LEARN · EXPLAIN · REVISE

Under perfect independence in a table with all four cells t, find χ².

Read the idea, work independently, then explain what changed.

高三選擇性必修第三册(A版).pdf · 8.3 · PDF 129 / printed page 124

TOPIC 01

Contingency tables and independence tests

Compute expected counts and a chi-square statistic and state a bounded conclusion.

Compute expected counts and a chi-square statistic and state a bounded conclusion.
Justify the method and check the conditions in a new situation.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Worked example

t=13

Observations must be independent; expected counts support the supplied large-sample approximation. Rejection is evidence of association, not causation.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Use the paired-data summaries, preserving which observations belong together.

Hint 2

Use this intermediate relation.

O_{ij}=E_{ij}=t

Worked solution

The requested value is 0.

Checks and common pitfalls: Observed counts exactly equal independence expectations.

Think first. Reveal a hint when the class is ready.

Teacher preparation and assessment

Compute expected counts and a chi-square statistic and state a bounded conclusion.
Which condition is essential in contingency tables and independence tests?
Can a larger sample increase the statistic when proportions stay fixed?

Defining relation: Compute expected counts and a chi-square statistic and state a bounded conclusion.
$E_{ij}=R_iC_j/N;\quad\chi^2=\sum(O-E)^2/E$
Conditions: Observations must be independent; expected counts support the supplied large-sample approximation. Rejection is evidence of association, not causation.