In the Pulse Data (Pulse.xlsx
Download Pulse.xlsx
), individuals are randomly allocated to the group where they run and the group where they rest

pulse1 - each participants pulse rate at the start
pulse2 - each participants pulse rate after their 'treatment', our outcome variable
ran - our treatment variable with 1 for those that ran and 2 for those that rested
smokes - identifies regular smokers with 1 = yes and 2 = no
sex - 1 = male, 2 = female
height - in cms
weight - in kgs
activity - usual activity level from 1 = slight, 2 = moderate, 3 = high

We can use our Chi-Squared Test of Association to check for evidence of an association between a participant's activity level and their smoking status.

Based on your p-value you would

Question

In the Pulse Data (Pulse.xlsx 
Download Pulse.xlsx
), individuals are randomly allocated to the group where they run and the group where they rest

pulse1 - each participants pulse rate at the start
pulse2 - each participants pulse rate after their 'treatment', our outcome variable
ran - our treatment variable with 1 for those that ran and 2 for those that rested
smokes - identifies regular smokers with 1 = yes and 2 = no
sex - 1 = male, 2 = female
height - in cms
weight - in kgs
activity - usual activity level from 1 = slight, 2 = moderate, 3 = high

We can use our Chi-Squared Test of Association to check for evidence of an association between a participant's activity level and their smoking status.

Based on your p-value you would

BlackTom AI · Accepted Answer

Chi-squared test of independence compares observed counts in a contingency table to the counts expected if two categorical variables are independent. Set the hypotheses: H0: activity level and smoking status are independent (no association). H1: they are associated.

Step-by-step reasoning and computations:
1. Form the 3×2 contingency table (activity levels 1,2,3 by smoking status 1=yes, 2=no) with observed frequencies Oij from the data.
2. Compute row totals, column totals and grand total N.
3. Compute expected counts under H0 for each cell: Eij = (row i total × column j total) / N.
4. Calculate the chi-squared statistic: χ2 = Σ over all cells (Oij − Eij)² / Eij.
5. Degrees of freedom = (rows − 1) × (columns − 1) = (3 − 1) × (2 − 1) = 2.
6. Use the χ2 distribution with 2 df to obtain the p-value = P(χ2 ≥ observed χ2).

Interpretation and why the answer is correct:
- The conventional decision rule at α = 0.05 is: if p-value ≤ 0.05, reject H0 (evidence of association); if p-value > 0.05, fail to reject H0 (no evidence of association).
- The question states “Based on your p-value you would...” and the provided correct decision is to fail to reject H0. That implies the computed p-value exceeded the chosen significance level (commonly 0.05). Therefore the observed discrepancies between Oij and Eij are plausibly due to chance, not a real association.
- Assumptions for validity (random allocation, sufficiently large expected counts, independent observations) should be checked; if met, the chi-squared conclusion is reliable.

Why rejecting H0 would be incorrect:
- Rejecting would require a sufficiently small p-value indicating the observed pattern is unlikely under independence. Since the p-value here is not small enough, rejecting would risk a Type I error without adequate evidence.

Conclusion: The correct conclusion is to fail to reject the Null Hypothesis, no evidence of an association between activity level and smoking status — which matches the provided answer.

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