Confidence Intervals and Hypothesis Tests — Brinell Hardness: 04-BS-2, December 2018, 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.

Given data
Quantity	Value
ΣX	1,800.0
ΣX²	129,654.0
n	25

Find. (a) 99% CI for the true mean and true σ. (b) Test H₀: μ=74.0 at α=0.05. (c) Test H₀: σ=1.2 at α=0.05.

Approach. Compute the sample mean and variance from the raw sums, then use the t-distribution (unknown σ) for the mean interval/test and the χ² distribution for the variance/sd interval/test, all with df = n−1 = 24.

Sample statistics. $$\bar x=\dfrac{\Sigma X}{n}=\dfrac{1800.0}{25}=72.0,\qquad s^2=\dfrac{\Sigma X^2-(\Sigma X)^2/n}{n-1}=\dfrac{129654.0-1800.0^2/25}{24}=\dfrac{54.0}{24}$$ $$\boxed{\bar x=72.0,\quad s^2=2.25\ (s=1.5)}$$
(a)(i) 99% CI for the true mean. Unknown σ, use $t_{0.005,24}=2.797$ (Appendix 2): $$\bar x\pm t_{0.005,24}\dfrac{s}{\sqrt n}=72.0\pm2.797\dfrac{1.5}{\sqrt{25}}=72.0\pm0.839$$ $$\boxed{71.161\lt \mu\lt 72.839}$$
(a)(ii) 99% CI for the true standard deviation. With df=24, $\chi^2_{0.995,24}=9.89$ and $\chi^2_{0.005,24}=45.55$ (Appendix 3): $$\dfrac{(n-1)s^2}{\chi^2_{0.005,24}}\lt \sigma^2\lt \dfrac{(n-1)s^2}{\chi^2_{0.995,24}} \ \Rightarrow\ \dfrac{54.0}{45.55}\lt \sigma^2\lt \dfrac{54.0}{9.89}$$ i.e. $1.185\lt \sigma^2\lt 5.460$; taking square roots: $$\boxed{1.089\lt \sigma\lt 2.337}$$
(b) Test H₀: μ=74.0 vs H₁: μ≠74.0, α=0.05. σ unknown → t-test, df=24: $$t=\dfrac{\bar x-74.0}{s/\sqrt n}=\dfrac{72.0-74.0}{1.5/5}=\dfrac{-2.0}{0.3}=-6.667$$ Critical value $t_{0.025,24}=2.064$ (two-tailed). Since $|t|=6.667\gt 2.064$: $$\boxed{\text{Reject }H_0\text{ -- the mean is significantly different from 74.0}}$$
(c) Test H₀: σ=1.2 vs H₁: σ≠1.2, α=0.05. $$\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{24(2.25)}{1.2^2}=\dfrac{54.0}{1.44}=37.5$$ Critical values (df=24, two-tailed): $\chi^2_{0.975,24}=12.40$ and $\chi^2_{0.025,24}=39.36$. Since $12.40\lt 37.5\lt 39.36$: $$\boxed{\text{Fail to reject }H_0\text{ -- }\sigma\text{ is not significantly different from 1.2}}$$

Question 5 — final results
Part	Quantity	Result
	&bar;x, s²	72.0, 2.25
(a)(i)	99% CI for μ	(71.161, 72.839)
(a)(ii)	99% CI for σ	(1.089, 2.337)
(b)	t (H₀: μ=74.0)	−6.667 → reject H₀
(c)	χ² (H₀: σ=1.2)	37.5 → fail to reject H₀

Topic: Estimation and hypothesis testing for a normal population mean and standard deviation, σ unknown.

Key relations: $\bar x\pm t_{\alpha/2,n-1}\dfrac{s}{\sqrt n}$; $\dfrac{(n-1)s^2}{\chi^2_{\alpha/2,n-1}}\lt \sigma^2\lt \dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2,n-1}}$; $t=\dfrac{\bar x-\mu_0}{s/\sqrt n}$; $\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}$.

Why this works: With σ unknown and estimated by s, $\dfrac{\bar X-\mu}{S/\sqrt n}$ follows a Student-t (not normal) because s itself is a random variable adding extra variability; $(n-1)S^2/\sigma^2$ follows a χ² distribution by the sampling-distribution theory of the sample variance under normality.

Common pitfall: Using $z$ instead of $t$ when σ is unknown; swapping which tail of the (asymmetric) χ² distribution is the lower vs. upper limit when building the variance CI.

Source: Walpole et al., Probability & Statistics for Engineers and Scientists, Ch. 9 (Estimation) & Ch. 10 (Hypothesis Tests, One Sample).

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

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