04-BS-2, May 2017, 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. Sample of $n=14$ scaffold maximum loads, $\sum X=4{,}900.0$ kg, $\sum X^2=1{,}716{,}872.0$ kg$^2$; $X$ assumed normal.

Find. (a) 99% CI for the true mean $\mu$ and true standard deviation $\sigma$; (b) two-sided test $H_0:\mu=360$ at $\alpha=0.05$; (c) two-sided test $H_0:\sigma=10$ at $\alpha=0.05$.

Approach. Compute $\bar x$ and $s^2$ from the sums, use the $t$-distribution for the mean's CI/test (population variance unknown) and the $\chi^2$-distribution for the variance's CI/test.

Sample statistics. $\bar x=\dfrac{\sum X}{n}=\dfrac{4900}{14}=350.0$ kg. $S_{xx}=\sum X^2-\dfrac{(\sum X)^2}{n}=1{,}716{,}872-\dfrac{4900^2}{14}=1{,}872.0$. $s^2=\dfrac{S_{xx}}{n-1}=\dfrac{1872}{13}=144.0$, so $s=12.0$ kg.
(a)(i) 99% CI for the mean. With $df=13$, $t_{0.005,13}=3.012$. Margin $=t\cdot\dfrac{s}{\sqrt n}=3.012\left(\dfrac{12}{\sqrt{14}}\right)=9.661$. So $\boxed{340.3\text{ kg}<\mu<359.7\text{ kg}}$.
(a)(ii) 99% CI for the standard deviation. Using $\chi^2_{0.995,13}=3.565$ and $\chi^2_{0.005,13}=29.82$: $\sigma_{\text{lo}}=\sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.005,13}}}=\sqrt{\dfrac{1872}{29.82}}=7.92$ kg, $\sigma_{\text{hi}}=\sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.995,13}}}=\sqrt{\dfrac{1872}{3.565}}=22.92$ kg. So $\boxed{7.92\text{ kg}<\sigma<22.92\text{ kg}}$.
(b) Test $H_0:\mu=360$ vs $H_1:\mu\ne360$. $t=\dfrac{\bar x-360}{s/\sqrt n}=\dfrac{350-360}{12/\sqrt{14}}=-3.118$. Critical value $t_{0.025,13}=2.160$. Since $|t|=3.118>2.160$, $\boxed{\text{reject }H_0\text{: the mean is significantly different from 360 kg}}$.
(c) Test $H_0:\sigma=10$ vs $H_1:\sigma\ne10$. $\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{13(144)}{100}=18.72$. Critical values (two-sided, $\alpha=0.05$, $df=13$): $\chi^2_{0.975,13}=5.01$ and $\chi^2_{0.025,13}=24.74$. Since $5.01<18.72<24.74$, $\boxed{\text{fail to reject }H_0\text{: }\sigma\text{ is not significantly different from 10 kg}}$.

Question 5 — final results
Part	Quantity	Value
	$\bar x$; $s$	350.0 kg; 12.0 kg
(a)(i)	99% CI for μ	(340.34, 359.66) kg
(a)(ii)	99% CI for σ	(7.92, 22.92) kg
(b)	t-stat; conclusion (μ=360)	−3.118; reject H₀
(c)	χ²-stat; conclusion (σ=10)	18.72; fail to reject H₀

Topic: Confidence intervals and hypothesis tests for the mean and variance of a normal population, small sample.

Key relations: $\dfrac{\bar X-\mu}{S/\sqrt n}\sim t_{n-1}$ gives both the CI and the test for $\mu$; $\dfrac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}$ gives both the CI and the test for $\sigma^2$ (hence $\sigma$).

Why this works: with $\sigma$ unknown and estimated from the same sample, standardizing the mean introduces extra variability captured exactly by the $t$-distribution (heavier tails than normal); the $\chi^2$ distribution arises because a sum of squared standard normals is $\chi^2$ by definition, and $(n-1)S^2/\sigma^2$ is exactly such a sum after removing one degree of freedom for estimating $\bar x$.

Common pitfall: using the SAME single $\chi^2$ critical value for both ends of the variance CI/test (the chi-square is not symmetric, so the low and high critical values are genuinely different numbers, unlike the $\pm t$ used for the mean).

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 9 (Estimation) and Ch. 10 (Hypothesis Testing).

Question 5 of 8

Question 5 (20 marks)