Confidence Intervals and Hypothesis Tests — Brinell Hardness: 04-BS-2, December 2014, Q5

Question 5: Confidence Intervals and Hypothesis Tests — Brinell Hardness (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=12 Brinell hardness measurements, $\Sigma X=2{,}460$, $\Sigma X^2=504{,}476$, X assumed Normal.

Find. (a) 99% CI for μ and σ; (b) test H₀: μ=208 at α=0.05; (c) test H₀: σ=3.0 at α=0.05.

Approach. Compute $\bar x$ and $s^2$ from the sums, then use the t-distribution (df=n−1) for the mean's CI/test and the χ² distribution (df=n−1) for the standard deviation's CI/test.

Sample statistics. $\bar x=\dfrac{2460}{12}=205.0$. The corrected sum of squares is $S_{xx}=\Sigma X^2-\dfrac{(\Sigma X)^2}{n}=504476-\dfrac{2460^2}{12}=176.0$, so $$s^2=\dfrac{S_{xx}}{n-1}=\dfrac{176.0}{11}=16.0,\qquad s=\boxed{4.0}.$$
(a)(i) 99% CI for the mean. With df=11, $t_{0.005,11}=3.106$: $$\bar x\pm t_{0.005,11}\dfrac{s}{\sqrt n}=205.0\pm3.106\dfrac{4.0}{\sqrt{12}}=205.0\pm3.586=\boxed{(201.41,\ 208.59)}.$$
(a)(ii) 99% CI for the standard deviation. With $\chi^2_{0.005,11}=26.76$ and $\chi^2_{0.995,11}=2.603$: $$\left(\sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.005,11}}},\ \sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.995,11}}}\right)=\left(\sqrt{\dfrac{176.0}{26.76}},\ \sqrt{\dfrac{176.0}{2.603}}\right)=\boxed{(2.565,\ 8.222)}.$$
(b) Test H₀: μ=208 vs H₁: μ≠208. $$t=\dfrac{\bar x-208}{s/\sqrt n}=\dfrac{205.0-208}{4.0/\sqrt{12}}=\boxed{-2.598}.$$ The critical value is $t_{0.025,11}=2.201$. Since $|t|=2.598>2.201$, reject H₀: the true mean IS significantly different from 208 at α=0.05.
(c) Test H₀: σ=3.0 vs H₁: σ≠3.0. $$\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{11(16.0)}{3.0^2}=\boxed{19.556}.$$ The two-tailed critical bounds are $\chi^2_{0.975,11}=3.816$ and $\chi^2_{0.025,11}=21.92$. Since $3.816<19.556<21.92$, fail to reject H₀: the true standard deviation is NOT significantly different from 3.0 at α=0.05.

Final results — Question 5
Part	Quantity	Result
(a)(i)	99% CI for μ	(201.41, 208.59)
(a)(ii)	99% CI for σ	(2.565, 8.222)
(b)	t-stat vs t-crit; conclusion	−2.598 vs 2.201; reject H₀
(c)	χ² vs (3.816, 21.92); conclusion	19.556; fail to reject H₀

Topic: Confidence intervals and hypothesis tests for the mean and variance of a Normal population from summary sums.

Key relations: $\bar x\pm t_{\alpha/2,n-1}\,s/\sqrt n$ for μ; $\left(\sqrt{(n-1)s^2/\chi^2_{\alpha/2}},\sqrt{(n-1)s^2/\chi^2_{1-\alpha/2}}\right)$ for σ; test statistics $t=(\bar x-\mu_0)/(s/\sqrt n)$ and $\chi^2=(n-1)s^2/\sigma_0^2$.

Why this works: for a Normal population, $\bar X$ is Normal (hence t after studentizing by the estimated s) and $(n-1)S^2/\sigma^2$ is exactly χ² with n−1 df — both are classical exact pivotal quantities, not large-sample approximations.

Common pitfall: using the same (symmetric) ± form for the σ interval as for μ — the χ² distribution is skewed, so its two critical values are NOT symmetric about the mean and must be looked up separately from the upper and lower tails.

Source: Walpole, Myers, Myers & Ye, Ch. 9–10 (Estimation and hypothesis testing).

Question 5 of 8: Confidence Intervals and Hypothesis Tests — Brinell Hardness

Question 5: Confidence Intervals and Hypothesis Tests — Brinell Hardness (20 marks)