23-Ind-A6 Systems Simulation · December 2017

Question 4 of 8: Comparing Two Alternatives: Paired vs. Independent $t$-Test

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4 — Comparing Two Alternatives: Paired vs. Independent $t$-Test (10 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. Route-completion times (minutes), 10 replications each:

Find. (a) whether a paired test is the right tool here; (b) whether the proposed routing significantly reduces route time at 90% confidence, treating the samples as independent with equal variance (as instructed).

(a) Should a paired $t$-test be used? It depends entirely on whether the two systems' 10 replications share Common Random Numbers (CRN). If replication $i$ of the Original run and replication $i$ of the Proposed run were driven by the same underlying random-number streams (the same simulated traffic/weather/timing draws for "day $i$"), the two samples are correlated pair-by-pair, and a paired $t$-test on the 10 differences is not only valid but preferable — it cancels the shared random variation and gives a narrower CI / more powerful test for the same 10 replications than an independent-samples analysis. If independent streams were used for each system (no CRN), there is no genuine correspondence between "replication 3 Original" and "replication 3 Proposed", and pairing would be invalid; an independent two-sample test is then the correct (and only defensible) choice. This paper does not state that CRN was used, so part (b) below follows the question's own explicit instruction and treats the two samples as independent.

Approach (b). Pooled-variance two-sample $t$-test, $H_0:\mu_{Orig}=\mu_{Prop}$, $\alpha=0.10$.

1. Sample means and pooled variance. $$\bar x_{Orig}=180.30,\ \ \bar x_{Prop}=176.20\ \text{min};\qquad s_p^2 = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} = 51.98.$$
2. Test statistic. $$t = \frac{\bar x_{Orig}-\bar x_{Prop}}{\sqrt{s_p^2(1/n_1+1/n_2)}} = \frac{180.30-176.20}{\sqrt{51.98(1/10+1/10)}} = \boxed{1.272},\qquad df=18.$$
3. Critical value and decision. Two-sided, $\alpha=0.10$: $t_{0.05,18}=1.734$ (one-sided $t_{0.10,18}=1.330$ gives the same call). Since $t=1.272$ is below BOTH the two-sided and the (more favourable) one-sided critical value, fail to reject $H_0$ — the 4.1-minute sample-mean improvement is not statistically significant at 90% confidence.