23-Ind-B1 Reliability and Maintainability · May 2015

Question 5 of 10: Marathon Training — Simple Linear Regression, ANOVA, Intervals

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

Question 5 (Section B.1): Marathon Training — Simple Linear Regression, ANOVA, Intervals (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. Six races, completion time (minutes) vs. race number:

Find. (a) $\hat y=a+bx$; (b) ANOVA $F$-test for the slope; (c) 95% PI at $x=8$; (d) $t$-test on $a$; (e) qualitative comparison of prediction uncertainty; (f) 95% CI for the mean of 30 future runners at $x=8$.

Approach. Compute least-squares $\hat a,\hat b$ from $S_{xx},S_{xy}$, run the regression ANOVA $F$-test, then use the standard $t$-based interval formulas at $x_0=8$ for a single new observation, for the slope/intercept tests, and for the mean of $m=30$ future observations.

1. (a) Least-squares fit. $\bar x=3.5$, $\bar y=225$, $S_{xx}=\sum(x_i-\bar x)^2=17.5$, $S_{xy}=\sum(x_i-\bar x)(y_i-\bar y)=-130$. $$\hat b=\frac{S_{xy}}{S_{xx}}=\frac{-130}{17.5}=-7.4286,\qquad \hat a=\bar y-\hat b\bar x=225-(-7.4286)(3.5)=251.0$$ $$\boxed{\hat y = 251.0 - 7.4286x}$$
2. (b) ANOVA $F$-test for the linear relationship. Residual sum: $SSE=\sum(y_i-\hat y_i)^2=134.29$. Using the given $SST=1100$: $SSR=SST-SSE=965.71$. $$F=\frac{SSR/1}{SSE/(n-2)}=\frac{965.71}{134.29/4}=\frac{965.71}{33.57}=\boxed{28.77}$$ $F_{0.05,1,4}=7.709$; since $28.77\gt 7.709$, reject $H_0:\beta=0$ — a significant linear relationship exists between race number and time.
3. (c) 95% prediction interval at $x_0=8$. $\hat y_0=251.0-7.4286(8)=191.57$. With $s=\sqrt{MSE}=\sqrt{33.57}=5.794$ and $t_{0.025,4}=2.776$: $$se_{pred}=s\sqrt{1+\tfrac1n+\tfrac{(x_0-\bar x)^2}{S_{xx}}}=5.794\sqrt{1+\tfrac16+\tfrac{(4.5)^2}{17.5}}=8.833$$ $$PI=191.57\pm 2.776(8.833)=\boxed{(167.0,\ 216.1)\text{ min}}$$
4. (d) Is the intercept $a$ significantly different from 0? $se_{\hat a}=s\sqrt{\tfrac1n+\tfrac{\bar x^2}{S_{xx}}}=5.794\sqrt{\tfrac16+\tfrac{12.25}{17.5}}=5.394$. $$t=\frac{251.0-0}{5.394}=\boxed{46.5}$$ Since $46.5\gg t_{0.025,4}=2.776$ ($p\lt 0.001$), the intercept is significantly different from 0.
5. (e) Race 10 vs. race 8, no new calculation. The prediction-interval width grows with $(x_0-\bar x)^2/S_{xx}$; race 10 is farther from $\bar x=3.5$ than race 8 ($6.5^2=42.25$ vs. $4.5^2=20.25$), so the interval for race 10 would be visibly wider, and it also extrapolates further beyond the observed range (races 1–6) — both push the prediction toward less certainty.
6. (f) 95% CI for the mean of $m=30$ new runners at $x_0=8$. The standard error for the mean of $m$ future observations replaces the "$1$" in step (c) with "$1/m$": $$se_{mean}=s\sqrt{\tfrac1{30}+\tfrac16+\tfrac{(4.5)^2}{17.5}}=6.750$$ $$CI=191.57\pm 2.776(6.750)=\boxed{(172.8,\ 210.3)\text{ min}}$$ — narrower than the single-runner PI in (c), as expected when averaging over 30 runners.