Unit 3 · Topic 3.15 Beta

Carrying Out a Chi-Square Test for Homogeneity or Independence

Carrying out a chi-square test: find each expected count, add the components ((O − E)²)/E to get χ² , use the chi-square distribution with (r − 1)(c − 1) degrees of freedom for the right-tail p-value, compare it with α , and conclude about the association or the distributions in context.

Practice 3: Analyze DataPractice 4: Interpret Results

Part 1 · Hook

Why this matters

Of 300 randomly chosen students, those who keep a phone in the bedroom seem to sleep less: 35% of them sleep under 7 hours, against 20% of the others. The expected counts are ready. Now one number,

\chi^2

, has to sum up how far the whole table is from "no association", and a new distribution has to say whether that number is surprising.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Row total 200, column total 90, grand total 300. What is the expected count?

$60$
$90$
$66.7$
$45$

Show the answer

$\frac{200 \times 90}{300} = 60$ .

Correct: $60$ :
$90$ :
$66.7$ :
$45$ :

2. A test gives a p-value of 0.016 at $\alpha = 0.05$ . The decision is

reject $H_0$
fail to reject $H_0$
accept $H_0$
accept $H_a$ as proven

Show the answer

$0.016 < 0.05$ , so reject $H_0$ .

Correct: reject $H_0$ :
fail to reject $H_0$ :
accept $H_0$ :
accept $H_a$ as proven:

Part 4 · See it

See it first

Right-skewed chi-square density curve with 2 degrees of freedom; a dashed line marks the observed statistic 8.21 and the thin shaded tail to its right is the p-value, 0.0165. — The chi-square distribution with 2 degrees of freedom. The observed $\chi^2 = 8.21$ cuts off a right tail of area 0.0165, the p-value. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

For each cell, compare the observed count with the expected countthe term $\frac{(O - E)^2}{E}$ is that cell's contribution: large where the table departs from $H_0$
Add the terms over every cell $\chi^2 = \sum \frac{(O - E)^2}{E}$ , which is 0 when the table matches $H_0$ and grows as it departs
If $H_0$ is true, $\chi^2$ follows a chi-square distribution with $\text{df} = (r - 1)(c - 1)$ the p-value is the area to the right of the observed $\chi^2$ under that curve
Compare the p-value with $\alpha$ conclude about an association (or a difference in distributions) in context, then describe it from the largest terms

Part 6 · Key ideas

Key ideas

$\chi^2 = \sum \frac{(O - E)^2}{E}$ over every cell; each term is a chi-square component.
Degrees of freedom: $\text{df}$ $= (\text{rows} - 1)(\text{columns} - 1)$ .
The p-value is always the right-tail area: only large $\chi^2$ values are evidence against $H_0$ .
Conclude: "There is (is not) convincing evidence of an association between ... and ... for ..." (independence) or "that the distribution of ... differs among ..." (homogeneity).

Part 7 · Misconception

A common mistake

The wrong idea: The cell with the largest observed count contributes the most to $\chi^2$ .

What actually happens: What matters is the gap between observed and expected, relative to expected: $\frac{(O - E)^2}{E}$ . A big count close to its expected count adds almost nothing.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Phones in the bedroom and sleep

A random sample of 300 students at a large high school is asked whether they keep their phone in the bedroom at night and how many hours they usually sleep. A chi-square test for independence is planned at

\alpha = 0.05

; the conditions are met. (Invented data.)

Observed counts: phone in bedroom by usual sleep
Phone in bedroom	Under 7 hours	7 to 8 hours	Over 8 hours	Total
Yes	70	90	40	200
No	20	50	30	100
Total	90	140	70	300

1. What is the value of the chi-square statistic? Give two decimal places.

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

Expected counts: 60, 93.3, 46.7 (yes) and 30, 46.7, 23.3 (no). $\chi^2$ $= \frac{(70 - 60)^2}{60}$ $+ \frac{(90 - 93.3)^2}{93.3}$ $+ \dots$ $+ \frac{(30 - 23.3)^2}{23.3}$ $= 8.21$ .

Answer: 8.21

2. How many degrees of freedom does the test have?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

$\text{df} = (2 - 1)(3 - 1) = 2$ .

Answer: 2

3. What is the p-value? Give four decimal places.

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

$P(\chi^2 \ge 8.21)$ with $\text{df} = 2$ is 0.0165, the area in the right tail.

Answer: 0.0165

4. State the conclusion of the test in context.

Write the conclusion in context.

Show the answer

Reject $H_0$ : convincing evidence of an association, in context, for the population sampled.

A full-credit answer: Because the p-value of 0.0165 is less than 0.05, we reject the null hypothesis. There is convincing evidence of an association between keeping a phone in the bedroom and usual hours of sleep for students at the school.

5. Which cell contributes most to the chi-square statistic, and what does it show?

No phone, under 7 hours: 20 observed, 30 expected
Phone, 7 to 8 hours: the largest observed count in the table, 90
Phone, under 7 hours: 70 observed against 60 expected
No phone, over 8 hours: 30 observed against 23.3 expected

Show the answer

Each cell's term $\frac{(O - E)^2}{E}$ shows where the table departs most from $H_0$ .

Correct: No phone, under 7 hours: 20 observed, 30 expected: Right: its term is $\frac{(20 - 30)^2}{30} = 3.33$ , the largest of the six.
Phone, 7 to 8 hours: the largest observed count in the table, 90: A large count is not the same as a large gap. Here 90 is close to the expected 93.3, a term of only 0.12.
Phone, under 7 hours: 70 observed against 60 expected: This term is $\frac{(70 - 60)^2}{60} = 1.67$ . The same gap of 10 counts for more where only 30 were expected.
No phone, over 8 hours: 30 observed against 23.3 expected: This term is 1.90, second largest. The no phone, under 7 hours cell is larger.

6. In one cell of a two-way table, the observed count is 30 and the expected count is 22.4. Write the expression for this cell's contribution to $\chi^2$ , then give its value to two decimal places.

Type the expression as on the exam, with the values substituted.

Show the answer

$\frac{(30 - 22.4)^2}{22.4}$ $= \frac{57.76}{22.4}$ $= 2.58$ .

Answer: $\frac{(30 - 22.4)^{2}}{22.4}$

7. A two-way table classifies students by class year (4 categories) and favorite lunch option (5 categories). How many degrees of freedom does a chi-square test for independence have?

Show the answer

$\text{df} = (r - 1)(c - 1)$ .

19: That is the number of cells minus 1. Use $(r - 1)(c - 1)$ .
7: That adds $(4 - 1) + (5 - 1)$ . Multiply instead.
20: That is the number of cells.
Correct: 12: Right: $(4 - 1)(5 - 1)$ $= 3 \times 4$ $= 12$ .

Part 9 · Summary

Summary

Carrying out a chi-square test: find each expected count, add the components

\frac{(O - E)^2}{E}

to get

\chi^2

, use the chi-square distribution with

(r - 1)(c - 1)

degrees of freedom for the right-tail p-value, compare it with

\alpha

, and conclude about the association or the distributions in context. The largest components show where the association lies.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic