23-Ind-B1 Reliability and Maintainability · December 2016
Question 6 of 11: BMI vs. Age — Simple Linear Regression, ANOVA F -Test, Prediction Interval
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2016 — 98-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: three sections — Section A: do 2 of 4 (30 marks); Section B: do 2 of 3 (30 marks); Section C: do 2 of 4 (40 marks) — a 6-question, 100-mark paper as printed. All eleven questions across the three sections are solved below for completeness.
Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — joint distributions and covariance (ch. 5), point/interval estimation (ch. 8), hypothesis testing incl. two-sample and goodness-of-fit tests (ch. 9–10), simple linear regression (ch. 11), design and analysis of single-factor and factorial experiments (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — two-way factorial ANOVA and $2^k$ designs (ch. 5, 6–7).
Question 6 (Section B.2): BMI vs. Age — Simple Linear Regression, ANOVA F-Test, Prediction Interval (15 marks)
Find. (a) fitted line $y=ax+b$; (b) overall $F$-test; (c) $t$-test on the slope; (d) $\hat y$ at $x=49.0$, and is $(49.0, 32.5)$ anomalous?
Approach. Least-squares via $S_{xx},S_{yy},S_{xy}$, then the standard regression ANOVA (which for simple regression satisfies $F=t^2$ on the slope), then a prediction interval to judge the new point.
(a) Fit the line. $\bar x=38.256$, $\bar y=23.608$, $S_{xx}=560.342$, $S_{xy}=143.363$. $$b_1=\frac{S_{xy}}{S_{xx}}=0.2558,\qquad b_0=\bar y-b_1\bar x=13.820$$ $$\boxed{\hat y=13.820+0.2558\,x}$$
(b) Overall significance ($F$-test). $S_{yy}=251.506$. Fitted residuals give $SSE=214.827$, so $SSR=S_{yy}-SSE=36.679$, $MSE=SSE/(n-2)=30.690$. $$F=\frac{SSR/1}{MSE}=\frac{36.679}{30.690}=\boxed{1.195}$$ $F_{0.05,1,7}=5.591$; since $1.195\lt5.591$, the regression is not significant — age does not significantly explain BMI in this sample.
(c) Slope significance. $se(b_1)=\sqrt{MSE/S_{xx}}=0.2340$. $$t=\frac{b_1}{se(b_1)}=\frac{0.2558}{0.2340}=\boxed{1.093}$$ $t_{0.025,7}=2.365$; since $|t|\lt t_{crit}$, the slope is not significant (consistent with (b): $t^2=1.195=F$).
(d) Prediction at age 49 and anomaly check. $\hat y(49)=13.820+0.2558(49)=\boxed{26.36}$. The 95% PI for a new individual at $x_0=49$: $$se_{pred}=\sqrt{MSE\left(1+\tfrac19+\tfrac{(49-38.256)^2}{560.342}\right)}=6.010,\qquad \hat y\pm t_{0.025,7}\,se_{pred}=\boxed{(11.32,\ 41.39)}$$ Since $32.5$ falls comfortably inside $(11.32,41.39)$, $$\boxed{\text{no, this is not a statistical anomaly}}$$
