Question 8 of 8: Chi-Square Test of Homogeneity Across Assembly Lines

Question 8: Chi-Square Test of Homogeneity Across 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.

Given.

Find. Test $H_0: p_A=p_B=p_C=p_D$ (homogeneity of substandard proportions across the 4 lines) at $\alpha=0.05$.

Approach. A $2\times4$ contingency-table chi-square test of homogeneity: pool all four lines to estimate the common proportion under $H_0$, compute expected substandard/standard counts per line under that pooled proportion, and sum $(O-E)^2/E$ over all 8 cells.

1. (a) Pooled proportion under $H_0$. Total units $=12{,}000+11{,}500+12{,}600+11{,}900=48{,}000$; total substandard $=500+435+574+591=2{,}100$. $$\hat p=\dfrac{2{,}100}{48{,}000}=\boxed{0.04375}$$
2. Expected counts. Expected substandard for line $i$ is $E_i=n_i\hat p$: $E_A=525.0,\ E_B=503.1,\ E_C=551.3,\ E_D=520.6$; the corresponding expected standard counts are $n_i(1-\hat p)$.
3. Chi-square statistic. Summing $(O-E)^2/E$ over both the substandard and standard cells for all four lines, $$\chi^2=\sum_{i}\left[\dfrac{(O_{i,\text{sub}}-E_{i,\text{sub}})^2}{E_{i,\text{sub}}}+\dfrac{(O_{i,\text{std}}-E_{i,\text{std}})^2}{E_{i,\text{std}}}\right]=\boxed{21.82}$$ with $df=(4-1)=3$.
4. Decision. $\chi^2_{0.05,3}=7.815$ from the chi-square table. Since $21.82>7.815$: reject $H_0$ — the proportion of substandard units is NOT the same across all four assembly lines (Line D, at $591/11{,}900=4.97\%$, and Line C, at $574/12{,}600=4.56\%$, run visibly higher than Line A's $4.17\%$ and Line B's $3.78\%$).