23-Ind-B1 Reliability and Maintainability · December 2014

Question 4 of 9: Multiple Linear Regression — Pull Strength on Wire Length and Die Height

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

Question 4 (Section B.1): Multiple Linear Regression — Pull Strength on Wire Length and Die Height (25 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. $n=10$ observations of pull strength $y$, wire length $x_1$, and die height $x_2$:

Find. (a) $X'X$, $X'Y$; (b) fitted model $\hat y=b_0+b_1x_1+b_2x_2$; (c) ANOVA and overall significance; (d) which $b_i$ are individually significant; (e) reconcile (c) and (d).

Approach. Build the $10\times3$ design matrix $X=[\mathbf{1}\ \ x_1\ \ x_2]$, form $X'X$ and $X'Y$ directly from the data, invert $X'X$ to get $b=(X'X)^{-1}X'Y$, then partition variability via $SST=SSR+SSE$ for the overall $F$-test and use $(X'X)^{-1}$'s diagonal for each coefficient's standard error and $t$-test.

1. Part (a) — $X'X$ and $X'Y$. With $X=[\mathbf 1\ \ x_1\ \ x_2]$, direct summation over the 10 rows gives $$X'X=\begin{bmatrix}10 & 64 & 2152\\ 64 & 522 & 15114\\ 2152 & 15114 & 701854\end{bmatrix},\qquad X'Y=\begin{bmatrix}218.86\\ 1669.46\\ 53576.60\end{bmatrix}.$$
2. Part (b) — checking the given $(X'X)^{-1}$, then solving for $b$. Inverting the $X'X$ above directly gives $$(X'X)^{-1}=\begin{bmatrix}0.550947 & -0.049502 & -0.000623\\ -0.049502 & 0.009536 & -0.000054\\ -0.000623 & -0.000054 & 0.0000045\end{bmatrix},$$ which agrees with the printed matrix in magnitude but not in every sign (the source prints element (2,3) as $+0.0000536$ against $-0.0000536$ at (3,2), an internal asymmetry that already flags a transcription slip in a matrix that must be symmetric by construction). Substituting the printed matrix into $b=(X'X)^{-1}X'Y$ gives $b\approx(236.6,\ 29.6,\ 0.467)$ — physically impossible for this data (it predicts a pull strength above 470 units at the smallest $x_1,x_2$, twenty times the largest observed $y$), so the printed inverse does not belong to this 10-row data set. Using the inverse recomputed directly from this paper's own $X'X$ instead, $$b = (X'X)^{-1}X'Y = \boxed{(4.544,\ 2.216,\ 0.01469)},\qquad \hat y = 4.544 + 2.216\,x_1 + 0.01469\,x_2.$$ This is confirmed correct independently of the inverse dispute: the resulting fit's own residual sum of squares recomputes to $7.624$, matching the question's supplied $SSE=7.62$ to rounding.
3. Part (c) — ANOVA for overall significance. With $\bar y=21.886$, $SST=\sum(y_i-\bar y)^2=698.274$ and the given $SSE=7.62$, so $SSR=SST-SSE=690.654$. With $k=2$ predictors, $df_R=2$, $df_E=n-k-1=7$: $$MSR=\frac{SSR}{2}=345.33,\quad MSE=\frac{SSE}{7}=1.0886,\quad F_0=\frac{MSR}{MSE}=\boxed{317.2}.$$ Choosing $\alpha=0.05$ (the conventional engineering default, appropriate here since no cost asymmetry between Type I/II errors is stated), $F_{0.05,2,7}=4.74$. Since $F_0=317.2 \gg 4.74$, the regression is highly significant — reject $H_0:\beta_1=\beta_2=0$. $R^2=SSR/SST=0.989$: the model explains 98.9% of the variability in pull strength.
4. Part (d) — individual coefficient significance. Using $MSE=1.0886$ and the diagonal of the (recomputed) $(X'X)^{-1}$, each coefficient's standard error is $se(b_i)=\sqrt{MSE\cdot[(X'X)^{-1}]_{ii}}$: $$se(b_0)=0.774,\ se(b_1)=0.1019,\ se(b_2)=0.002211,\qquad t_0=5.87,\ t_1=21.75,\ t_2=6.64.$$ Against $t_{0.025,7}=2.365$, all three $|t_i|$ exceed the critical value, so every coefficient — intercept, wire length, and die height — is individually significant at $\alpha=0.05$.
5. Part (e) — relating (c) and (d). The two conclusions are fully consistent: the overall $F$-test says the model as a whole explains significant variation, and the individual $t$-tests confirm both predictors contribute meaningfully on their own. This is the "clean" case — it is possible for an overall $F$-test to be significant while one or more individual $t$-tests are not (a symptom of multicollinearity, since $x_1$ and $x_2$ can share explanatory power), but that did not happen here: wire length and die height are contributing largely independent information about pull strength.