04-BS-2, May 2016, 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 data — good items / rejects by machine
Machine	A	B	C	D
Good items	25	23	25	27
Rejects	16	10	13	11

Find. Whether the rejection rate differs significantly across the four machines, using a $2\times4$ chi-square test of homogeneity at $\alpha=0.05$.

Approach. Build the contingency table of good/reject counts, compute row×column/grand-total expected counts under $H_0$ (equal rejection rate), form the χ² statistic, and compare against the critical value with df$=(2-1)(4-1)=3$.

Row, column, and grand totals. Totals per machine: A=41, B=33, C=38, D=38 (grand total=150). Row totals: good=100, rejects=50.
Expected counts under H0 (proportional rejection rate). $E_{\text{good},j}=\dfrac{100\times n_j}{150}$, $E_{\text{rej},j}=\dfrac{50\times n_j}{150}$: A: (27.33, 13.67); B: (22.00, 11.00); C: (25.33, 12.67); D: (25.33, 12.67).
Chi-square statistic. $\chi^2=\sum\dfrac{(O-E)^2}{E}=1.076$ (summing all 8 cells).
Compare to the critical value. $\chi^2_{0.05,3}=7.815$. Since $1.076<7.815$, $\boxed{\text{fail to reject }H_0}$ — the rejection rate does not differ significantly among the four machines.

Question 6 — final results
Quantity	Value
χ² statistic	1.076
Critical value χ²_0.05,3	7.815
Conclusion	fail to reject H₀ (rates not significantly different)

Topic: Chi-square test of homogeneity for a $2\times k$ contingency table.

Key relations: $\chi^2=\sum\dfrac{(O_{ij}-E_{ij})^2}{E_{ij}}$, $E_{ij}=\dfrac{(\text{row total}_i)(\text{column total}_j)}{\text{grand total}}$, df$=(r-1)(c-1)$.

Why this works: under $H_0$ every machine shares the same true rejection rate, so the expected counts scale each machine's own total proportionally to the pooled good/reject split; large deviations of observed from expected accumulate into a statistic that is approximately χ² distributed for reasonably large expected cell counts.

Common pitfall: using df$=k-1$ (as in a single-row goodness-of-fit test) instead of $(r-1)(c-1)=3$ for this two-row table, and forgetting to sum BOTH the good-item and reject cells' contributions.

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)