Heat-Producing Capacity of Anthracite (Estimation and Testing): 04-BS-2, December 2019, Q5

Question 5: Heat-Producing Capacity of Anthracite (Estimation and Testing) (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, $\Sigma X=189.0$, $\Sigma X^2=1706.0$, X normally distributed.

Given data
n	ΣX	ΣX²
21	189.0	1706.0

Find. (a) 99% CI for the true mean and true standard deviation. (b) test H₀: μ=8.8 at α=0.05. (c) test H₀: σ=0.4 at α=0.05.

Approach. Compute the sample mean and variance from the sums, then use the t-distribution (df=n−1) for the mean's CI/test and the chi-square distribution (df=n−1) for the standard deviation's CI/test.

Sample statistics. $$\bar X=\dfrac{\Sigma X}{n}=\dfrac{189.0}{21}=9.0, \qquad s^2=\dfrac{\Sigma X^2-n\bar X^2}{n-1}=\dfrac{1706.0-21(9.0)^2}{20}=\dfrac{5.0}{20}=0.25$$ $$\boxed{\bar X=9.0, \qquad s^2=0.25,\ \ s=0.5}$$
(a)(i) 99% CI for the true mean. With df=20, $t_{0.005,20}=2.845$: $$\bar X\pm t_{0.005,20}\dfrac{s}{\sqrt n}=9.0\pm 2.845\dfrac{0.5}{\sqrt{21}}=9.0\pm0.3105$$ $$\boxed{8.690\lt\mu\lt9.311}$$
(a)(ii) 99% CI for the true standard deviation. With df=20, $\chi^2_{0.995,20}=7.43$ and $\chi^2_{0.005,20}=40.00$: $$\dfrac{(n-1)s^2}{\chi^2_{0.005,20}}\lt\sigma^2\lt\dfrac{(n-1)s^2}{\chi^2_{0.995,20}} \ \Rightarrow\ \dfrac{20(0.25)}{40.00}\lt\sigma^2\lt\dfrac{20(0.25)}{7.43}$$ $$0.125\lt\sigma^2\lt0.673 \ \Rightarrow\ \boxed{0.354\lt\sigma\lt0.820}$$
(b) Test H₀: μ=8.8 vs H₁: μ≠8.8, α=0.05. $$t=\dfrac{\bar X-\mu_0}{s/\sqrt n}=\dfrac{9.0-8.8}{0.5/\sqrt{21}}=1.833$$ The two-sided critical value is $t_{0.025,20}=2.086$. Since $|t|=1.833\lt2.086$, $$\boxed{\text{Fail to reject } H_0\text{: the mean is not significantly different from }8.8}$$
(c) Test H₀: σ=0.4 vs H₁: σ≠0.4, α=0.05. $$\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{20(0.25)}{0.16}=31.25$$ The two-sided critical region is bounded by $\chi^2_{0.975,20}=9.59$ and $\chi^2_{0.025,20}=34.17$. Since $9.59\lt31.25\lt34.17$, $$\boxed{\text{Fail to reject } H_0\text{: the standard deviation is not significantly different from }0.4}$$

Question 5 — summary of results
Part	Quantity	Result
—	sample mean, variance	9.0, 0.25
(a)(i)	99% CI for μ	(8.690, 9.311)
(a)(ii)	99% CI for σ	(0.354, 0.820)
(b)	t-statistic (H₀: μ=8.8)	1.833 — fail to reject
(c)	χ²-statistic (H₀: σ=0.4)	31.25 — fail to reject

Topic: Interval estimation and hypothesis testing for the mean and standard deviation of a normal population from a small sample.

Key relations: mean CI/test uses $t=(\bar X-\mu_0)/(s/\sqrt n)$, df=n−1; variance CI/test uses $\chi^2=(n-1)s^2/\sigma_0^2$, df=n−1.

Why this works: for a normal population, $(\bar X-\mu)/(s/\sqrt n)$ follows Student's t with n−1 df, and $(n-1)s^2/\sigma^2$ follows a chi-square distribution with n−1 df — both classical results of sampling theory.

Common pitfall: using the z (normal) table instead of the t table for a small sample (n=21 here), and forgetting that the chi-square distribution is not symmetric, so the two critical values for a two-sided test are NOT ± of each other.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 8–9 (Estimation, Hypothesis Testing).

Question 5 of 8: Heat-Producing Capacity of Anthracite (Estimation and Testing)

Question 5: Heat-Producing Capacity of Anthracite (Estimation and Testing) (20 marks)