Insulating-Material Breakdown Voltage — Two-Sample Tests: 04-BS-2, May 2014, Q7

Question 7: Insulating-Material Breakdown Voltage — Two-Sample Tests (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. n_A=n_B=12; m_A=26.25, s_A=0.6; m_B=25.95, s_B=0.7 (kV/mm); both populations assumed Normal, independent samples.

Find. (a) test H₀: σ_A²=σ_B² via the F-test; (b) test H₀: μ_A=μ_B via the appropriate two-sample t-test.

Approach. Test the variance ratio first (F-test), because its outcome decides which mean test to use: if the variances are not significantly different, pool them into a common estimate and use the pooled two-sample t-test with df=n_A+n_B−2 (Welch's unequal-variance t-test would instead be needed if the F-test rejected equal variances).

(a) F-test for equal variances. $H_0:\sigma_A^2=\sigma_B^2$ vs $H_1:\sigma_A^2\ne\sigma_B^2$. Placing the larger sample variance on top, $$F=\dfrac{s_B^2}{s_A^2}=\dfrac{0.7^2}{0.6^2}=\dfrac{0.49}{0.36}=\boxed{1.361}.$$ The two-sided critical value with $\nu_1=\nu_2=11$ is $F_{0.025,11,11}=3.474$. Since $1.361<3.474$, fail to reject H₀: the two manufacturers' breakdown-voltage variability is not significantly different. This assumption (equal population variances) justifies pooling in part (b).
(b) Pooled two-sample t-test for the means. $H_0:\mu_A=\mu_B$ vs $H_1:\mu_A\ne\mu_B$. The pooled variance is $$s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{11(0.36)+11(0.49)}{22}=\dfrac{9.35}{22}=0.4250,\qquad s_p=\boxed{0.6519}.$$ The test statistic is $$t=\dfrac{m_A-m_B}{s_p\sqrt{1/n_A+1/n_B}}=\dfrac{26.25-25.95}{0.6519\sqrt{2/12}}=\boxed{1.127},$$ with df=22, $t_{0.025,22}=2.074$. Since $|1.127|<2.074$, fail to reject H₀: the mean breakdown voltages of the two manufacturers' materials are not significantly different.

Final results — Question 7
Part	Quantity	Result
(a)	F, F_crit(0.025,11,11)	1.361 < 3.474 → fail to reject H₀
(b)	s_p, t, t_crit(df=22)	0.6519; 1.127 < 2.074 → fail to reject H₀

Topic: Two-sample tests for comparing variances (F) and means (pooled t) of two independent Normal populations.

Key relations: $F=S_1^2/S_2^2\sim F_{n_1-1,n_2-1}$ under $H_0$; pooled $S_p^2=\dfrac{(n_A-1)S_A^2+(n_B-1)S_B^2}{n_A+n_B-2}$; $T=\dfrac{\bar X_A-\bar X_B}{S_p\sqrt{1/n_A+1/n_B}}\sim t_{n_A+n_B-2}$.

Why this works: pooling is valid only when the two population variances can be assumed equal — which is exactly what the preliminary F-test in part (a) checks; testing variance before mean (rather than jumping straight to a mean comparison) is standard practice precisely because the correct mean-test procedure depends on that answer.

Common pitfall: running the two-sample mean test without first checking the variance-equality assumption, or placing the smaller variance on top of the F ratio (the convention is always larger-over-smaller so the statistic exceeds 1, matching the tabulated upper-tail critical value).

Source: Walpole, Myers, Myers & Ye, Ch. 10 (Two-Sample Tests of Hypotheses).

Question 7 of 8: Insulating-Material Breakdown Voltage — Two-Sample Tests

Question 7: Insulating-Material Breakdown Voltage — Two-Sample Tests (20 marks)