SE4, SE5 - Interpreting association

SE4, SE5 - Interpreting association

Use cell contributions to describe the pattern behind a significant association, distinguish evidence from causation and apply Yates' continuity correction to a two-by-two table. You will compare corrected and uncorrected tests without overstating their conclusions.

Locating the pattern of association

Can two valid calculations on the same table lead to different 5% decisions? A continuity correction can make that happen. First, learn what either statistic can tell you about the pattern in a table.

The overall chi-squared statistic answers whether the table departs substantially from independence. A cell contribution, (O−E)2/E(O-E)^2/E, helps locate that departure. A large contribution identifies a cell that differs strongly relative to its expected count. Squaring removes the sign: you must also compare OO with EE to say whether the count is above or below expectation.

Worked example 1: identifying excesses and shortages

Question. A random commuter sample gives this table. All six expected counts under independence are 20. A test has already found significant association.

AreaCycleBusCar
Central351510
Outer52530

Identify the main sources of association.

Plan. Calculate contributions, locate the largest and restore the direction using O−EO-E.

Central contributions=(11.25,1.25,5)Outer contributions=(11.25,1.25,5)\begin{aligned} \text{Central contributions}&=(11.25,1.25,5)\\ \text{Outer contributions}&=(11.25,1.25,5) \end{aligned}

For example, (35−20)2/20=11.25(35-20)^2/20=11.25. Both Cycle cells have the largest contributions.

Answer. Cycling is more common than expected under independence among central residents (35 rather than 20) and less common among outer residents (5 rather than 20). The Car cells also contribute: fewer central and more outer residents use cars than predicted under independence.

Check. The contributions sum to 2(11.25+1.25+5)=352(11.25+1.25+5)=35. The two Cycle cells account for 22.5/3522.5/35, approximately 64% of that total, supporting their description as the main source.

Swap only the two row profiles. The statistic stays the same, while the contextual direction reverses.

Worked example 2: a contribution has no sign

Question. The Central row is now (5,25,30)(5,25,30), and Outer is (35,15,10)(35,15,10). Expectations remain 20 throughout. Describe the largest contributions.

Plan. Reuse the squared discrepancies but read their signs from the new observations.

(5−20)220=11.25=(35−20)220\frac{(5-20)^2}{20}=11.25=\frac{(35-20)^2}{20}

Answer. Cycling is less common than expected among central residents and more common among outer residents. These are still the largest contributions.

Check. Swapping row labels does not change any squared discrepancy or their sum. It must change the verbal description of which area has the excess.

This is an exploratory description of the overall association, not a separate significance test of each cell. Do not claim that every cell with a large contribution is independently “significant”.

Watch out

Wrong: “The largest contribution is positive, so this category is above expectation.” Every contribution is nonnegative. Use the signed difference O−EO-E to decide above or below.

Association does not establish causation

A significant table shows evidence that the variables are associated in the population sampled, provided the sampling and model assumptions are appropriate. It does not show that changing one variable would cause a change in the other.

An observational survey might find an association between holding a gym membership and reporting regular exercise. People already keen on exercise may buy memberships. Income, health or access could affect both variables. The table alone cannot separate those explanations from a causal effect of membership.

Confounding

A confounding variable is another factor associated with the variables under investigation that can help explain their observed relationship. Random sampling helps represent a population; random allocation of treatments has a different role in supporting causal comparisons.

A small p-value is the probability, under the null model, of a result at least as extreme as the one observed. It is not the probability that the null hypothesis is true, nor a measure of the practical size of the effect. A large sample can detect a small association. A non-significant result may reflect limited information rather than genuine independence.

Appropriate wording is “There is sufficient evidence to suggest an association between membership and reported exercise”, with the population and significance level where relevant. “Membership makes people exercise” goes beyond the evidence. Sampling only volunteers at one gym would also limit generalisation to all adults.

Key idea

A test detects evidence against independence. Causal conclusions and population-wide claims need support from the study design, not just a small p-value.

Yates' correction for two-by-two tables

Observed counts jump in whole numbers, whereas a chi-squared reference distribution is continuous. Yates' continuity correction adjusts for this discreteness in a 2×22\times2 association table, which has one degree of freedom. For this course, use it for a 2×22\times2 test unless the question explicitly requests an uncorrected calculation. Do not apply it to a 2×32\times3 or larger table.

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