Kenneth ChengMathematical modeling
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Contingency tables and independence tests

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

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.

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.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Can a larger sample increase the statistic when proportions stay fixed?

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

Independent-observation model: every expected count=10; χ²=1.6. Scaling the table scales χ²; association is not a causal conclusion.O = [12, 8; 8, 12]E per cell = 10χ² = 1.6

Independent-observation model: every expected count=10; χ²=1.6. Scaling the table scales χ²; association is not a causal conclusion.

Explain: Calculate two valid cases and explain the change using the defining relation.

Transfer: Scale every cell of a 2×2 table and explain the effect on expected counts and χ².

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.

02 / Standard#Worked example

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

t=13t=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.
Oij=Eij=tO_{ij}=E_{ij}=t
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    χ2=0\chi^2=0

  3. Observed counts exactly equal independence expectations.

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.

03 / Transfer#Worked example

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

χ2=1+4/5\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<3.8411<t/(t+1)+1<2<3.841
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    insufficient evidence against independence under the stated model\text{insufficient evidence against independence under the stated model}

  3. Absence of sufficient evidence is not proof of no association; sample power and assumptions still matter.

The requested relation or conclusion is shown below.

do not reject independence at this level\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
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

04 / Foundation#Your turn

Find the first row total.

O=(30151530)O=\begin{pmatrix}30&15\\15&30\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=30+15=45R_1=30+15=45

  3. Marginal totals sum categories without changing their pairing.

The requested value is 45.

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

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

05 / Foundation#Your turn

Under independence, find E11.

O=(32161632)O=\begin{pmatrix}32&16\\16&32\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.
E11=R1C1/NE_{11}=R_1C_1/N
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    E11=(48)2/96=24E_{11}=(48)^2/96=24

  3. Use both relevant margins and the full sample size.

The requested value is 24.

Checks and common pitfalls: Use both relevant margins and the full sample size.

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

06 / Foundation#Your turn

Calculate χ² for the table.

O=(34171734)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

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

  2. Apply the stated relation and retain its conditions.

    χ2=4[(17/2)2/(51/2)]=34/3\chi^2=4[(17/2)^2/(51/2)]=34/3

  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.

07 / Foundation#Your turn

Use the supplied 5% critical value 3.841 (one degree of freedom) to assess independence.

O=(36181836)O=\begin{pmatrix}36&18\\18&36\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.
χ2=2u/3\chi^2=2u/3
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    χ2=12>3.841\chi^2=12>3.841

  3. The rule provides evidence of association; it does not estimate a causal effect.

The requested relation or conclusion is shown below.

reject independence at the stated 5% rule\text{reject independence at the stated 5\% rule}

Checks and common pitfalls: The rule provides evidence of association; it does not estimate a causal effect.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

08 / Standard#Your turn

Use the supplied 5% critical value 3.841 (one degree of freedom) to assess independence.

O=(38191938)O=\begin{pmatrix}38&19\\19&38\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.
χ2=2u/3\chi^2=2u/3
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    χ2=12.6666666667>3.841\chi^2=12.6666666667>3.841

  3. The rule provides evidence of association; it does not estimate a causal effect.

The requested relation or conclusion is shown below.

reject independence at the stated 5% rule\text{reject independence at the stated 5\% rule}

Checks and common pitfalls: The rule provides evidence of association; it does not estimate a causal effect.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

09 / Standard#Your turn

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

t=20t=20
  • 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.
Oij=Eij=tO_{ij}=E_{ij}=t
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    χ2=0\chi^2=0

  3. Observed counts exactly equal independence expectations.

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.

10 / Standard#Your turn

A before/after category table records the same t people twice. Can a standard independent-observation chi-square analysis treat the records as separate people?

t=21t=21
  • 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
Check whether rows share participants.
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    paired records are dependent\text{paired records are dependent}

  3. The study needs a method for paired categorical outcomes rather than treating repeated people as independent.

The requested relation or conclusion is shown below.

the independent-observation assumption fails\text{the independent-observation assumption fails}

Checks and common pitfalls: The study needs a method for paired categorical outcomes rather than treating repeated people as independent.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

11 / Standard#Your turn

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

χ2=1+12/13\chi^2=1+12/13
  • 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.
1<t/(t+1)+1<2<3.8411<t/(t+1)+1<2<3.841
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    insufficient evidence against independence under the stated model\text{insufficient evidence against independence under the stated model}

  3. Absence of sufficient evidence is not proof of no association; sample power and assumptions still matter.

The requested relation or conclusion is shown below.

do not reject independence at this level\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
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

12 / Transfer#Your turn

A before/after category table records the same t people twice. Can a standard independent-observation chi-square analysis treat the records as separate people?

t=23t=23
  • 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
Check whether rows share participants.
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    paired records are dependent\text{paired records are dependent}

  3. The study needs a method for paired categorical outcomes rather than treating repeated people as independent.

The requested relation or conclusion is shown below.

the independent-observation assumption fails\text{the independent-observation assumption fails}

Checks and common pitfalls: The study needs a method for paired categorical outcomes rather than treating repeated people as independent.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

13 / Transfer#Your turn

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

χ2=1+14/15\chi^2=1+14/15
  • 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<3.8411<t/(t+1)+1<2<3.841
Worked solution

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

  2. Apply the stated relation and retain its conditions.

    insufficient evidence against independence under the stated model\text{insufficient evidence against independence under the stated model}

  3. Absence of sufficient evidence is not proof of no association; sample power and assumptions still matter.

The requested relation or conclusion is shown below.

do not reject independence at this level\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
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

Focus on one question

End-of-lesson check and correction

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    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.

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    Curriculum and source notes ↗