04-BS-2, May 2015, Q5 | Nivaar Exam Prep

Question 5 (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. $n=20$ nitrogen-content measurements $X$ (lbs), $\sum X=158.00$, $\sum X^2=1{,}251.24$; $X$ assumed Normal.

Find. (a) 99% CI for $\mu$ and for $\sigma$; (b) test $H_0:\mu=8.00$ at $\alpha=0.05$; (c) test $H_0:\sigma=0.2$ at $\alpha=0.05$.

Approach. Compute the sample mean and variance from the given sums, then use the $t$ distribution (df $=n-1$) for the mean-based inferences and the $\chi^2$ distribution (df $=n-1$) for the variance-based inferences.

Sample statistics. $$\bar X=\frac{\sum X}{n}=\frac{158.00}{20}=\boxed{7.90\text{ lb}}$$ $$s^2=\frac{\sum X^2-(\sum X)^2/n}{n-1}=\frac{1251.24-158.00^2/20}{19}=\frac{3.04}{19}=\boxed{0.16}\quad (s=0.40\text{ lb})$$
(a)(i) 99% CI for the mean. $t_{0.005,19}=2.861$. $$\bar X\pm t_{0.005,19}\frac{s}{\sqrt n}=7.90\pm2.861\frac{0.40}{\sqrt{20}}=7.90\pm0.256$$ $$\boxed{7.644\text{ lb} < \mu < 8.156\text{ lb}}$$
(a)(ii) 99% CI for the standard deviation. $\chi^2_{0.995,19}=6.844$, $\chi^2_{0.005,19}=38.582$. $$\frac{(n-1)s^2}{\chi^2_{0.005,19}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{0.995,19}} \implies \frac{19(0.16)}{38.582} < \sigma^2 < \frac{19(0.16)}{6.844}$$ $$0.0788 < \sigma^2 < 0.4442 \implies \boxed{0.281\text{ lb} < \sigma < 0.666\text{ lb}}$$
(b) Test $H_0:\mu=8.00$ vs $H_1:\mu\ne8.00$. $$t=\frac{\bar X-\mu_0}{s/\sqrt n}=\frac{7.90-8.00}{0.40/\sqrt{20}}=\boxed{-1.118}$$ Critical value $t_{0.025,19}=2.093$. Since $|{-1.118}|<2.093$, fail to reject $H_0$: the true mean is not significantly different from 8.00 lb.
(c) Test $H_0:\sigma=0.2$ vs $H_1:\sigma\ne0.2$. $$\chi^2=\frac{(n-1)s^2}{\sigma_0^2}=\frac{19(0.16)}{0.2^2}=\boxed{76.0}$$ Critical values $\chi^2_{0.975,19}=8.907$, $\chi^2_{0.025,19}=32.852$. Since $76.0>32.852$, reject $H_0$: the true standard deviation IS significantly different from 0.2 lb (the sample's $s=0.40$ lb is double the hypothesized value).

Question 5 — final results
Quantity	Value
$\bar X$, $s$	7.90 lb, 0.40 lb
99% CI for $\mu$	(7.644, 8.156) lb
99% CI for $\sigma$	(0.281, 0.666) lb
(b) $t$, conclusion	−1.118; fail to reject $\mu=8.00$
(c) $\chi^2$, conclusion	76.0; reject $\sigma=0.2$

Topic: Confidence intervals and hypothesis tests for the mean and variance of a Normal population (single sample).

Key relations: $\bar X\pm t_{\alpha/2,n-1}\,s/\sqrt n$ for $\mu$; $\left(\dfrac{(n-1)s^2}{\chi^2_{\alpha/2,n-1}},\ \dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2,n-1}}\right)$ for $\sigma^2$; $t=(\bar X-\mu_0)/(s/\sqrt n)$ and $\chi^2=(n-1)s^2/\sigma_0^2$ for the respective tests.

Why this works: $(\bar X-\mu)/(s/\sqrt n)$ follows Student's $t$ with $n-1$ df because $s$ (not $\sigma$) is used, adding uncertainty beyond the Normal; $(n-1)s^2/\sigma^2$ follows $\chi^2_{n-1}$ by definition of the sample variance's sampling distribution under Normality.

Common pitfall: Reversing the two $\chi^2$ critical values when building the variance CI (the smaller $\chi^2$ percentile goes with the LARGER bound on $\sigma^2$, since $\chi^2$ is in the denominator); using $z$ instead of $t$ for a small sample.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 8–9 (Estimation & Hypothesis Testing).

Question 5 of 8

Question 5 (20 marks)