Shear Modulus of Steel — Two-Sample Comparison: 04-BS-2, December 2019, Q7

Question 7: Shear Modulus of Steel — Two-Sample Comparison (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. $n_A=10$, $m_A=80.1$, $s_A=1.0$; $n_B=9$, $m_B=79.8$, $s_B=1.2$ (units GPa); assume both underlying populations are normal and independent.

Find. (a) 95% CI for $\sigma_B/\sigma_A$. (b) test H₀: $\sigma_A=\sigma_B$ at α=0.05. (c) test H₀: $\mu_A=\mu_B$ at α=0.05.

Approach. (a)–(b) use the F-distribution for the ratio of two sample variances (df$_B$=8, df$_A$=9). Since (b) shows the variances are not significantly different, (c) uses the pooled two-sample t-test with df=$n_A+n_B-2=17$.

(a) 95% CI for σ_B/σ_A. Assuming both populations are normal and independent, the CI for the variance ratio is $$\dfrac{s_B^2/s_A^2}{F_{0.025,\,8,9}}\lt\dfrac{\sigma_B^2}{\sigma_A^2}\lt\left(\dfrac{s_B^2}{s_A^2}\right)F_{0.025,\,9,8}$$ With $s_B^2/s_A^2=1.44/1.0=1.44$, $F_{0.025,8,9}=3.23$, $F_{0.025,9,8}=3.39$: $$\dfrac{1.44}{3.23}\lt\dfrac{\sigma_B^2}{\sigma_A^2}\lt1.44(3.39) \ \Rightarrow\ 0.446\lt\dfrac{\sigma_B^2}{\sigma_A^2}\lt4.882$$ Taking square roots, $$\boxed{0.668\lt\dfrac{\sigma_B}{\sigma_A}\lt2.209}$$
(b) Test H₀: σ_A=σ_B vs H₁: σ_A≠σ_B, α=0.05. $$F=\dfrac{s_B^2}{s_A^2}=\dfrac{1.44}{1.0}=1.44 \qquad (\text{df}=8,9)$$ The critical value is $F_{0.025,8,9}=3.23$. Since $1.44\lt3.23$, $$\boxed{\text{Fail to reject } H_0\text{: the two process standard deviations are not significantly different}}$$
(c) Test H₀: μ_A=μ_B vs H₁: μ_A≠μ_B, α=0.05. Since (b) found no significant difference in variance, use the pooled t-test: $$s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{9(1.0)+8(1.44)}{17}=\dfrac{20.52}{17}=1.207 \qquad s_p=1.099$$ $$t=\dfrac{m_A-m_B}{s_p\sqrt{\frac{1}{n_A}+\frac{1}{n_B}}}=\dfrac{80.1-79.8}{1.099\sqrt{0.1+0.1111}}=\dfrac{0.3}{0.505}=0.594$$ df=17, and $t_{0.025,17}=2.110$. Since $|t|=0.594\lt2.110$, $$\boxed{\text{Fail to reject } H_0\text{: the mean shear modulus of the two processes is not significantly different}}$$

Question 7 — summary of results
Part	Quantity	Result
(a)	95% CI for σ_B/σ_A	(0.668, 2.209)
(b)	F-statistic (H₀: σ_A=σ_B)	1.44 — fail to reject
(c)	pooled t-statistic (H₀: μ_A=μ_B)	0.594 — fail to reject

Topic: F-distribution for comparing two variances; pooled two-sample t-test for comparing two means.

Key relations: $F=s_1^2/s_2^2\sim F_{n_1-1,n_2-1}$ under equal variances; pooled t uses $s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}$ and $t=\dfrac{\bar X_A-\bar X_B}{s_p\sqrt{1/n_A+1/n_B}}$, df=$n_A+n_B-2$.

Why this works: the ratio of two independent scaled chi-square variables (each sample variance, scaled by its own σ²) follows an F-distribution; pooling the two sample variances into a single estimate is valid precisely because (b) established they estimate a common σ².

Common pitfall: testing the means with a pooled t-test WITHOUT first checking equal variances (an unequal-variance/Welch t-test would be needed if (b) had rejected), and mixing up which sample's variance goes in the numerator vs. denominator of F (convention: larger sample variance on top gives F≥1).

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 9–10.

Question 7 of 8: Shear Modulus of Steel — Two-Sample Comparison

Question 7: Shear Modulus of Steel — Two-Sample Comparison (20 marks)