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

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. $X\sim\text{Normal}$; $n=19$, $\Sigma X=589.0$, $\Sigma X^2=18{,}286.0$.

Find. (a) 99% CI for $\mu$ and for $\sigma$. (b) test $H_0:\mu=30.0$ at $\alpha=0.05$. (c) test $H_0:\sigma=1.1$ at $\alpha=0.05$.

Approach. Compute $\bar x$ and $s^2$ from the sums, then use the t-distribution (small $n$, $\sigma$ unknown) for the mean and the chi-square distribution for the standard deviation, both with $df=n-1=18$.

1. Sample statistics. $$\bar x=\dfrac{\Sigma X}{n}=\dfrac{589.0}{19}=31.0\ \text{MPa}$$ $$s^2=\dfrac{\Sigma X^2-\left(\Sigma X\right)^2/n}{n-1}=\dfrac{18{,}286.0-589.0^2/19}{18}=\dfrac{27.0}{18}=1.5\ \text{MPa}^2,\quad s=1.2247\ \text{MPa}$$
2. (a)(i) 99% CI for the true mean. With $t_{0.005,18}=2.878$: $$\bar x\pm t_{0.005,18}\dfrac{s}{\sqrt n}=31.0\pm 2.878\dfrac{1.2247}{\sqrt{19}}=31.0\pm 0.809$$ $$\boxed{30.191\ \text{MPa}\lt\mu\lt 31.809\ \text{MPa}}$$
3. (a)(ii) 99% CI for the true standard deviation. With $\chi^2_{0.995,18}=6.265$ and $\chi^2_{0.005,18}=37.156$: $$\dfrac{(n-1)s^2}{\chi^2_{0.005,18}}\lt\sigma^2\lt\dfrac{(n-1)s^2}{\chi^2_{0.995,18}}\ \Rightarrow\ \dfrac{27.0}{37.156}\lt\sigma^2\lt\dfrac{27.0}{6.265}$$ $$0.727\lt\sigma^2\lt 4.310\quad\Rightarrow\quad\boxed{0.852\ \text{MPa}\lt\sigma\lt 2.076\ \text{MPa}}$$
4. (b) Test $H_0:\mu=30.0$ vs. $H_1:\mu\ne 30.0$, $\alpha=0.05$. $$t=\dfrac{\bar x-\mu_0}{s/\sqrt n}=\dfrac{31.0-30.0}{1.2247/\sqrt{19}}=3.559$$ Critical value $t_{0.025,18}=2.101$. Since $|3.559|\gt 2.101$, reject $H_0$. $$\boxed{\text{Reject } H_0:\ \mu\text{ is significantly different from }30.0\text{ MPa}}$$
5. (c) Test $H_0:\sigma=1.1$ vs. $H_1:\sigma\ne 1.1$, $\alpha=0.05$. $$\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{18(1.5)}{1.1^2}=22.31$$ Critical bounds $\chi^2_{0.975,18}=8.231$ and $\chi^2_{0.025,18}=31.526$. Since $8.231\lt 22.31\lt 31.526$, fail to reject $H_0$. $$\boxed{\text{Fail to reject } H_0:\ \sigma\text{ is not significantly different from }1.1\text{ MPa}}$$