04-BS-2, May 2016, 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 data (n=12 ladders)
ΣX	2,700.0 kg
ΣX²	609,975.0 kg²
n	12

Given. $n=12$, $\Sigma x=2{,}700.0$ kg, $\Sigma x^2=609{,}975.0$ kg$^2$, X normally distributed.

Find. (a) 99% CI for μ and for σ; (b) test $H_0:\mu=220$ vs $H_1:\mu\ne220$ at $\alpha=0.05$; (c) test $H_0:\sigma=10$ vs $H_1:\sigma\ne10$ at $\alpha=0.05$.

Approach. Compute the sample mean and variance from the sums, then use the $t$-distribution (df=$n-1$) for the mean's confidence interval/test and the χ² distribution (df=$n-1$) for the standard deviation's confidence interval/test.

Sample statistics. $\bar x=\dfrac{2700}{12}=225$ kg. $S_{xx}=\Sigma x^2-\dfrac{(\Sigma x)^2}{n}=609{,}975-\dfrac{2700^2}{12}=2{,}475$. $s^2=\dfrac{S_{xx}}{n-1}=\dfrac{2475}{11}=225$, so $s=15$ kg.
(a)(i) 99% CI for μ, df=11. $t_{0.005,11}=3.106$. $\bar x\pm t\,\dfrac{s}{\sqrt n}=225\pm3.106\dfrac{15}{\sqrt{12}}=225\pm13.45$, so $\boxed{211.55\text{ kg}<\mu<238.45\text{ kg}}$.
(a)(ii) 99% CI for σ, df=11. $\chi^2_{0.995,11}=26.757$, $\chi^2_{0.005,11}=2.603$. $\sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.995}}}<\sigma<\sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.005}}}$: $\sqrt{\dfrac{2475}{26.757}}=9.62$, $\sqrt{\dfrac{2475}{2.603}}=30.83$, so $\boxed{9.62\text{ kg}<\sigma<30.83\text{ kg}}$.
(b) t-test for the mean. $t=\dfrac{\bar x-\mu_0}{s/\sqrt n}=\dfrac{225-220}{15/\sqrt{12}}=\dfrac{5}{4.330}=1.155$. Critical value $t_{0.025,11}=2.201$. Since $|1.155|<2.201$, $\boxed{\text{fail to reject }H_0}$ — the mean is not significantly different from 220 kg.
(c) χ²-test for the variance/std. dev. $\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{11(225)}{10^2}=24.75$. Critical values $\chi^2_{0.025,11}=3.816$, $\chi^2_{0.975,11}=21.92$. Since $24.75>21.92$, $\boxed{\text{reject }H_0}$ — the true standard deviation IS significantly different (greater) than 10 kg.

Question 5 — final results
Part	Quantity	Value
—	Sample mean, variance, sd	225 kg, 225 kg², 15 kg
(a)(i)	99% CI for μ	(211.55, 238.45) kg
(a)(ii)	99% CI for σ	(9.62, 30.83) kg
(b)	t-test, μ₀=220	t=1.155 < 2.201 → fail to reject
(c)	χ²-test, σ₀=10	χ²=24.75 > 21.92 → reject

Topic: Confidence intervals and hypothesis tests for a normal mean and variance from summary sums.

Key relations: $S_{xx}=\Sigma x^2-(\Sigma x)^2/n$; $t$-interval for $\mu$: $\bar x\pm t_{\alpha/2,n-1}s/\sqrt n$; χ²-interval for $\sigma$ uses $(n-1)s^2/\sigma^2\sim\chi^2_{n-1}$.

Why this works: with $\sigma$ unknown, standardizing $\bar X$ by the sample $s$ introduces the extra variability of estimating $s$, which the $t$-distribution (heavier tails than $Z$) correctly accounts for; the quantity $(n-1)s^2/\sigma^2$ follows a χ² distribution because it is a sum of squared standardized normal deviations, giving an asymmetric interval for σ.

Common pitfall: using $z$ instead of $t$ for a small sample with unknown σ, and swapping the upper/lower χ² percentiles when building the variance interval (the SMALLER χ² percentile goes with the LARGER bound on σ).

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

Question 5 of 8

Question 5 (20 marks)