04-BS-2, December 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. $n=21$ measurements, $\sum X_i=1{,}743.0$ MPa, $\sum X_i^2=144{,}749.0$ MPa². X assumed normal.

Find. (a) 99% CI for $\mu$ and $\sigma$; (b) test $H_0:\mu=82.0$ vs $\alpha=0.05$; (c) test $H_0:\sigma=1.6$ vs $\alpha=0.05$.

Approach. Compute the sample mean and variance from the sums, use the $t$-distribution ($n-1$ df) for the mean and the chi-square distribution ($n-1$ df) for the standard deviation, both for the confidence intervals and the hypothesis tests.

Sample statistics. $\bar X=\dfrac{1{,}743.0}{21}=83.0$ MPa. $s^2=\dfrac{\sum X_i^2-(\sum X_i)^2/n}{n-1}=\dfrac{144{,}749.0-1{,}743.0^2/21}{20}=\dfrac{144{,}749.0-144{,}669.0}{20}=\dfrac{80}{20}=\boxed{s^2=4.0}$, so $\boxed{s=2.0\text{ MPa}}$.
(a)(i) 99% CI for the true mean. $\bar X\pm t_{0.005,20}\dfrac{s}{\sqrt n}=83.0\pm(2.845)\dfrac{2.0}{\sqrt{21}}=83.0\pm1.242$, giving $\boxed{81.76\text{ MPa}<\mu<84.24\text{ MPa}}$.
(a)(ii) 99% CI for the true standard deviation. $\dfrac{(n-1)s^2}{\chi^2_{0.005,20}}<\sigma^2<\dfrac{(n-1)s^2}{\chi^2_{0.995,20}}$, i.e. $\dfrac{80}{39.997}<\sigma^2<\dfrac{80}{7.434}$, so $2.000<\sigma^2<10.762$. Taking square roots: $\boxed{1.414\text{ MPa}<\sigma<3.280\text{ MPa}}$.
(b) Test H₀: μ = 82.0 MPa (α = 0.05, two-tail, t-test). $t=\dfrac{\bar X-\mu_0}{s/\sqrt n}=\dfrac{83.0-82.0}{2.0/\sqrt{21}}=\dfrac{1.0}{0.4364}=\boxed{2.291}$. Critical value $t_{0.025,20}=2.086$. Since $|t|=2.291>2.086$, $\boxed{\text{reject }H_0}$ — the mean IS significantly different from 82.0 MPa at the 5% level.
(c) Test H₀: σ = 1.6 MPa (α = 0.05, two-tail, chi-square test). $\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{20(4.0)}{1.6^2}=\dfrac{80}{2.56}=\boxed{31.25}$. Critical values $\chi^2_{0.975,20}=9.591$ and $\chi^2_{0.025,20}=34.170$. Since $9.591<31.25<34.170$, $\boxed{\text{fail to reject }H_0}$ — the standard deviation is NOT significantly different from 1.6 MPa at the 5% level.

Question 5 — final results
Part	Quantity	Value
—	Sample mean, sample sd (n=21)	83.0 MPa, 2.0 MPa
(a)(i)	99% CI for μ	(81.76, 84.24) MPa
(a)(ii)	99% CI for σ	(1.414, 3.280) MPa
(b)	t (H₀: μ=82.0), t-crit	2.291, 2.086 → reject H₀
(c)	χ² (H₀: σ=1.6), crit range	31.25, (9.591, 34.170) → fail to reject H₀

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

Key relations: $\dfrac{\bar X-\mu}{s/\sqrt n}\sim t_{n-1}$; $\dfrac{(n-1)s^2}{\sigma^2}\sim\chi^2_{n-1}$; CI for $\mu$: $\bar X\pm t_{\alpha/2,n-1}s/\sqrt n$; CI for $\sigma^2$: $\left(\dfrac{(n-1)s^2}{\chi^2_{\alpha/2,n-1}},\dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2,n-1}}\right)$.

Why this works: for a normal population with unknown $\sigma$, the standardized sample mean follows Student's $t$ (accounting for the extra uncertainty of estimating $\sigma$ from the same sample), while the scaled sample variance follows a chi-square distribution — both derived from the normal sampling theory of $\bar X$ and $s^2$.

Common pitfall: swapping which chi-square percentile bounds the upper vs lower limit of the variance CI (the SMALLER $\chi^2$ percentile, $\chi^2_{1-\alpha/2}$, gives the LARGER variance bound, since variance and $\chi^2$ appear as a reciprocal-like ratio).

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

Question 5 of 8

Question 5 (20 marks)