23-Ind-B1 Reliability and Maintainability · December 2019

Question 2 of 9: Curriculum Change — Two-Sample Comparison

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

Question 2 (Section A.2): Curriculum Change — Two-Sample Comparison (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. Two independent samples of $n=6$ students each (different cohorts, before/after the curriculum change):

Find. (a) Equality of the two population variances; (b) whether the rise is genuinely 6 (points); (c) 95% CI for $\mu_{\text{old}}$; (d) whether that CI speaks to 78.4 being an anomaly.

Approach. These are independent samples from two different student cohorts (not the same six students measured twice), so use an $F$-test for the variance-equality pre-check, then the corresponding two-sample $t$-test for the mean shift; the CI in (c) is a standard one-sample $t$-interval for $\mu_{\text{old}}$.

1. Sample statistics. $$\bar x_{\text{old}}=68.433,\ s_{\text{old}}=6.442\ (s_{\text{old}}^2=41.495)\qquad \bar x_{\text{new}}=74.017,\ s_{\text{new}}=2.929\ (s_{\text{new}}^2=8.582)$$
2. (a) $F$-test for equal variances, $\alpha=0.05$. $$F=\frac{s_{\text{old}}^2}{s_{\text{new}}^2}=\frac{41.495}{8.582}=\boxed{4.835}$$ Critical value $F_{0.025,5,5}=7.146$. Since $F=4.835\lt7.146$ ($p=0.109$), fail to reject $H_0:\sigma_{\text{old}}^2=\sigma_{\text{new}}^2$ — treat the variances as equal, which justifies a pooled two-sample $t$-test next.
3. (b) Test whether the rise is genuinely 6 (points), $\alpha=0.05$. The raw scores are already percentage-style marks (out of 100), and the observed shift $\bar x_{\text{new}}-\bar x_{\text{old}}=5.583$ matches an additive 6-mark rise far more closely than a relative 6% multiplicative rise would ($1.06\times68.433=72.54$, an implied shift of only 4.11 — a much worse match to the data); this answer therefore reads “risen by about 6 percent” as “risen by about 6 percentage points” on the marks scale, flagged as an assumption. With equal variances supported by (a), pool: $$s_p^2=\frac{5(41.495)+5(8.582)}{10}=25.038,\qquad SE=s_p\sqrt{\tfrac16+\tfrac16}=2.889,\qquad df=10$$ $$H_0:\mu_{\text{new}}-\mu_{\text{old}}=6\qquad H_1:\mu_{\text{new}}-\mu_{\text{old}}\ne6$$ $$t=\frac{5.583-6}{2.889}=\boxed{-0.144}$$ Critical value $t_{0.025,10}=2.228$. Since $|t|=0.144\ll2.228$ ($p=0.888$), fail to reject $H_0$ — the data are fully consistent with a genuine 6-point average rise; there is no evidence against the faculty's claimed figure.
4. (c) 95% CI for $\mu_{\text{old}}$. $t_{0.025,5}=2.571$: $$\bar x_{\text{old}}\pm t_{0.025,5}\frac{s_{\text{old}}}{\sqrt6}=68.433\pm2.571\left(\frac{6.442}{\sqrt6}\right)=68.433\pm6.760$$ $$\boxed{(61.67,\ 75.19)}$$
5. (d) Does (c) speak to whether 78.4 is an anomaly? (essay). No — the interval in (c) is a confidence interval for the population mean $\mu_{\text{old}}$; it quantifies sampling uncertainty in estimating that mean from $n=6$ observations, and it shrinks toward zero width as $n$ grows. It says nothing about how spread out individual old-curriculum grades are. The right tool to judge whether a single new observation is unusual is a prediction interval for one future value, $\bar x\pm t_{0.025,5}\,s\sqrt{1+\tfrac1n}=68.433\pm17.89=(50.55,86.32)$, or equivalently a $z$/$t$-score of the point against the sample mean and SD: $(78.4-68.433)/6.442=1.55$ — well within about 1.5 SD of the mean and inside the prediction interval, so 78.4 is not an anomaly under the old curriculum, but that conclusion comes from a prediction interval, not the CI computed in (c).