Question 5 of 8: Confidence Intervals and Hypothesis Tests on Young's Modulus
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 5: Confidence Intervals and Hypothesis Tests on Young's Modulus (20 marks)
Find. (a) 99% CI for μ and σ; (b) test H₀: μ=31.0 at α=0.05; (c) test H₀: σ=1.6 at α=0.05.
Approach. Compute $\bar X$ and $s^2$ from the sums, then use the $t$-distribution (df=n−1) for the mean and the $\chi^2$-distribution for the variance, since $n$ is small and $X$ is assumed normal.
(a)(i) 99% CI for μ. With df$=20$, $t_{0.005,20}=2.845$:
$$\bar X\pm t_{0.005,20}\dfrac{s}{\sqrt n}=30.0\pm2.845\times\dfrac{2.0}{\sqrt{21}}=30.0\pm1.242.$$
$$\boxed{28.76\text{ MPa}<\mu<31.24\text{ MPa}}$$
(a)(ii) 99% CI for σ. Using $\dfrac{(n-1)s^2}{\chi^2_{0.005,20}}<\sigma^2<\dfrac{(n-1)s^2}{\chi^2_{0.995,20}}$ with $\chi^2_{0.005,20}=40.00$ and $\chi^2_{0.995,20}=7.43$:
$$\dfrac{20\times4.0}{40.00}<\sigma^2<\dfrac{20\times4.0}{7.43}\ \Rightarrow\ 2.00<\sigma^2<10.77.$$
Taking square roots, $\boxed{1.41\text{ MPa}<\sigma<3.28\text{ MPa}}$.
(b) Test H₀: μ=31.0 vs H₁: μ≠31.0, α=0.05.
$$t=\dfrac{\bar X-31.0}{s/\sqrt n}=\dfrac{30.0-31.0}{2.0/\sqrt{21}}=\boxed{-2.291},\qquad t_{0.025,20}=\pm2.086.$$
Since $|t|=2.291>2.086$, reject H₀: the mean is significantly different from 31.0 MPa at the 5% level.
(c) Test H₀: σ=1.6 vs H₁: σ≠1.6, α=0.05.
$$\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{20\times4.0}{1.6^2}=\boxed{31.25},\qquad \chi^2_{0.975,20}=9.59,\ \ \chi^2_{0.025,20}=34.17.$$
Since $9.59<31.25<34.17$, the test statistic falls inside the acceptance region, so we fail to reject H₀: σ is not significantly different from 1.6 MPa.
