LEARN · EXPLAIN · REVISE

The displayed test statistic is below the supplied 5% threshold 3.841. State the defensible conclusion.

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 / Transfer#Worked example

\chi^2=1+4/5

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.

1<t/(t+1)+1<2<3.841

Worked solution

Use the paired-data summaries, preserving which observations belong together.
Apply the stated relation and retain its conditions.
$\text{insufficient evidence against independence under the stated model}$
Absence of sufficient evidence is not proof of no association; sample power and assumptions still matter.

The requested relation or conclusion is shown below.

\text{do not reject independence at this level}

Checks and common pitfalls: Absence of sufficient evidence is not proof of no association; sample power and assumptions still matter.

Reasoning checklist · self / teacher assessment

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.