Chi-Square Test of Homogeneity — Four Assembly Lines: 04-BS-2, December 2018, Q6

Question 6: Chi-Square Test of Homogeneity — Four Assembly Lines (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.

Find. Whether the proportion substandard is the same across all 4 lines (α=0.05).

Approach. A 2×4 chi-square test of homogeneity: pool all 4 lines to get one overall (defective) proportion under H₀, compute expected counts per cell, and sum $(O-E)^2/E$ over all 8 cells (defective and non-defective × 4 lines), df=(2−1)(4−1)=3.

Pooled proportion under H₀. Total units $=12{,}000+11{,}400+12{,}500+11{,}900=47{,}800$; total substandard $=600+691+673+536=2{,}500$. $$\bar p=\dfrac{2500}{47800}=0.05230$$
Expected counts and chi-square statistic. Each line’s expected substandard count is $E_i=n_i\bar p$ and expected good count is $n_i(1-\bar p)$. Computing $(O-E)^2/E$ for all 8 cells:

Observed vs. expected substandard counts
Line	n_i	O (defective)	E (defective)	(O−E)²/E, defective	(O−E)²/E, good
A	12,000	600	627.6	1.213	0.067
B	11,400	691	596.2	15.077	0.833
C	12,500	673	653.8	0.564	0.031
D	11,900	536	622.4	11.988	0.663

Sum and decision. Summing all 8 cell contributions: $$\chi^2=1.213+0.067+15.077+0.833+0.564+0.031+11.988+0.663$$ $$\boxed{\chi^2=30.42\ \ (\text{df}=3)}$$ Critical value $\chi^2_{0.05,3}=7.81$ (Appendix 3). Since $30.42\gg7.81$: $$\boxed{\text{Reject }H_0\text{ -- the substandard proportion is NOT the same across the four assembly lines}}$$ (Lines B and D deviate most from the pooled average, driving most of the statistic.)

Question 6 — final results
Quantity	Result
Pooled proportion &bar;p	0.0523
χ² statistic (df=3)	30.42
Critical value χ²_0.05,3	7.81
Conclusion	Reject H₀ — lines differ

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

Key relations: $\chi^2=\sum\dfrac{(O-E)^2}{E}$, df = (rows−1)(cols−1); pooled proportion $\bar p=\dfrac{\sum\text{defects}}{\sum n}$ used to build every expected cell under H₀.

Why this works: Under H₀ all four lines share one true defect rate, so the best estimate of that shared rate is the overall pooled proportion; each cell’s squared deviation from its expectation, scaled by that expectation, approximately follows a χ² distribution once summed over the independent cells (Pearson’s theorem).

Common pitfall: Summing only the “defective” column contributions and forgetting the paired “good unit” cells (each row of a 2×k table contributes 2 cells, not 1); using the raw counts as if they were a goodness-of-fit test against a fixed proportion rather than a homogeneity test with a proportion estimated from the data.

Source: Walpole et al., Probability & Statistics for Engineers and Scientists, Ch. 10 (Chi-Square Tests).

Question 6 of 8: Chi-Square Test of Homogeneity — Four Assembly Lines

Question 6: Chi-Square Test of Homogeneity — Four Assembly Lines (20 marks)