23-Ind-B1 Reliability and Maintainability · December 2019

Question 5 of 9: Simple Linear Regression — OR I vs. Statistics Marks

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

Notes on this paper

National Exams — December 2019 — 17-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 (20 marks); Section B: do 2 of 3 (40 marks); Section C: do 1 of 2 (20 marks) — a 9-question, 80-mark paper as actually structured by the section table and each question's own printed marks (the front page's own summary line states “Exam: 5 Questions. Total marks: 100”, which is internally inconsistent with 20+40+20=80 from its own section breakdown and every question's printed mark value). All nine questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — probability density functions and moments (ch. 4), point/interval estimation (ch. 8), two-sample hypothesis testing and the sign test (ch. 9–10, 16), simple linear regression (ch. 11), single-factor ANOVA and multiple comparisons (ch. 13). 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) — Bartlett's test, Tukey's HSD, $2^k$ factorial designs (ch. 3, 6).

Question 5 (Section B.1): Simple Linear Regression — OR I vs. Statistics Marks (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. $n=10$ paired student marks:

OR I ($x$)	86	75	69	74	90	94	83	86	71	65
Statistics ($y$)	80	81	75	81	92	95	80	81	76	72

Find. (a) Regression line $\hat y=b_0+b_1x$; (b) $F$-test of overall regression significance ($SST=460.1$ given); (c) $t$-test of the intercept; (d) $r$; (e) a 95% prediction interval for $y$ at $x_0=60$.

Approach. Compute $S_{xx},S_{xy},S_{yy}$ from the raw data (cross-checking $S_{yy}$ against the given $SST=460.1$), fit by least squares, then run the standard regression-ANOVA $F$-test, the intercept $t$-test, and the prediction interval.

Sums of squares and the fitted line. $$\bar x=79.3,\ \bar y=81.3,\qquad S_{xx}=860.1,\quad S_{xy}=545.1,\quad S_{yy}=460.1$$ $S_{yy}=460.1$ computed directly from the raw data matches the exam's given $SST=460.1$ exactly — the data is internally consistent, so it is used with full confidence below. $$b_1=\frac{S_{xy}}{S_{xx}}=\frac{545.1}{860.1}=0.6338,\qquad b_0=\bar y-b_1\bar x=81.3-0.6338(79.3)=31.04$$ $$\boxed{\hat y=31.04+0.6338\,x}$$
(b) $F$-test of overall regression significance. $$SSR=b_1S_{xy}=0.6338(545.1)=345.46,\qquad SSE=SST-SSR=460.1-345.46=114.64$$ $$MSR=\frac{SSR}{1}=345.46,\qquad MSE=\frac{SSE}{n-2}=\frac{114.64}{8}=14.33$$ $$F=\frac{MSR}{MSE}=\frac{345.46}{14.33}=\boxed{24.11}$$ Critical value $F_{0.05,1,8}=5.318$. Since $F=24.11\gt5.318$ ($p=0.0012$), reject $H_0:\beta_1=0$ — the regression is highly significant; OR I mark is a real linear predictor of Statistics mark.
(c) $t$-test of the intercept, $H_0:\beta_0=0$. $$SE(b_0)=\sqrt{MSE\left(\frac1n+\frac{\bar x^2}{S_{xx}}\right)}=\sqrt{14.33\left(\frac{1}{10}+\frac{79.3^2}{860.1}\right)}=10.31$$ $$t=\frac{b_0}{SE(b_0)}=\frac{31.04}{10.31}=\boxed{3.01}$$ Critical value $t_{0.025,8}=2.306$. Since $|t|=3.01\gt2.306$ ($p=0.017$), reject $H_0$ — the intercept is statistically significant (i.e. it is not defensible to force the line through the origin).
(d) Correlation coefficient. $$r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{545.1}{\sqrt{(860.1)(460.1)}}=\boxed{0.867}$$ A strong positive linear correlation; $r^2=0.751$, matching $SSR/SST=345.46/460.1=0.751$ — about 75% of the variation in Statistics marks is explained by OR I mark.
(e) Prediction interval for a new student scoring $x_0=60$ in OR I. “Range of likely values for their mark” asks for a prediction interval for one new individual, not a confidence interval for the mean response, so the $\left(1+\tfrac1n\right)$ term is included. $$\hat y_0=31.04+0.6338(60)=69.07$$ $$SE_{\text{pred}}=\sqrt{MSE\left(1+\frac1n+\frac{(x_0-\bar x)^2}{S_{xx}}\right)}=\sqrt{14.33\left(1+\tfrac{1}{10}+\tfrac{(60-79.3)^2}{860.1}\right)}=4.687$$ $$\hat y_0\pm t_{0.025,8}\,SE_{\text{pred}}=69.07\pm2.306(4.687)=69.07\pm10.81$$ $$\boxed{(58.26,\ 79.88)}$$

Final Results
Quantity	Value
(a) Regression line	$\hat y=31.04+0.6338x$
(b) $F$ (overall significance)	24.11 vs. crit. 5.318 — regression significant
(c) $t$ (intercept)	3.01 vs. crit. $\pm2.306$ — intercept significant
(d) $r$	0.867 ($r^2=0.751$)
(e) 95% PI at $x_0=60$	$(58.26,\ 79.88)$

Topic: Simple linear regression — least squares, the regression-ANOVA $F$-test, coefficient $t$-tests, correlation, and prediction intervals.

Key relations: $b_1=S_{xy}/S_{xx}$; $F=MSR/MSE$ with $MSR=SSR/1$, $MSE=SSE/(n-2)$; $t=\hat\beta/SE(\hat\beta)$; $r=S_{xy}/\sqrt{S_{xx}S_{yy}}$; prediction interval uses $1+\tfrac1n+\tfrac{(x_0-\bar x)^2}{S_{xx}}$ inside the square root.

Why this works: The overall $F$-test in (b) and the slope's own $t$-test are mathematically equivalent for simple linear regression ($F=t_{\beta_1}^2$), so a significant $F$ here already implies a significant slope; testing the intercept in (c) separately answers a different question (is a nonzero baseline needed).

Common pitfall: Using the confidence-interval-for-the-mean-response formula (which omits the leading “1+”) when the question asks for the range of a single new observation, as in part (e) — that omission understates the true spread substantially.

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th ed., §11.2–11.5 (simple linear regression, hypothesis tests, prediction intervals).

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