Correlation and Simple Linear Regression — Study Hours vs. Marks: 04-BS-2, December 2018, Q8

Question 8: Correlation and Simple Linear Regression — Study Hours vs. 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.

Given data
Quantity	Value
n	28
ΣX	4,200.0
ΣX²	654,300.0
ΣY	1,960.0
ΣY²	148,000.0
ΣXY	307,284.0

Find. (a) r. (b) 95% CI for ρ. (c) least-squares normal equations and b₀, b₁. (d) SSE and the 95% CI for β₁.

Approach. Build the corrected sums of squares/products $S_{xx}, S_{yy}, S_{xy}$ once and reuse them for r, the regression slope, and the error sum of squares; use Fisher’s z-transformation for the ρ interval (since r is not normally distributed) and the t-distribution (df=n−2) for the β₁ interval.

Corrected sums (used throughout). $$S_{xx}=\Sigma X^2-\dfrac{(\Sigma X)^2}{n}=654300-\dfrac{4200^2}{28}=24300.0$$ $$S_{yy}=\Sigma Y^2-\dfrac{(\Sigma Y)^2}{n}=148000-\dfrac{1960^2}{28}=10800.0$$ $$S_{xy}=\Sigma XY-\dfrac{(\Sigma X)(\Sigma Y)}{n}=307284-\dfrac{4200\times1960}{28}=13284.0$$
(a) Correlation coefficient. $$r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{13284.0}{\sqrt{24300.0\times10800.0}}=\dfrac{13284.0}{16200.0}$$ $$\boxed{r=0.82}$$ a strong positive linear association between study hours and marks.
(b) 95% CI for ρ via Fisher’s z-transformation. $$z_r=\dfrac{1}{2}\ln\!\left(\dfrac{1+r}{1-r}\right)=\dfrac{1}{2}\ln\!\left(\dfrac{1.82}{0.18}\right)=1.1568,\qquad \sigma_{z}=\dfrac{1}{\sqrt{n-3}}=\dfrac{1}{\sqrt{25}}=0.20$$ 95% CI in z: $1.1568\pm1.96(0.20)=1.1568\pm0.392\Rightarrow(0.765,\,1.549)$. Transforming back with $r=\dfrac{e^{2z}-1}{e^{2z}+1}$: $$\boxed{0.644\lt \rho\lt 0.914}$$
(c) Normal equations and least-squares estimates. The normal equations for $\hat Y=b_0+b_1X$ are $$\Sigma Y=nb_0+b_1\Sigma X,\qquad \Sigma XY=b_0\Sigma X+b_1\Sigma X^2$$ Solved directly from the corrected sums: $$b_1=\dfrac{S_{xy}}{S_{xx}}=\dfrac{13284.0}{24300.0}=0.5467,\qquad b_0=\bar Y-b_1\bar X=70.0-0.5467(150.0)=70.0-82.0$$ $$\boxed{b_0=-12.0,\qquad b_1=0.5467}$$ so $\hat Y=-12.0+0.5467X$ — each extra hour of study is associated with about 0.55 additional marks out of 100.
(d) Error sum of squares and 95% CI for β₁. $$SSE=S_{yy}-b_1S_{xy}=10800.0-0.5467(13284.0)=10800.0-7261.9$$ $$\boxed{SSE=3538.08}$$ then $s^2=\dfrac{SSE}{n-2}=\dfrac{3538.08}{26}=136.08$, and the standard error of the slope is $$se(b_1)=\sqrt{\dfrac{s^2}{S_{xx}}}=\sqrt{\dfrac{136.08}{24300.0}}=0.07483$$ With $t_{0.025,26}=2.056$: $$b_1\pm t_{0.025,26}\,se(b_1)=0.5467\pm2.056(0.07483)=0.5467\pm0.1538$$ $$\boxed{0.393\lt \beta_1\lt 0.701}$$

Question 8 — final results
Part	Quantity	Result
(a)	r	0.82
(b)	95% CI for ρ	(0.644, 0.914)
(c)	b₀, b₁	−12.0, 0.5467
(d)	SSE; 95% CI for β₁	3538.08; (0.393, 0.701)

Topic: Pearson correlation, Fisher’s z-transform, and simple linear regression by least squares.

Key relations: $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$; $z_r=\tfrac12\ln\!\left(\tfrac{1+r}{1-r}\right)$, $\sigma_z=1/\sqrt{n-3}$; $b_1=S_{xy}/S_{xx}$, $b_0=\bar Y-b_1\bar X$; $SSE=S_{yy}-b_1S_{xy}$; $se(b_1)=\sqrt{s^2/S_{xx}}$.

Why this works: r itself is a bounded, skewed statistic near ±1, so its own sampling distribution is not normal — Fisher’s z-transform maps it onto an (approximately) normal scale where a standard z-interval is valid, then the inverse transform maps the interval back to the r-scale. The least-squares normal equations are just the calculus conditions that minimize $\Sigma(Y_i-b_0-b_1X_i)^2$; solving them algebraically collapses to the corrected-sums formulas used above.

Common pitfall: Building a CI for ρ directly with a normal approximation on r (ignores its skew near ±1); confusing $S_{xx}$ (used for both r and b₁) with the raw $\Sigma X^2$ (forgetting the $-(\Sigma X)^2/n$ correction).

Source: Walpole et al., Probability & Statistics for Engineers and Scientists, Ch. 11 (Simple Linear Regression & Correlation).

Question 8 of 8: Correlation and Simple Linear Regression — Study Hours vs. Marks

Question 8: Correlation and Simple Linear Regression — Study Hours vs. Marks (20 marks)