Question 4 of 8

Question 4 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. 3×3 contingency table of observed member counts by Age Bracket (row) × Insurance Policy (column):

Find. Whether Insurance Policy preference is independent of Age Bracket, at $\alpha=0.05$.

Approach. Chi-square test of independence: compute expected counts under independence $E_{ij}=(\text{row}_i)(\text{col}_j)/n$, form $\chi^2=\sum(O-E)^2/E$, and compare to the critical value with $(r-1)(c-1)$ degrees of freedom.

1. State the hypotheses. $H_0$: Insurance Policy preference is independent of Age Bracket. $H_1$: they are not independent.
2. Compute the expected counts. $E_{ij}=\dfrac{\text{row total}_i\times\text{column total}_j}{540}$, e.g. $E_{11}=\dfrac{165\times180}{540}=55.0$.

   Expected counts under $H_0$

   	Policy A	Policy B	Policy C
   25–40	55.00	48.89	61.11
   41–56	66.67	59.26	74.07
   Over 56	58.33	51.85	64.81

3. Compute the test statistic. Degrees of freedom $=(3-1)(3-1)=4$. $$\chi^2=\sum_{i,j}\frac{(O_{ij}-E_{ij})^2}{E_{ij}}=\boxed{23.05}$$
4. Compare and conclude. Critical value $\chi^2_{0.05,4}=9.488$. Since $23.05 > 9.488$, reject $H_0$ at $\alpha=0.05$: there is significant evidence that Insurance Policy preference is NOT independent of Age Bracket — preference varies systematically with age (notably: Policy A is favoured disproportionately by the Over-56 group, Policy B by the 25–40 group).