Inferential and relational statistical techniques to include Spearman’s…

The idea

Inferential and relational statistics test whether a sample pattern provides evidence against a null hypothesis. Spearman’s rank correlation measures the strength and direction of a monotonic association between paired ranked observations, such as river velocity and distance downstream. Chi-square compares observed and expected frequencies for categorical data, either for association in a contingency table or for a stated expected distribution. For a significance test, choose the test and significance level before interpreting results, calculate the statistic, and use the correct critical-value table. Spearman tables are entered with n, the number of valid pairs; Chi-square uses degrees of freedom, (rows − 1)(columns − 1) for association. If the result is significant, reject the null hypothesis; otherwise fail to reject it. Statistical association does not by itself prove causation.

What earns the marks

• State a testable null hypothesis before analysis and identify the paired ranked data or categorical frequencies being tested.
• For Spearman's rank, rank each variable consistently, calculate the coefficient and use n, the number of valid pairs, to select the critical value.
• For Chi-square association, calculate expected frequencies and degrees of freedom as (rows − 1)(columns − 1), then compare the statistic with the correct critical value.
• State the chosen significance level, reject the null hypothesis only when the result is significant, and otherwise report that there is insufficient evidence to reject it.
• Interpret strength, direction and context without treating statistical association as proof of causation.

Common mistakes and how to put them right

• Writing a vague null hypothesis.

  Put it right: Name the variables or categories and state that there is no statistically significant association, correlation or difference as appropriate.

• Treating a negative Spearman coefficient as weak.

  Put it right: The sign gives direction while the absolute value indicates strength.

• Using n − 2 as Spearman's degrees of freedom.

  Put it right: Use the AQA critical-value table row for n valid pairs, while degrees of freedom apply to Chi-square.

• Saying the null hypothesis is accepted when a result is not significant.

  Put it right: Report that you fail to reject it because the sample gives insufficient evidence.

• Using Chi-square on raw continuous measurements.

  Put it right: It requires observed frequencies in meaningful categories and suitable expected counts.

Exam tips

• Write the null hypothesis and significance level before calculating so the decision rule is clear.
• For Spearman, check tied-rank handling and count the valid pairs; for Chi-square, show observed and expected frequencies and the degrees-of-freedom calculation.
• Use 'significant at p ≤ 0.05' or 'not significant at p ≤ 0.05' as appropriate, then reject or fail to reject the null hypothesis.
• Finish with a contextual interpretation and state that a significant association alone does not demonstrate causation.