Correlation and Simple Linear Regression — Advertising vs. Sales: 04-BS-2, Undated paper, Q8

Question 8: Correlation and Simple Linear Regression — Advertising vs. Sales (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 data (n=25 dealerships)
ΣX	275.0
ΣX²	3,079.0
ΣY	900.0
ΣY²	32,496.0
ΣXY	9,960.0

Given. $n=25$; X = advertising level ($\$10{,}000$ units), Y = sales level ($\$300{,}000$ units); sums as tabulated above.

Find. (a) $r$. (b) 95% CI for $\rho$. (c) the normal equations and least-squares estimates $b_0,b_1$. (d) $SSE$ and the 95% CI for $\beta_1$.

Approach. Reduce the raw sums to the corrected sums of squares/cross-products $S_{xx},S_{yy},S_{xy}$; every later quantity (r, regression slope/intercept, SSE, CIs) is built from these three numbers.

Corrected sums of squares. $$S_{xx}=\Sigma X^2-\dfrac{(\Sigma X)^2}{n}=3{,}079.0-\dfrac{275.0^2}{25}=54.0$$ $$S_{yy}=\Sigma Y^2-\dfrac{(\Sigma Y)^2}{n}=32{,}496.0-\dfrac{900.0^2}{25}=96.0$$ $$S_{xy}=\Sigma XY-\dfrac{(\Sigma X)(\Sigma Y)}{n}=9{,}960.0-\dfrac{275.0(900.0)}{25}=60.0$$
(a) Coefficient of correlation. $$r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{60.0}{\sqrt{54.0(96.0)}}=\dfrac{60.0}{72.0}$$ $$\boxed{r=0.8333}$$
(b) 95% CI for ρ, via Fisher's z-transform. $$z'=\dfrac12\ln\left(\dfrac{1+r}{1-r}\right)=\dfrac12\ln\left(\dfrac{1.8333}{0.1667}\right)=1.199,\qquad \sigma_{z'}=\dfrac{1}{\sqrt{n-3}}=\dfrac{1}{\sqrt{22}}=0.2132$$ $$z'\pm 1.96(0.2132)=1.199\pm0.418\ \Rightarrow\ (0.781,\,1.617)$$ Back-transforming with $\rho=\dfrac{e^{2z}-1}{e^{2z}+1}$: $$\boxed{0.653\lt\rho\lt0.924}$$
(c) Normal equations and least-squares estimates. The normal equations of the least-squares line $\hat Y=b_0+b_1X$ are $$nb_0+b_1\Sigma X=\Sigma Y,\qquad b_0\Sigma X+b_1\Sigma X^2=\Sigma XY$$ Solving (equivalently, $b_1=S_{xy}/S_{xx}$ and $b_0=\bar Y-b_1\bar X$): $$b_1=\dfrac{S_{xy}}{S_{xx}}=\dfrac{60.0}{54.0}=1.1111$$ $$b_0=\dfrac{900.0}{25}-1.1111\left(\dfrac{275.0}{25}\right)=36.0-12.222=23.778$$ $$\boxed{\hat Y=23.778+1.1111X}$$
(d) Error sum of squares and CI for β₁. $$SSE=S_{yy}-b_1S_{xy}=96.0-1.1111(60.0)=29.333$$ $$MSE=\dfrac{SSE}{n-2}=\dfrac{29.333}{23}=1.2754,\qquad s_{b_1}=\sqrt{\dfrac{MSE}{S_{xx}}}=\sqrt{\dfrac{1.2754}{54.0}}=0.1537$$ With $t_{0.025,23}=2.069$: $$b_1\pm t_{0.025,23}\,s_{b_1}=1.1111\pm2.069(0.1537)=1.1111\pm0.318$$ $$\boxed{0.793\lt\beta_1\lt1.429}$$

Question 8 — final results
Part	Quantity	Result
	S_xx, S_yy, S_xy	54.0, 96.0, 60.0
(a)	r	0.8333
(b)	95% CI for ρ	(0.653, 0.924)
(c)	b₀, b₁	23.778, 1.1111
(d)	SSE, 95% CI for β₁	29.333, (0.793, 1.429)

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

Key relations: $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$; $z'=\tfrac12\ln\dfrac{1+r}{1-r}\sim N\!\left(\zeta,\dfrac{1}{n-3}\right)$; $b_1=\dfrac{S_{xy}}{S_{xx}}$, $b_0=\bar Y-b_1\bar X$; $SE(b_1)=\sqrt{MSE/S_{xx}}$.

Why this works: Fisher's transform maps the bounded, skewed sampling distribution of $r$ onto an (approximately) Normal one for $z'$, allowing a symmetric CI that is then back-transformed. Least squares minimizes $SSE=\Sigma(Y-\hat Y)^2$; differentiating gives exactly the two normal equations shown, and $MSE$ estimates the error variance $\sigma^2$ needed for every regression standard error.

Common pitfall: Confusing $r$ with $r^2$; forgetting the $n-3$ (not $n-1$ or $n-2$) degrees-of-freedom-like divisor in Fisher's transform; using $n-1$ instead of $n-2$ for $MSE$ in a simple regression (one degree of freedom is spent on each of $b_0,b_1$).

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 11–12.

Question 8 of 8: Correlation and Simple Linear Regression — Advertising vs. Sales

Question 8: Correlation and Simple Linear Regression — Advertising vs. Sales (20 marks)