Question 7 of 8: Two-Sample Tests — Comparing Two Processes
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, December 2015 — 04-BS-2 Probability and Statistics (2 hours, closed book except one hand-written information sheet and an approved calculator). Statistical tables of the Normal, t, chi-square and F distributions are supplied with the exam. The rubric states "Any 5 questions constitute a complete paper"; every question is answered.
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed. (Pearson) — used throughout for normal/binomial/Poisson/hypergeometric probability, confidence intervals, hypothesis tests, and simple linear regression.
Check — source defect in Question 4. As literally printed, K/y cannot integrate
to a finite value over (1, ∞), so the density as given is not valid. This solution
reconstructs the intended density as f(y) = 1/y² for y > 1 — the unique
minimal correction (recovering one dropped exponent) that makes the function integrate to 1 on the stated
domain, and it is the standard "infinite mean" example from this course's own textbook (Walpole). This choice is disclosed here rather than presented as certain.
Question 7: Two-Sample Tests — Comparing Two Processes (20 marks)
Approach. Compare variances with the F-test first (needed to decide whether to pool); if variances aren't significantly different, use the pooled two-sample t-test for the means.
(a) F-test for equal variances.Assumption: both processes' shear-modulus measurements are independent, normally distributed random samples. Put the larger sample variance on top:
$$F=\dfrac{s_B^2}{s_A^2}=\dfrac{1.1^2}{0.9^2}=\dfrac{1.21}{0.81}=\boxed{1.494},\qquad F_{0.025,\,8,\,9}=4.10.$$
Since $F=1.494<4.10$, we fail to reject H₀: the two standard deviations are not significantly different, so pooling the variances for part (b) is justified.
(b) Pooled two-sample t-test for the means.
$$s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{9(0.81)+8(1.21)}{17}=\dfrac{16.97}{17}=0.9982,\qquad s_p=0.9991.$$
$$t=\dfrac{\bar X_A-\bar X_B}{s_p\sqrt{1/n_A+1/n_B}}=\dfrac{80.1-79.8}{0.9991\sqrt{1/10+1/9}}=\dfrac{0.3}{0.4591}=\boxed{0.653},\qquad t_{0.025,\,17}=2.110.$$
Since $|t|=0.653<2.110$, we fail to reject H₀: the mean shear modulus of Process A is not significantly different from that of Process B.
