Two-Proportion Test and Contingency Table χ² Test: 04-BS-2, December 2014, Q6

Question 6: Two-Proportion Test and Contingency Table χ² Test (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) new process: 16/400 substandard; current process: 24/500 substandard. (B) a 2×3 contingency table of favour/against counts across job-responsibility levels A, B, C.

Find. (A) whether the new process has a significantly LOWER substandard proportion (one-tailed, α=0.05). (B) whether the "against" proportions differ significantly across A, B, C (α=0.05).

Approach. (A) is a one-tailed two-proportion z-test using the pooled proportion under H₀: p₁=p₂. (B) is a χ² test of homogeneity of proportions on a 2×3 contingency table, comparing observed counts to row-total×column-total/grand-total expected counts.

(A) One-tailed two-proportion z-test. $\hat p_1=16/400=0.040$ (new), $\hat p_2=24/500=0.048$ (current). Under $H_0: p_1=p_2$, the pooled estimate is $\bar p=\dfrac{16+24}{400+500}=0.04444$, so $$z=\dfrac{\hat p_1-\hat p_2}{\sqrt{\bar p(1-\bar p)\left(\frac{1}{400}+\frac{1}{500}\right)}}=\boxed{-0.579}.$$ H₁ is one-tailed ($p_1<p_2$, "new is better"), so the critical value is $-z_{0.05}=-1.645$. Since $-0.579>-1.645$ (not in the rejection region), fail to reject H₀: there is NOT enough evidence at α=0.05 that the new process yields a lower substandard proportion.
(B) Contingency-table χ² test of homogeneity. Column totals are 496, 444, 543 (grand total 1,483); row totals are 1,351 (favour) and 132 (against). Expected counts under independence are (row total)(column total)/grand total:

Expected counts under H₀: proportions are equal across A, B, C
	A	B	C
Favour (expected)	451.85	404.48	494.67
Against (expected)	44.15	39.52	48.33

χ² statistic and conclusion. $$\chi^2=\sum\dfrac{(O-E)^2}{E}=\boxed{5.647},\qquad df=(2-1)(3-1)=2.$$ The critical value is $\chi^2_{0.05,2}=5.991$. Since $5.647<5.991$, fail to reject H₀: the proportions of members against the new fee schedule are NOT significantly different across the three job-responsibility levels at α=0.05.

Final results — Question 6
Part	Quantity	Result
(A)	z-stat vs critical; conclusion	−0.579 vs −1.645; fail to reject H₀
(B)	χ²-stat vs critical (df=2); conclusion	5.647 vs 5.991; fail to reject H₀

Topic: Two-sample proportion tests and the χ² test of homogeneity for a contingency table.

Key relations: $z=\dfrac{\hat p_1-\hat p_2}{\sqrt{\bar p(1-\bar p)(1/n_1+1/n_2)}}$; $\chi^2=\sum(O-E)^2/E$ with $E_{ij}=$ row total × column total / grand total, $df=(r-1)(c-1)$.

Why this works: pooling the two sample proportions under H₀ gives the correct standard error for the difference when the null hypothesis assumes a common p; the contingency-table expected counts are exactly what independence between row and column classifications predicts, so squared standardized deviations from them sum to a χ² statistic.

Common pitfall: using each sample's own (unpooled) $\hat p_i(1-\hat p_i)$ in the two-proportion test's standard error instead of the pooled $\bar p$ under H₀, and using a two-tailed critical value for a one-tailed "is the new process better" claim.

Source: Walpole, Myers, Myers & Ye, Ch. 10 (Two-sample tests) and Ch. 9 (Goodness-of-fit and contingency tables).

Question 6 of 8: Two-Proportion Test and Contingency Table χ² Test

Question 6: Two-Proportion Test and Contingency Table χ² Test (20 marks)