23-Ind-B1 Reliability and Maintainability · May 2015

Question 8 of 10: Bearing Wear vs. Viscosity and Load — Multiple Linear Regression by Matrices

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

Notes on this paper

National Exams — May 2015 — 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 questions (10 marks); Section B: do 2 of 3 (20 marks); Section C: do 1 of 3 (20 marks) — a 5-question, 50-mark paper as printed. All ten 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 (ch. 5), binomial/geometric probability (ch. 3), normal distribution and sampling distributions (ch. 4–7), point/interval estimation (ch. 8–9), hypothesis testing incl. sample-size design (ch. 9–10), simple linear regression and ANOVA (ch. 11–13), Bartlett's and Tukey's tests, single-degree-of-freedom contrasts and 2³ factorial designs (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).

Question 8 (Section C.1): Bearing Wear vs. Viscosity and Load — Multiple Linear Regression by Matrices (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. $n=6$ bearings:

$x_1$ (viscosity)	1.6	15.5	22.0	43.0	33.0	40.0
$x_2$ (load)	851	816	1058	1201	1357	1115
$y$ (wear)	193	230	172	91	113	125

Find. (a) $b_0,b_1,b_2$; (b) ANOVA $F$-test for the whole model; (c) $\hat y$, 95% CI (mean) and 95% PI (observed) at $x_1=20,x_2=1000$.

Approach. Build the design matrix $X=[1\ x_1\ x_2]$, compute $\hat\beta=(X'X)^{-1}X'y$ directly from the raw data (cross-checking the exam's own partial $(X'X)^{-1}$), then run the regression ANOVA $F$-test and the standard interval formulas at the requested point $x_0$.

(a) Fit by matrices. Computing $(X'X)^{-1}$ directly from the 6 rows above gives $$\left(X'X\right)^{-1}=\begin{pmatrix}8.595096 & 0.080958 & -0.009867\\ 0.080958 & 0.002102 & -0.000127\\ -0.009867 & -0.000127 & 0.0000123\end{pmatrix}$$ — matching the exam's given partial matrix once symmetrized. Then $\hat\beta=(X'X)^{-1}X'y$ gives $$\boxed{\hat y = 350.99 - 1.272\,x_1 - 0.154\,x_2}$$
(b) ANOVA $F$-test. $SST=\sum(y_i-\bar y)^2=14{,}112.0$; using $SSE=1950.42$ (recomputed value matches the given $1950.422$), $SSR=SST-SSE=12{,}161.58$. With $k=2$ predictors, $n-k-1=3$: $$F=\frac{SSR/k}{SSE/(n-k-1)}=\frac{12{,}161.58/2}{1950.42/3}=\frac{6080.79}{650.14}=\boxed{9.35}$$ $F_{0.05,2,3}=9.552$; since $9.35\lt 9.552$ ($p=0.051$), the model is not quite significant at the conventional $\alpha=0.05$ level — a borderline result driven by the very small sample ($n=6$, only 3 error df).
(c) Prediction at $x_1=20,\ x_2=1000$. $x_0=(1,20,1000)'$, $\hat y_0=x_0'\hat\beta=171.65$. $s=\sqrt{MSE}=\sqrt{650.14}=25.50$, $t_{0.025,3}=3.182$. Mean-response variance factor $x_0'(X'X)^{-1}x_0=0.1969$; observed-value factor adds 1: $$CI_{mean}=171.65\pm 3.182(25.50)\sqrt{0.1969}=\boxed{(135.9,\ 207.4)}$$ $$PI_{obs}=171.65\pm 3.182(25.50)\sqrt{1.1969}=\boxed{(83.0,\ 260.3)}$$

Summary
Quantity	Value
Fit	$\hat y=350.99-1.272x_1-0.154x_2$
$F$ (model)	9.35 < $F_{crit}=9.55$ ($p=0.051$) — borderline, not significant at $\alpha=0.05$
$\hat y_0$ at (20, 1000)	171.6
95% CI (mean wear)	(135.9, 207.4)
95% PI (observed wear)	(83.0, 260.3)

Check — the given (X'X)⁻¹ needs recomputing, not plugging in directly Multiplying the exam's rounded matrix entries (6 decimal places, e.g. $0.000012$ for the $(3,3)$ entry) directly by $X'y$ gives wildly wrong coefficients (an SSE in the hundreds of thousands) because $x_1$ and $x_2$ are strongly correlated here, which amplifies rounding error. Recomputing $(X'X)^{-1}$ at full machine precision from the raw data and cross-checking against the given $SSE=1950.422$ (reproduced to 3 decimals) confirms the fit above is correct.

Topic: Multiple linear regression by matrices — parameter estimation, the overall-model ANOVA $F$-test, and CI/PI at a new point.

Key relations: $\hat\beta=(X'X)^{-1}X'y$; $F=\dfrac{SSR/k}{SSE/(n-k-1)}$; interval half-width at $x_0$: $t_{\alpha/2,n-k-1}\cdot s\sqrt{c+x_0'(X'X)^{-1}x_0}$, $c=0$ for mean response, $c=1$ for a new observation.

Why this works: $(X'X)^{-1}$ carries both the coefficient estimates and (via $x_0'(X'X)^{-1}x_0$) the estimation-uncertainty needed for every interval at any new point $x_0$ — the same matrix does both jobs.

Common pitfall: With only $n=6$ observations and $k=2$ predictors, the model has just 3 error df, so the ANOVA $F$-test has very low power — a borderline result like $F=9.35$ vs. $F_{crit}=9.55$ should be reported as marginal, not silently rounded to "significant."

Source: Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed.), ch. 2–3 (multiple regression by matrices, CI/PI at a point).

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