04-BS-2, May 2017, Q6 | Nivaar Exam Prep

Question 6 (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. A 2×4 contingency table of good/reject counts across machines A–D (table above), $\alpha=0.05$.

Find. Whether the proportion of rejects differs significantly across the four machines.

Approach. Chi-square test of homogeneity: compute row/column/grand totals, get expected counts under the null (equal reject-proportion), sum $(O-E)^2/E$ over all 8 cells, compare with $\chi^2_{0.05,3}$.

Totals. Column totals: A=294, B=283, C=287, D=336. Row totals: good=1,000, rejects=200. Grand total = 1,200.
Expected counts under H₀: equal reject rate. $E_{\text{good},j}=1000\cdot\dfrac{n_j}{1200}$, $E_{\text{rej},j}=200\cdot\dfrac{n_j}{1200}$: good $(245.0,\,235.83,\,239.17,\,280.0)$, rejects $(49.0,\,47.17,\,47.83,\,56.0)$.
Chi-square statistic. $\chi^2=\displaystyle\sum\frac{(O-E)^2}{E}$ over all 8 cells $=\dfrac{(230-245)^2}{245}+\dfrac{(240-235.83)^2}{235.83}+\cdots=8.896$, with $df=(4-1)(2-1)=3$.
Decision. Critical value $\chi^2_{0.05,3}=7.815$. Since $8.896>7.815$, $\boxed{\text{reject }H_0}$: the proportion of rejects DOES differ significantly among the four machines — machine A's reject rate (64/294=21.8%) stands out well above D's (56/336=16.7%).

Question 6 — final results
Quantity	Value
χ² statistic (df=3)	8.896
Critical value χ²_0.05,3	7.815
Conclusion	Reject H₀ — reject proportions differ

Topic: Chi-square test of homogeneity of proportions across several populations.

Key relations: $\chi^2=\sum_{\text{cells}}\dfrac{(O-E)^2}{E}\sim\chi^2_{(r-1)(c-1)}$ under $H_0$, with $E_{ij}=\dfrac{(\text{row total}_i)(\text{column total}_j)}{\text{grand total}}$.

Why this works: under the null that all four machines share one common reject rate, each cell's expected count is what that pooled rate predicts given the machine's own output volume; large squared deviations relative to $E$ accumulate evidence against a single shared rate, and the sum of such standardized squared deviations is asymptotically chi-square with degrees of freedom equal to the number of free row/column combinations.

Common pitfall: using $df=4$ (number of machines) instead of the correct $(r-1)(c-1)=3$ for a $2\times4$ table, or computing $(O-E)^2/E$ using only one of the two rows (good/reject) instead of summing across all 8 cells.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 10 (Chi-Square Tests).

Question 6 of 8

Question 6 (20 marks)