the matrix for Table 10.4.2, then enter the table values by row, pressing ENTER after each. Press 2nd QUIT. Press STAT, arrow over to TESTS, arrow down to C: χ²-TEST and press ENTER. You should see Observed: [A] and Expected: [B]. Arrow down to Calculate and press ENTER. The test statistic is 10.1287 and the p-value is 0.0175. Run the procedure a second time, choosing Draw instead of Calculate, to see the shaded right tail.
Notice that the conclusion is only that the distributions are not the same. We cannot use the test for homogeneity to draw any conclusions about how they differ. Looking at Table 10.4.2 you may well suspect the difference sits in the "with parents" column, where women outnumber men 88 to 49 — but that is a hypothesis for a further study, not something this test established. A single p-value about a whole table cannot be spent on a claim about one column of it.
10.4.6 Before and After as Two Populations
The two "populations" being compared need not be two different groups of people. The same population measured at two different times works the same way.
Ivy League schools receive many applications, but only some can be accepted. At the schools listed in Table 10.4.3, two types of applications are accepted: regular and early decision. We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, and draw a conclusion about the test of homogeneity.
| Application Type Accepted | Brown | Columbia | Cornell | Dartmouth | Penn | Yale |
|---|---|---|---|---|---|---|
| Regular | 2,115 | 1,792 | 5,306 | 1,734 | 2,685 | 1,245 |
| Early Decision | 577 | 627 | 1,228 | 444 | 1,195 | 761 |
Solution — comparing two acceptance distributions across six schools
Step 1 — State the hypotheses.
\(H_{0}\): The distribution of regular applications accepted is the same as the distribution of early decision applications accepted.
\(H_{a}\): The distribution of regular applications accepted is not the same as the distribution of early decision applications accepted.
Step 2 — Find the degrees of freedom. There are six schools, so six columns.
$$ df = \text{number of columns} - 1 = 6 - 1 = 5 $$Step 3 — Check the condition. The smallest expected count is about 492, far above five.
Step 4 — Calculate the test statistic. Summing over all twelve cells gives \(\chi^{2} \approx 430.06\).
Step 5 — Find the p-value. \(P(\chi^{2} > 430.06)\) with \(df = 5\) is far smaller than any significance level in use — on the order of \(10^{-91}\).
Step 6 — Decide and conclude. Since \(\alpha >\) p-value, reject \(H_{0}\).
Answer: \(\chi^{2} \approx 430.06\) with \(df = 5\) and a p-value effectively zero. There is sufficient evidence to conclude that the distribution of regular acceptances across these six schools is not the same as the distribution of early decision acceptances.
Both before and after a recent earthquake, surveys were conducted asking voters which of the three candidates they planned on voting for in the upcoming city council election. Has there been a change since the earthquake? Use a level of significance of 0.05. Table 10.4.4 shows the results of the survey.
| Perez | Chung | Stevens | |
|---|---|---|---|
| Before | 167 | 128 | 135 |
| After | 214 | 197 | 225 |
Solution — a test that does not reject
Step 1 — State the hypotheses.
\(H_{0}\): The distribution of voter preferences was the same before and after the earthquake.
\(H_{a}\): The distribution of voter preferences was not the same before and after the earthquake.
Step 2 — Find the degrees of freedom.
$$ df = \text{number of columns} - 1 = 3 - 1 = 2 $$Distribution for the test: \(\chi_{2}^{2}\).
Step 3 — Check the condition. The smallest expected count is about 131, well above five.
Step 4 — Calculate the test statistic. Using a calculator or computer, \(\chi^{2} = 3.2603\).
Step 5 — Write the probability statement. \(p\text{-value} = P(\chi^{2} > 3.2603) = 0.1959\).
Step 6 — Compare \(\alpha\) and the p-value. Here \(\alpha = 0.05\) and the p-value is 0.1959, so \(\alpha <\) p-value.
Step 7 — Make a decision. Since \(\alpha <\) p-value, do not reject \(H_{0}\).
Answer: \(\chi^{2} = 3.2603\) with \(df = 2\) and a p-value of 0.1959. At a 5% level of significance, from the data, there is insufficient evidence to conclude that the distribution of voter preferences was not the same before and after the earthquake.
Using the TI-83, 83+, 84, 84+ Calculator
For Table 10.4.4, press MATRIX, arrow to EDIT, press 1: [A], then 2 ENTER 3 ENTER to size it. Enter the table values by row, pressing ENTER after each, then press 2nd QUIT. Press STAT, arrow to TESTS, arrow down to C: χ²-TEST and press ENTER. Arrow down to Calculate and press ENTER. The test statistic is 3.2603 and the p-value is 0.1959. Run it again with Draw to see the graph.
The "after" survey collected 636 responses to the "before" survey's 430, and every candidate's raw count went up. None of that moves the test. Homogeneity compares the shares within each row, not the totals, so a survey half again as large but distributed the same way produces a small statistic. That property is what lets you compare two groups of different sizes at all — and it is why a rising raw count is never by itself evidence of a shift.
Problem Set 10.4
Problem 1. State the null and alternative hypotheses for a test for homogeneity, in words.
Solution
Step 1 — State the null hypothesis. \(H_{0}\): The distributions of the two populations are the same.
Step 2 — State the alternative hypothesis. \(H_{a}\): The distributions of the two populations are not the same.
Step 3 — Note what is absent from both statements. Neither hypothesis names a particular category or gives a direction; both are statements about the entire distribution across all response categories at once.
Answer: \(H_{0}\): the two populations have the same distribution; \(H_{a}\): the two populations do not have the same distribution — stated in words, with no direction and no named category.
Problem 2. Explain the difference between the question a goodness-of-fit test answers and the question a test for homogeneity answers.
Solution
Step 1 — Describe the goodness-of-fit question. A goodness-of-fit test decides whether ONE population fits a distribution that is given in advance, as percentages or counts handed to you before the data were collected.
Step 2 — Describe the homogeneity question. A test for homogeneity decides whether TWO populations have the same distribution as each other, when that distribution is not known in advance.
Step 3 — Highlight the key contrast. Goodness-of-fit compares a sample to an external, published standard; homogeneity compares two samples to each other, with no external standard at all.
Answer: Goodness-of-fit asks "does one population match this given distribution?" while homogeneity asks "do two populations share the same (unknown) distribution?"
Problem 3. A test for homogeneity compares two groups across 7 response categories. What is the number of degrees of freedom?
Solution
Step 1 — Apply the homogeneity formula.
$$ df = \text{number of columns} - 1 = 7 - 1 = 6 $$Step 2 — Confirm with the independence formula. A homogeneity table has exactly two rows, one per population:
$$ df = (\text{rows} - 1)(\text{columns} - 1) = (2 - 1)(7 - 1) = 6 $$Answer: \(df = 6\).
Problem 4. Show why \(df = \text{number of columns} - 1\) is the same rule as \((\text{rows} - 1)(\text{columns} - 1)\) for this test.
Solution
Step 1 — Recall the structure of a homogeneity table. A test for homogeneity always compares exactly two populations, so the table has exactly two rows: \(\text{rows} = 2\).
Step 2 — Substitute into the independence formula.
$$ df = (\text{rows} - 1)(\text{columns} - 1) = (2 - 1)(\text{columns} - 1) $$Step 3 — Simplify. Since \(2 - 1 = 1\),
$$ df = (1)(\text{columns} - 1) = \text{columns} - 1 $$Step 4 — Interpret. The rows factor drops out because there are only ever two of them, so counting the response categories and subtracting one gives the same answer either way.
Answer: Because a homogeneity table always has two rows, \((\text{rows}-1)(\text{columns}-1) = (2-1)(\text{columns}-1) = \text{columns} - 1\), making the two rules identical.
Problem 5. A researcher surveys 150 city residents and 150 rural residents about their primary mode of transportation, with four categories. State the degrees of freedom.
Solution
Step 1 — Count the columns. There are four transportation categories, so four columns.
Step 2 — Apply the degrees-of-freedom rule.
$$ df = \text{number of columns} - 1 = 4 - 1 = 3 $$Answer: \(df = 3\).
Problem 6. A test for homogeneity gives \(\chi^{2} = 8.4\) with \(df = 3\) and a p-value of 0.0384. State the decision at the 5% significance level and write the conclusion in context for a comparison of two schools' grade distributions.
Solution
Step 1 — Compare \(\alpha\) with the p-value. Here \(\alpha = 0.05\) and the p-value is 0.0384, so \(\alpha >\) p-value.
Step 2 — Make the decision. Since \(\alpha >\) p-value, reject \(H_{0}\).
Step 3 — Write the conclusion in context. At the 5% level of significance, there is sufficient evidence to conclude that the grade distributions of the two schools are not the same.
Answer: Reject \(H_{0}\); there is sufficient evidence at the 5% level that the two schools' grade distributions differ.
Problem 7. A test for homogeneity rejects \(H_{0}\). A student concludes that the first group prefers category B more than the second group does. Explain what is wrong with that conclusion.
Solution
Step 1 — Recall what the chi-square statistic measures. The statistic sums \(\frac{(O-E)^{2}}{E}\) over every cell of the whole table, collapsing everything into a single number about the entire distributions.
Step 2 — Explain why no single category can be singled out. Because the contributions from all cells are pooled, a significant result tells you the distributions differ somewhere, but it cannot say which column caused it or in which direction.
Step 3 — State what would be needed for such a claim. Concluding that one group prefers category B more than the other requires a separate analysis — for example, comparing just those two proportions in a follow-up study.
Answer: The student's conclusion overreaches: rejecting \(H_{0}\) shows only that the overall distributions differ, not which category differs or how, since the test pools information from all cells into one number.
Problem 8. A 2-by-5 table has an expected count of 4.2 in one cell. State what must be done before the test is run and what happens to the degrees of freedom.
Solution
Step 1 — Identify the problem. The requirement is that every expected count be at least five, and 4.2 falls below five, so the condition fails.
Step 2 — Apply the repair. Merge the failing column into an adjacent category before running the test.
Step 3 — Track the effect on degrees of freedom. Merging reduces the number of columns by one, so the degrees of freedom drop by one, from \(5 - 1 = 4\) to \(4 - 1 = 3\).
Answer: Merge the category with expected count 4.2 into an adjacent one first; the degrees of freedom decrease by one, from 4 to 3.
Problem 9. Explain why a test for homogeneity can compare two groups of very different sizes.
Solution
Step 1 — Recall what the test actually compares. The test compares the shares (proportions) within each row, not the raw totals.
Step 2 — See why row size cancels out. Each expected count is computed as \(E = \frac{(\text{row total})(\text{column total})}{n}\), so a larger group simply scales its row proportionally; if the within-row pattern is identical, the statistic stays small regardless of how unequal the group sizes are.
Step 3 — Connect to the design. In a homogeneity study the researcher chooses the sample sizes anyway, so groups of very different sizes are normal, and the comparison remains valid because only the distributions matter.
Answer: Because the test compares the proportions within each population rather than the raw counts, group sizes need not be equal — a larger group contributes proportionally scaled counts without biasing the result.
Problem 10. A study samples one group of 500 shoppers and records both their preferred payment method and their age bracket. Name the correct test and explain why it is not a test for homogeneity.
Solution
Step 1 — Examine how many populations were sampled. Only one group was sampled: 500 shoppers.
Step 2 — Examine how many characteristics were recorded. Two characteristics were recorded on each person: payment method and age bracket.
Step 3 — Match the design to the correct test. One population plus two characteristics recorded per individual is the signature of a test of independence.
Step 4 — Say why it fails the homogeneity definition. A test for homogeneity requires sampling two populations separately, with sample sizes chosen by the researcher, and recording one characteristic on each person — none of which holds here.
Answer: This calls for a test of independence, because one population was sampled and two characteristics were recorded per person; a homogeneity test instead requires two separately sampled populations with one characteristic each.
Key Terms
test for homogeneity — a chi-square test that decides whether two populations have the same distribution across a categorical variable, when that distribution is not known in advance.
homogeneous — having the same distribution; the null hypothesis of this test is that the two populations are homogeneous.
degrees of freedom for homogeneity — \(\text{number of columns} - 1\); the same as \((\text{rows}-1)(\text{columns}-1)\), because a homogeneity table always has exactly two rows.
two populations — the two groups being compared, sampled separately and with sample sizes chosen by the researcher; this is what distinguishes a homogeneity test from a test of independence, which samples one population and records two characteristics.
expected-count condition — the requirement that every expected value be at least five; it constrains the expected counts, never the observed ones, and a failing category is merged into an adjacent one.