LEARN · EXPLAIN · REVISE

Calculate χ² for the table.

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 / Foundation#Your turn

O=\begin{pmatrix}34&17\\17&34\end{pmatrix}

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

All expected cells are 3u/2.

Worked solution

Use the paired-data summaries, preserving which observations belong together.
Apply the stated relation and retain its conditions.
$\chi^2=4[(17/2)^2/(51/2)]=34/3$
All four nonnegative cell contributions must be included.

The requested value is 11.3333333333.

Checks and common pitfalls: All four nonnegative cell contributions must be included.

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.