Lesson 25: Chi-Square Test for Independence

(a) Warm-Up

In Lesson 24 you ran a chi-square test for homogeneity: you took several separate samples — one from each of several populations — and asked whether one categorical variable was distributed the same way across all of them.

Now picture a different study. A researcher takes one random sample of 350 college students and, for each student, records two things: how much caffeine they consume (None, Moderate, High) and the quality of their sleep (Good, Poor). Nobody was sorted into groups in advance — every student is just one dot, tagged on two variables at once.

The question changes too. It is no longer "are these groups the same?" It is "are these two variables associated?" Does caffeine level go with sleep quality, or are they unrelated?

Here is the beautiful part: the arithmetic is identical to homogeneity. Same expected counts, same χ² formula, same degrees of freedom, same calculator routine. What changes is the design, the hypotheses, and the conclusion. Let's see exactly how.

(b) Core Concept

When this test applies

Use the chi-square (χ²) test for independence when:

• You have ONE random sample from a single population, and
• Each individual is classified on TWO categorical variables, and
• You want to know whether those two variables are associated (related) in the population.

Contrast that with last lesson: homogeneity used several samples (several populations) and one categorical variable. Same math, different design — we'll nail this down in a comparison box below.

The data live in a two-way table (also called a contingency table). Here is our caffeine-and-sleep sample of 350 students:

Notice there is just one grand total (350) — one sample, split two ways. That single-sample structure is the fingerprint of an independence test.

The hypotheses

The hypotheses are stated in terms of association, not proportions:

• H₀: Caffeine consumption and sleep quality are independent (no association) in the population of college students.
• Hₐ: Caffeine consumption and sleep quality are associated in the population of college students.

There is always exactly one χ² alternative, and it is always two-sided in spirit — "associated" — never "greater than."

Expected counts: what we'd see if H₀ were true

If the two variables were truly independent, the proportion with poor sleep should be the same regardless of caffeine level. We turn that idea into expected counts with the same formula from Lesson 24:

E = (row total × column total) / grand total

Let's fully work the cell for No caffeine + Good sleep:

E = (row total × column total) / grand total
E = (100 × 180) / 350
E = 18000 / 350
E = 51.43

So if caffeine and sleep were unrelated, we'd expect about 51.43 of these students to be no-caffeine/good-sleep. We actually observed 60. Filling in every cell:

The χ² statistic

The test statistic measures how far the observed counts (O) stray from the expected counts (E):

χ² = Σ (O − E)² / E

Compute the contribution of a few cells so you see the machinery:

No caffeine / Good:   (60 − 51.43)² / 51.43 = (8.57)²/51.43  = 1.43
High caffeine / Poor: (70 − 48.57)² / 48.57 = (21.43)²/48.57 = 9.45

That High/Poor cell contributes a huge 9.45 — heavy caffeine users had far more poor sleep than independence predicts. Adding all six contributions:

1.43 + 1.51 + 2.14 + 2.27 + 8.93 + 9.45 = 25.74

χ² = 25.74

Degrees of freedom and the p-value

Degrees of freedom depend only on the shape of the table:

df = (r − 1)(c − 1) = (3 − 1)(2 − 1) = 2

The p-value is the upper-tail area of the χ² distribution — large χ² means strong evidence, so we always look to the right:

TI-84: χ²cdf(25.74, 1E99, 2) = 0.0000026   (≈ 2.6 × 10⁻⁶)

Interpret in context

p ≈ 0.0000026 is far below α = 0.05, so we reject H₀. There is very strong evidence of an association between caffeine consumption and sleep quality among college students.

Crucial wording: we found an association, not a cause. This was observational data — we cannot conclude that caffeine causes poor sleep. Poor sleepers might reach for more caffeine, or a lurking variable (stress, late-night study) could drive both.

Independence vs. Homogeneity — the one box to memorize

SAME ARITHMETIC. DIFFERENT DESIGN.

| | Homogeneity (L24) | Independence (L25) |

|---|---|---|

| How many samples? | Several samples / several populations | ONE sample / one population |

| How many variables? | ONE categorical variable | TWO categorical variables |

| The question | Is the distribution the same across groups? | Are the two variables associated? |

| H₀ | Distributions are the same across populations | The two variables are independent |

| Hₐ | At least one distribution differs | The two variables are associated |

| Margins fixed? | Row (or column) totals fixed by the researcher | Only the grand total is fixed |

| Expected counts | (row × col)/grand | (row × col)/grand ← identical |

| Statistic / df / p | χ²=Σ(O−E)²/E, (r−1)(c−1), χ²cdf | identical |

Decision rule: Did the researcher take one sample and measure two things (independence), or take several separate samples and measure one thing (homogeneity)?

On the TI-84 both tests use the same χ²-Test command — the calculator does not know or care which design you have. The distinction lives entirely in your hypotheses and conclusion.

Conditions (check every time)

1. Random: the data come from one random sample (or randomized experiment).
2. 10% condition: if sampling without replacement, n ≤ 10% of the population (so individuals are roughly independent).
3. Large counts: all expected counts ≥ 5. (Check the expected counts in matrix [B], never the observed counts.)

(c) Worked Examples

Example 1 — Expected counts only (warm-up to the full test)

Problem. A campus survey randomly samples 250 students and classifies each by pet ownership (Dog, Cat, None) and whether they exercise regularly (Yes, No). Find the expected count for Dog owners who exercise, and verify all conditions for the large-counts requirement.

Solution.

E(Dog, Yes) = (100 × 100) / 250 = 10000/250 = 40

Full expected-count table:

Interpretation. Every expected count (40, 60, 20, 30, 40, 60) is ≥ 5, so the Large Counts condition is satisfied. We could proceed to compute χ² with df = (3−1)(2−1) = 2.

Example 2 — Full PANIC independence test (AP-style)

Problem. A media company takes a random sample of 600 social-media users and records each user's preferred platform (Instagram, TikTok, YouTube) and age group (Teen, Adult). Is there an association between platform preference and age group? Test at α = 0.05.

P — Parameter / hypotheses.

• H₀: Platform preference and age group are independent in the population of social-media users.
• Hₐ: Platform preference and age group are associated in the population of social-media users.

A — Assumptions / conditions.

• Random: one random sample of 600 users. ✓
• 10%: 600 is clearly less than 10% of all social-media users. ✓
• Large counts: expected counts (below) are all 116.67 or 83.33 — all ≥ 5. ✓

Expected counts E = (row × col)/grand, e.g. E(Instagram, Teen) = (200×350)/600 = 116.67:

N — Name the procedure. Chi-square (χ²) test for independence.

I — Inference (compute).

A couple of contributions:

TikTok / Teen:  (150 − 116.67)² / 116.67 = (33.33)²/116.67 = 9.52
YouTube/ Adult: (120 −  83.33)² /  83.33 = (36.67)²/ 83.33 = 16.13

Summing all six contributions:

χ² = 0.10 + 0.13 + 9.52 + 13.33 + 11.52 + 16.13 = 50.74
df = (3 − 1)(2 − 1) = 2
p-value = χ²cdf(50.74, 1E99, 2) ≈ 9.6 × 10⁻¹²

TI-84: [A] = 3×2 matrix of counts → STAT → TESTS → χ²-Test
Output: χ² = 50.743, p = 9.58E-12, df = 2   (expected counts stored in [B])

C — Conclusion in context. Because p ≈ 9.6 × 10⁻¹² < 0.05, we reject H₀. There is very strong evidence of an association between preferred platform and age group among social-media users. (Because this is observational, we do not claim age causes platform choice.)

Example 3 — Checking conditions and reading cell contributions

Problem. A small study randomly samples 80 gamers and classifies each by favorite genre (Action, Strategy) and primary device (Console, PC). Confirm the conditions, find df, and identify which cell contributes most to χ².

Solution. Expected counts (e.g. E(Action,Console) = (40×35)/80 = 17.5):

All expected counts (17.5, 22.5, 17.5, 22.5) ≥ 5 ✓; one random sample ✓; 80 ≤ 10% of all gamers ✓.

Cell contributions (O−E)²/E:

Action/Console:   (25−17.5)²/17.5 = 3.21
Action/PC:        (15−22.5)²/22.5 = 2.50
Strategy/Console: (10−17.5)²/17.5 = 3.21
Strategy/PC:      (30−22.5)²/22.5 = 2.50
χ² = 3.21 + 2.50 + 3.21 + 2.50 = 11.43
df = (2−1)(2−1) = 1
p-value = χ²cdf(11.43, 1E99, 1) ≈ 0.0007

Interpretation. The two Console cells (each 3.21) contribute most, signaling that the genre–device mismatch is biggest there: Action players gravitate to Console more, Strategy players to PC, more than independence would predict. With p ≈ 0.0007 < 0.05, reject H₀ — genre and device are associated.

Example 4 — Classify the design: independence or homogeneity?

Problem. For each scenario, decide whether a χ² test for independence or homogeneity is appropriate.

1. A researcher randomly samples 500 adults and records each person's blood type and whether they donate blood.
2. A researcher takes separate random samples of 200 freshmen, 200 sophomores, and 200 juniors and records each student's preferred study location.
3. One random sample of 300 voters is taken; each is classified by political party and opinion on a ballot measure.

Solution.

1. Independence — one sample (500 adults), two variables (blood type, donor status). Only the grand total was fixed.
2. Homogeneity — three separate samples (one variable, study location, compared across three populations). The researcher fixed each group's size at 200.
3. Independence — one sample (300 voters), two variables (party, opinion).

Interpretation. The tell is always: one sample with two variables → independence; several preset samples with one variable → homogeneity. The χ² arithmetic is the same either way.

(d) Common Mistakes

1. Confusing the independence setup with homogeneity.

Wrong: writing H₀ as "the distributions are the same across groups" for a one-sample, two-variable study.

Right: an independence test has one sample classified on two variables. State H₀ as "the two variables are independent (not associated) in the population." If you see several preset sample sizes, it's homogeneity instead.

2. Using percentages or proportions instead of raw counts.

Wrong: entering 0.60, 0.40, … into matrix [A], or computing expected counts from percentages.

Right: χ² procedures require raw frequency counts. Convert any percentages back to counts first; expected counts ≥ 5 is a count condition.

3. Getting degrees of freedom wrong.

Wrong: df = n − 1, or df = (number of cells) − 1.

Right: df = (r − 1)(c − 1) using the number of rows and columns of the table, not the sample size. A 3×2 table gives df = 2, every time, regardless of n.

4. Claiming causation from an association.

Wrong: "Caffeine causes poor sleep." A significant χ² in observational data shows the variables are associated, not that one causes the other.

Right: "There is an association between caffeine consumption and sleep quality." Reserve causal language for randomized experiments.

5. Checking the observed counts instead of the expected counts.

Wrong: declaring the Large Counts condition met because all observed counts ≥ 5.

Right: the condition is on expected counts (matrix [B] on the TI-84). Observed counts can be small as long as every expected count is ≥ 5.

(e) Practice Problems

MC = 4 choices. Answers and full work in section (g).

1 (MC). A χ² test for independence requires:

A. Several independent random samples, one categorical variable

B. One random sample, two categorical variables

C. Two quantitative variables from one sample

D. A randomized experiment with two treatment groups

2 (MC). A two-way table has 4 rows and 3 columns. The degrees of freedom are:

A. 12 B. 11 C. 7 D. 6

3 (MC). In a 2×2 table, a row total is 80, a column total is 120, and the grand total is 200. The expected count for that cell is:

A. 32 B. 40 C. 48 D. 60

4 (MC). For a χ² test for independence, the p-value is found with:

A. χ²cdf(0, χ², df)

B. χ²pdf(χ², df)

C. χ²cdf(χ², 1E99, df)

D. 2 × χ²cdf(χ², 1E99, df)

5 (MC). A χ² test for independence on observational data yields p = 0.003. The best conclusion is:

A. One variable causes a change in the other

B. The variables are associated; no causal claim is justified

C. The variables are independent

D. The sample was too small to conclude anything

6 (MC). Which is the correct H₀ for a χ² test for independence?

A. The proportions are equal across all populations

B. The two variables are independent in the population

C. The two variables are associated in the population

D. The mean is the same in each group

7 (MC). Which study calls for a χ² test for independence (not homogeneity)?

A. Separate samples of 100 men and 100 women, each asked their favorite sport

B. One random sample of 250 people, each classified by region and vote choice

C. Comparing the mean income of three cities

D. A single proportion compared to 0.5

8 (MC). All of a table's observed counts exceed 5, but one expected count is 3.8. The Large Counts condition is:

A. Met, because observed counts matter

B. Met, because 3.8 rounds to 4

C. Not met, because all expected counts must be ≥ 5

D. Irrelevant for χ² tests

9 (Short answer). A random sample of 400 employees is classified by department (Sales, Engineering, Support) and whether they work remotely (Yes, No). Researchers want to know if department and remote-work status are associated. State H₀, Hₐ, the name of the procedure, and the degrees of freedom.

10 (Short answer · in context). A χ² test for independence on a random sample of patients gives χ² = 14.2, df = 3, p = 0.0027. The two variables are diet type and presence of a certain health condition. Write a complete conclusion in context at α = 0.05, and explain why a doctor should not tell patients that one diet "causes" the condition.

11 (Short answer · in context). A school randomly samples 300 students and records each student's grade level (9, 10, 11, 12) and primary mode of getting to school (Bus, Car, Walk). For the cell "Grade 9 / Bus," the row total is 90, the column total is 100, and the grand total is 300. Compute the expected count and state what it represents.

12 (Short answer). Explain in one or two sentences how you decide, just from a study's design, whether to run a test for independence or a test for homogeneity.

(f) FRQ Practice — 10 Points

Statistical Practices 3 (Analyze Data) & 4 (Interpret Results)

A public-health researcher takes a single random sample of 350 adults from a large city. Each adult is classified by whether they regularly drink coffee (Yes, No) and how often they report headaches (Rare, Occasional, Frequent). The results:

Do these data provide convincing evidence of an association between coffee drinking and headache frequency among adults in this city? Conduct an appropriate test at the α = 0.05 level. (10 points)

Model Response

P — Parameter / Hypotheses (in context).

• H₀: Coffee-drinking status and headache frequency are independent (not associated) among adults in this city.
• Hₐ: Coffee-drinking status and headache frequency are associated among adults in this city.

A — Assumptions / Conditions.

• Random: the 350 adults are one random sample. ✓
• 10%: 350 adults is less than 10% of all adults in a "large city." ✓
• Large Counts: all expected counts (below) are ≥ 5. ✓

Expected counts, E = (row × col)/grand:

E(Yes, Rare)       = (200 × 120)/350 = 68.57
E(Yes, Occasional) = (200 × 140)/350 = 80.00
E(Yes, Frequent)   = (200 ×  90)/350 = 51.43
E(No,  Rare)       = (150 × 120)/350 = 51.43
E(No,  Occasional) = (150 × 140)/350 = 60.00
E(No,  Frequent)   = (150 ×  90)/350 = 38.57

All six are ≥ 5. ✓

N — Name the procedure. Chi-square (χ²) test for independence.

I — Inference (test statistic, df, p-value).

χ² = Σ(O − E)²/E
   = (50−68.57)²/68.57 + (90−80)²/80   + (60−51.43)²/51.43
   + (70−51.43)²/51.43 + (50−60)²/60   + (30−38.57)²/38.57
   = 5.03 + 1.25 + 1.43 + 6.71 + 1.67 + 1.90
   = 17.99
df = (2 − 1)(3 − 1) = 2
p-value = χ²cdf(17.99, 1E99, 2) ≈ 0.00012

TI-84: [A] = 2×3 matrix → STAT → TESTS → χ²-Test
Output: χ² = 17.986, p = 1.243E-4, df = 2

C — Conclusion in context. Because p ≈ 0.00012 < α = 0.05, we reject H₀. There is convincing evidence of an association between coffee-drinking status and headache frequency among adults in this city. Because the data are observational, we cannot conclude that coffee drinking causes a change in headache frequency.

Point-by-Point Rubric (sums to 10)

Where students lose points.

• Writing H₀ as "the proportions are equal" (homogeneity language) → loses the hypotheses point.
• Naming the test "homogeneity" or "goodness-of-fit" → loses the procedure point.
• Using df = n − 1 = 349 or df = 5 (cells − 1) instead of (r−1)(c−1) = 2.
• Checking observed counts ≥ 5 instead of expected counts → loses the conditions point.
• Concluding coffee causes headaches → loses the conclusion point (causation from observational data).
• Stopping at "reject H₀" with no context, or omitting the comparison p < α.