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Find the first row total.

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

Revisit first: Correlation of paired data

TOPIC 01

Contingency tables and independence tests

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

What you will be able to explain

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

Try it. Leave your reasoning visible.

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

01 / Foundation#Worked example

Find the first row total.

O=(24121224)O=\begin{pmatrix}24&12\\12&24\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
Use this intermediate relation.
R1=O11+O12R_1=O_{11}+O_{12}
Worked solution
  1. Use the paired-data summaries, preserving which observations belong together.

  2. Apply the stated relation and retain its conditions.

    R1=24+12=36R_1=24+12=36
  3. Marginal totals sum categories without changing their pairing.

The requested value is 36.

Checks and common pitfalls: Marginal totals sum categories without changing their pairing.

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

Focus on one question

Teacher preparation and assessment

Question sequence

  • 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?

Board plan

  • Defining relation: Compute expected counts and a chi-square statistic and state a bounded conclusion.
    Eij=RiCj/N;χ2=∑(O−E)2/EE_{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.

Anticipated thinking

  • Failure to reject does not prove independence.

Assessment checklist

  • 1 mark: choose the correct representation and conditions.
  • 1 mark: establish the intermediate relation.
  • 1 mark: complete a connected calculation or proof.
  • 1 mark: interpret and check the conclusion.

No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.

Curriculum and source notes ↗