Carburetor Modification — Estimation & Hypothesis Testing: 04-BS-2, May 2014, Q5

Question 5: Carburetor Modification — Estimation & Hypothesis 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=12 vans (modified carburetor); ∑X=157.20, ∑X²=2,062.07; historical (unmodified) μ₀=12.2, σ₀=0.9 km/L; X assumed Normal.

Find. (a) 99% CI for the true mean and for the true standard deviation; (b)(i) test H₀: μ=12.2 vs H₁: μ≠12.2; (b)(ii) test H₀: σ=0.9 vs H₁: σ≠0.9.

Approach. Compute the sample mean and variance from the sums first. Since n=12 is small and σ is unknown, use the t-distribution (df=11) for the mean's CI and test, and the χ² distribution (df=11) for the variance's CI and test.

Sample mean and variance. $$\bar x=\dfrac{\sum x}{n}=\dfrac{157.20}{12}=\boxed{13.10}\ \text{km/L},$$ $$s^2=\dfrac{\sum x^2-(\sum x)^2/n}{n-1}=\dfrac{2062.07-157.20^2/12}{11}=\dfrac{2.75}{11}=\boxed{0.2500},\qquad s=\boxed{0.5000}\ \text{km/L}.$$
(a)(i) 99% CI for the true mean. With df=11, $t_{0.005,11}=3.106$, so the margin is $t\,s/\sqrt{n}=3.106(0.5)/\sqrt{12}=0.4483$: $$13.10-0.448<\mu<13.10+0.448\ \Rightarrow\ \boxed{12.65<\mu<13.55\ \text{km/L}}.$$
(a)(ii) 99% CI for the true standard deviation. Using $\chi^2_{0.995,11}=2.603$ and $\chi^2_{0.005,11}=26.757$, $$\sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.005}}}<\sigma<\sqrt{\dfrac{(n-1)s^2}{\chi^2_{0.995}}}\ \Rightarrow\ \sqrt{\dfrac{2.75}{26.757}}<\sigma<\sqrt{\dfrac{2.75}{2.603}}\ \Rightarrow\ \boxed{0.321<\sigma<1.028\ \text{km/L}}.$$
(b)(i) Test H₀: μ=12.2 vs H₁: μ≠12.2. The test statistic is $$t=\dfrac{\bar x-\mu_0}{s/\sqrt{n}}=\dfrac{13.10-12.2}{0.5/\sqrt{12}}=\boxed{6.235}.$$ The critical value is $t_{0.025,11}=2.201$. Since $|6.235|>2.201$, reject H₀: the modified mean fuel consumption is significantly different from (higher than) 12.2 km/L.
(b)(ii) Test H₀: σ=0.9 vs H₁: σ≠0.9. The test statistic is $$\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{11(0.25)}{0.9^2}=\dfrac{2.75}{0.81}=\boxed{3.395}.$$ The two-sided critical values are $\chi^2_{0.975,11}=3.816$ and $\chi^2_{0.025,11}=21.920$. Since $3.395<3.816$, the statistic falls in the lower rejection region, so reject H₀: the modified process's standard deviation is significantly smaller (more consistent) than 0.9 km/L.

Final results — Question 5
Part	Quantity	Result
	&bar;x, s², s	13.10, 0.2500, 0.5000
(a)(i)	99% CI for μ	(12.65, 13.55)
(a)(ii)	99% CI for σ	(0.321, 1.028)
(b)(i)	t (H₀: μ=12.2)	6.235 > 2.201 → reject H₀
(b)(ii)	χ² (H₀: σ=0.9)	3.395 < 3.816 → reject H₀

Topic: Small-sample interval estimation and hypothesis testing for a Normal mean (t) and variance (χ²).

Key relations: $\dfrac{\bar X-\mu}{S/\sqrt n}\sim t_{n-1}$; $\dfrac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}$.

Why this works: with σ unknown and n small, replacing σ by S introduces extra sampling variability that the Normal distribution understates; the t and χ² distributions (derived from a Normal sample's exact sampling theory) correctly widen the interval/critical region to compensate.

Common pitfall: using a two-sided z critical value (1.96) instead of the wider t critical value (2.201/3.106) for a sample this small — the extra t-tail width matters at n=12; also, the χ² distribution is not symmetric, so its two critical values must be looked up separately (χ²_0.975 ≠ χ²_0.025 mirrored about the mean).

Source: Walpole, Myers, Myers & Ye, Ch. 9 (Single-Sample Estimation and Tests).

Question 5 of 8: Carburetor Modification — Estimation & Hypothesis Testing

Question 5: Carburetor Modification — Estimation & Hypothesis Testing (20 marks)