Correlation and Simple Linear Regression — Gas Consumption: 04-BS-2, May 2013, Q8

Question 8: Correlation and Simple Linear Regression — Gas Consumption (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
n	24
ΣX, ΣX²	216.0, 2,151.0
ΣY, ΣY²	360.0, 6,228.0
ΣXY	3,516.0

Find. (a) r; (b) test H₀: ρ=0.6; (c) normal equations and b₀, b₁; (d) SSE and 95% CI for β₁.

Approach. Build the corrected sums of squares/cross-products $S_{xx}$, $S_{yy}$, $S_{xy}$ from the raw totals, get r and the least-squares slope/intercept from them, then use Fisher's z-transform to test ρ and the standard regression-SE formula for the CI on β₁.

Corrected sums. $$S_{xx}=\Sigma x^2-\dfrac{(\Sigma x)^2}{n}=2151.0-\dfrac{216.0^2}{24}=207.0,\qquad S_{yy}=\Sigma y^2-\dfrac{(\Sigma y)^2}{n}=6228.0-\dfrac{360.0^2}{24}=828.0,$$ $$S_{xy}=\Sigma xy-\dfrac{\Sigma x\Sigma y}{n}=3516.0-\dfrac{216.0(360.0)}{24}=276.0.$$
(a) Correlation coefficient. $$r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{276.0}{\sqrt{207.0(828.0)}}=\dfrac{276.0}{414.0}=\boxed{0.6667}.$$
(b) Fisher z-test, H₀: ρ=0.6 vs H₁: ρ≠0.6. $$z_r=\tfrac12\ln\!\dfrac{1+r}{1-r}=\tfrac12\ln\dfrac{1.6667}{0.3333}=0.8047,\qquad z_{\rho_0}=\tfrac12\ln\dfrac{1.6}{0.4}=0.6931,\qquad \mathrm{SE}=\dfrac{1}{\sqrt{n-3}}=\dfrac{1}{\sqrt{21}}=0.2182.$$ $$Z=\dfrac{z_r-z_{\rho_0}}{\mathrm{SE}}=\dfrac{0.8047-0.6931}{0.2182}=\boxed{0.511}.$$ Critical value $z_{0.025}=1.96$. Since $|Z|=0.511<1.96$, fail to reject H₀: the true correlation is not significantly different from 0.6 at α=0.05.
(c) Normal equations and least-squares estimates. The normal equations for $\hat y=b_0+b_1x$ are $$\Sigma y=nb_0+b_1\Sigma x\ \Rightarrow\ 360.0=24b_0+216.0\,b_1,\qquad \Sigma xy=b_0\Sigma x+b_1\Sigma x^2\ \Rightarrow\ 3516.0=216.0\,b_0+2151.0\,b_1.$$ Solving (equivalently, $b_1=S_{xy}/S_{xx}$ and $b_0=\bar y-b_1\bar x$): $$b_1=\dfrac{276.0}{207.0}=\boxed{1.333},\qquad b_0=\dfrac{360.0}{24}-1.333\left(\dfrac{216.0}{24}\right)=15.0-12.0=\boxed{3.00}.$$ So $\hat y=3.00+1.333x$ (thousand m³ of gas per additional resident: +1.333 kilo-m³/person).
(d) Error sum of squares and CI for β₁. $$\mathrm{SSE}=S_{yy}-b_1S_{xy}=828.0-1.333(276.0)=\boxed{460.0}.$$ $$s^2_{y\cdot x}=\dfrac{\mathrm{SSE}}{n-2}=\dfrac{460.0}{22}=20.91,\qquad \mathrm{SE}(b_1)=\sqrt{\dfrac{s^2_{y\cdot x}}{S_{xx}}}=\sqrt{\dfrac{20.91}{207.0}}=0.3178.$$ With $t_{0.025,22}=2.074$: $$b_1\pm t_{0.025,22}\,\mathrm{SE}(b_1)=1.333\pm2.074(0.3178)=1.333\pm0.659\ \Rightarrow\ \boxed{0.674<\beta_1<1.992}.$$

Question 8 — final results
Part	Quantity	Value
(a)	r	0.6667
(b)	Z, decision (ρ=0.6)	0.511, fail to reject H₀
(c)	b₀, b₁	3.00, 1.333
(d)	SSE; 95% CI for β₁	460.0; (0.674, 1.992)

Topic: Sample correlation, Fisher's z-transform test, and simple linear regression via least squares.

Key relations: $r=S_{xy}/\sqrt{S_{xx}S_{yy}}$; Fisher $z_r=\tfrac12\ln\frac{1+r}{1-r}$, approximately Normal with SE $=1/\sqrt{n-3}$; least-squares $b_1=S_{xy}/S_{xx}$, $b_0=\bar y-b_1\bar x$; $\mathrm{SE}(b_1)=\sqrt{s^2_{y\cdot x}/S_{xx}}$.

Why this works: r's own sampling distribution is skewed and not directly Normal except near ρ=0, so Fisher's z-transform is used to stabilize variance and normalize the distribution before testing a hypothesized ρ other than zero; the least-squares line minimizes SSE by construction, and SSE partitions total variability ($S_{yy}$) into the part explained by the regression ($b_1S_{xy}$) and the part left over.

Common pitfall: testing ρ=0.6 with the simple $r\sqrt{n-2}/\sqrt{1-r^2}\sim t_{n-2}$ statistic, which is valid only for testing $\rho=0$ — a nonzero null requires the Fisher z-transform used here.

Source: Walpole, Myers, Myers & Ye, Ch. 11 (Simple Linear Regression and Correlation).

Question 8 of 8: Correlation and Simple Linear Regression — Gas Consumption

Question 8: Correlation and Simple Linear Regression — Gas Consumption (20 marks)