Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, May 2016 — 04-BS-2, Probability and Statistics (2 hours, closed book except one hand-written information sheet and an approved calculator; any 5 of 8 questions constitute a complete paper — all 8 are answered here as a full study resource).
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed. (Pearson) — sampling distributions, binomial/Poisson/hypergeometric models, confidence intervals, hypothesis tests, correlation and simple linear regression.
Find. (a) test $H_0:\sigma_A=\sigma_B$ via an F-test; (b) test $H_0:\mu_A=\mu_B$, choosing pooled or Welch's $t$ per the result of (a).
Approach. Compare the two sample variances with an F-test (assuming both populations are approximately normal and independent) to decide whether pooling is justified, then run the corresponding two-sample $t$-test for the means.
(a) F-test for equal variances. Assumption: both tensile-strength populations are independently and approximately normally distributed. Put the larger variance on top: $F=\dfrac{s_B^2}{s_A^2}=\dfrac{7.8^2}{5.5^2}=\dfrac{60.84}{30.25}=2.011$. Critical value $F_{0.025,11,10}=3.665$ (two-tailed, $\alpha=0.05$). Since $2.011<3.665$, $\boxed{\text{fail to reject }H_0}$ — the variances are not significantly different, so pooling is justified.
(b) Pooled two-sample t-test for the means. $s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{10(30.25)+11(60.84)}{21}=46.27$, $s_p=6.802$. $t=\dfrac{m_A-m_B}{s_p\sqrt{1/n_A+1/n_B}}=\dfrac{520-528}{6.802\sqrt{1/11+1/12}}=\dfrac{-8}{2.840}=-2.817$. Critical value $t_{0.025,21}=2.080$. Since $|-2.817|>2.080$, $\boxed{\text{reject }H_0}$ — the mean tensile strengths ARE significantly different.
