Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-2 Probability and Statistics — National Examination, May 2018. Closed book (approved calculator + one hand-written information sheet); any 5 of 8 questions constitute a complete paper (only 5 marked) — all 8 are solved below as a complete study resource.
Reference texts: Walpole, Myers, Myers & Ye, Probability and Statistics for Engineers and Scientists (9th ed.), Pearson — used throughout for point-estimation, confidence-interval, and hypothesis-testing formulas; Montgomery & Runger, Applied Statistics and Probability for Engineers, for the regression and correlation-inference formulas in Question 8.
Given. A sample of $n=16$ bag weights, assumed normal, summarized by $\sum X=636.0$ kg and $\sum X^2=25{,}293.0$ kg².
Given data
Symbol
Meaning
Value
$n$
Sample size
16
$\sum X$
Sum of weights
636.0 kg
$\sum X^2$
Sum of squared weights
25,293.0 kg²
$\alpha$
Significance level (b, c)
0.05
Find. (a) 99% CI for $\mu$ and for $\sigma$; (b) test $H_0:\mu=40.0$; (c) test $H_0:\sigma=0.6$.
Approach. Compute $\bar x$ and $s^2$ from the sums first. The mean's CI/test uses the $t_{n-1}$ distribution (population $\sigma$ unknown); the standard deviation's CI/test uses the $\chi^2_{n-1}$ distribution of $(n-1)s^2/\sigma^2$.
Sample mean and variance.
$$\bar x=\frac{\sum X}{n}=\frac{636.0}{16}=39.75\text{ kg},\qquad s^2=\frac{\sum X^2-n\bar x^2}{n-1}=\frac{25{,}293.0-16(39.75)^2}{15}=0.800\text{ kg}^2,$$
so $s=\sqrt{0.800}=0.8944$ kg.
a)(i) 99% CI for $\mu$. With $df=15$, $t_{0.005,15}=2.947$:
$$\bar x \pm t_{0.005,15}\frac{s}{\sqrt n}=39.75\pm 2.947\frac{0.8944}{\sqrt{16}}=39.75\pm 0.659,$$
$$\boxed{39.09\text{ kg} < \mu < 40.41\text{ kg}}.$$
a)(ii) 99% CI for $\sigma$. $(n-1)s^2/\sigma^2\sim\chi^2_{15}$; with $\chi^2_{0.005,15}=32.80$ and $\chi^2_{0.995,15}=4.60$:
$$\frac{(n-1)s^2}{\chi^2_{0.005,15}}<\sigma^2<\frac{(n-1)s^2}{\chi^2_{0.995,15}}\ \Rightarrow\ \frac{12.0}{32.80}<\sigma^2<\frac{12.0}{4.60}\ \Rightarrow\ 0.3658<\sigma^2<2.6082,$$
$$\boxed{0.605\text{ kg} < \sigma < 1.615\text{ kg}}.$$
b) Test $H_0:\mu=40.0$ vs. $H_1:\mu\ne 40.0$, $\alpha=0.05$.
$$t=\frac{\bar x-40.0}{s/\sqrt n}=\frac{39.75-40.0}{0.8944/4}=\boxed{-1.118},\qquad t_{0.025,15}=2.131.$$
Since $|t|=1.118<2.131$, we fail to reject $H_0$ — the mean is not significantly different from 40.0 kg.
c) Test $H_0:\sigma=0.6$ vs. $H_1:\sigma\ne 0.6$, $\alpha=0.05$.
$$\chi^2=\frac{(n-1)s^2}{\sigma_0^2}=\frac{15(0.800)}{0.36}=\boxed{33.33},\qquad \chi^2_{0.975,15}=6.26,\quad \chi^2_{0.025,15}=27.49.$$
Since $33.33>27.49$, we reject $H_0$ — the true standard deviation is significantly larger than 0.6 kg.
