Estimation and Hypothesis Testing — Young's Modulus: 04-BS-2, May 2013, Q5

Question 5: Estimation and Hypothesis Testing — Young's Modulus (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.

Given data
n	17
ΣX	544.0 MPa
ΣX²	17,444.0 MPa²

Find. (a) 99% CI for μ and σ; (b) test H₀: μ=31.0 at α=0.05; (c) test H₀: σ=1.3 at α=0.05.

Approach. Compute the sample mean and variance from the sums, then use the t-distribution (df = n−1) for the mean's CI and t-test, and the χ²-distribution (df = n−1) for the standard deviation's CI and test, since n is small (17) and the population is stated Normal.

Sample statistics. $$\bar x=\dfrac{544.0}{17}=32.0\text{ MPa},\qquad s^2=\dfrac{\Sigma x^2-(\Sigma x)^2/n}{n-1}=\dfrac{17444.0-544.0^2/17}{16}=\dfrac{36.0}{16}=2.25\ \Rightarrow\ s=1.5\text{ MPa}.$$
(a)(i) 99% CI for the mean. With df = 16, $t_{0.005,16}=2.921$: $$\bar x\pm t_{0.005,16}\dfrac{s}{\sqrt{n}}=32.0\pm2.921\dfrac{1.5}{\sqrt{17}}=32.0\pm1.063\ \Rightarrow\ \boxed{30.94<\mu<33.06\text{ MPa}}.$$
(a)(ii) 99% CI for the standard deviation. With $(n-1)s^2=16(2.25)=36.0$ and $\chi^2_{0.005,16}=34.27$, $\chi^2_{0.995,16}=5.142$: $$\sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.005,16}}}<\sigma<\sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.995,16}}}\ \Rightarrow\ \sqrt{\dfrac{36.0}{34.27}}<\sigma<\sqrt{\dfrac{36.0}{5.142}}\ \Rightarrow\ \boxed{1.025<\sigma<2.646\text{ MPa}}.$$
(b) t-test for the mean, H₀: μ=31.0 vs H₁: μ≠31.0. $$t=\dfrac{\bar x-\mu_0}{s/\sqrt n}=\dfrac{32.0-31.0}{1.5/\sqrt{17}}=\boxed{2.749}.$$ Critical value $t_{0.025,16}=2.120$. Since $|t|=2.749>2.120$, reject H₀: the true mean is significantly different from 31.0 MPa at α=0.05.
(c) χ²-test for the standard deviation, H₀: σ=1.3 vs H₁: σ≠1.3. $$\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{16(2.25)}{1.3^2}=\dfrac{36.0}{1.69}=\boxed{21.30}.$$ Critical region: reject if $\chi^2<\chi^2_{0.975,16}=6.908$ or $\chi^2>\chi^2_{0.025,16}=28.85$. Since $6.908<21.30<28.85$, fail to reject H₀: the true standard deviation is not significantly different from 1.3 MPa at α=0.05.

Question 5 — final results
Part	Quantity	Value
—	x̄, s	32.0 MPa, 1.5 MPa
(a)(i)	99% CI for μ	(30.94, 33.06) MPa
(a)(ii)	99% CI for σ	(1.025, 2.646) MPa
(b)	t, decision (μ=31.0)	2.749, reject H₀
(c)	χ², decision (σ=1.3)	21.30, fail to reject H₀

Topic: Small-sample estimation and hypothesis testing for a Normal mean and standard deviation.

Key relations: $\dfrac{\bar X-\mu}{s/\sqrt n}\sim t_{n-1}$; $\dfrac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}$; two-sided CIs invert the corresponding test statistic at the α/2 and 1−α/2 critical points.

Why this works: with only 17 observations and an unknown population standard deviation, the sample standard deviation itself carries sampling error, so the t-distribution (heavier tails than Normal) is the exact sampling distribution of the standardized mean, and the χ² distribution is the exact sampling distribution of the scaled sample variance — both are exact (not approximate) whenever the population is genuinely Normal.

Common pitfall: using the Normal $z$ critical value instead of $t_{16}$ for a sample this small (understates the CI width and overstates test significance), or using symmetric ± bounds for the χ²-based CI on σ (the χ² distribution is not symmetric, so the two critical values are not mirror images).

Source: Walpole, Myers, Myers & Ye, Ch. 9 (Estimation), Ch. 10 (Hypothesis Testing for One Population).

Question 5 of 8: Estimation and Hypothesis Testing — Young's Modulus

Question 5: Estimation and Hypothesis Testing — Young's Modulus (20 marks)