Correlation and Simple Linear Regression: 04-BS-2, December 2015, Q8

Question 8: Correlation and Simple Linear Regression (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	21
ΣX	588.0
ΣX²	16,544.0
ΣY	168.0
ΣY²	1,357.0
ΣXY	4,731.0

Find. (a) covariance, r; (b) 95% CI for ρ; (c) normal equations, b₀, b₁; (d) SSE and 95% CI for β₁.

Approach. Reduce every quantity to the corrected sums of squares/cross-products $S_{xx}, S_{yy}, S_{xy}$, then read off covariance, correlation, and the least-squares slope/intercept directly; use the Fisher z-transform for the CI on ρ, and the regression-SE formula for the CI on β₁.

Corrected sums of squares/products. $$S_{xx}=\Sigma X^2-\dfrac{(\Sigma X)^2}{n}=16{,}544.0-\dfrac{588.0^2}{21}=80.0,\qquad S_{yy}=1{,}357.0-\dfrac{168.0^2}{21}=13.0,$$ $$S_{xy}=\Sigma XY-\dfrac{\Sigma X\Sigma Y}{n}=4{,}731.0-\dfrac{588.0\times168.0}{21}=27.0.$$
(a)(i)–(ii) Covariance and correlation. $$\text{Cov}(X,Y)=\dfrac{S_{xy}}{n-1}=\dfrac{27.0}{20}=\boxed{1.35},\qquad r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{27.0}{\sqrt{80.0\times13.0}}=\boxed{0.837}.$$
(b) 95% CI for ρ (Fisher z-transform). $$z_r=\tfrac12\ln\dfrac{1+r}{1-r}=\tfrac12\ln\dfrac{1.837}{0.163}=1.212,\qquad \sigma_z=\dfrac{1}{\sqrt{n-3}}=\dfrac{1}{\sqrt{18}}=0.2357.$$ $$z_r\pm1.96\sigma_z=1.212\pm0.462\ \Rightarrow\ (0.750,\ 1.674).$$ Transforming back with $\rho=\tanh(z)$: $$\boxed{0.635<\rho<0.932}.$$
(c) Normal equations and least-squares estimates. The normal equations of the least-squares line $\hat Y=b_0+b_1X$ are $$\Sigma Y = n\,b_0+b_1\Sigma X,\qquad \Sigma XY = b_0\Sigma X+b_1\Sigma X^2,$$ i.e. $168.0=21b_0+588.0\,b_1$ and $4{,}731.0=588.0\,b_0+16{,}544.0\,b_1$. Solving (equivalently, $b_1=S_{xy}/S_{xx}$ and $b_0=\bar Y-b_1\bar X$): $$b_1=\dfrac{S_{xy}}{S_{xx}}=\dfrac{27.0}{80.0}=\boxed{0.3375},\qquad b_0=\dfrac{168.0}{21}-0.3375\times\dfrac{588.0}{21}=8.0-9.45=\boxed{-1.45}.$$ So $\hat Y=-1.45+0.3375X$.
(d) Error sum of squares and 95% CI for β₁. $$SSE=S_{yy}-b_1S_{xy}=13.0-0.3375(27.0)=\boxed{3.8875},\qquad MSE=\dfrac{SSE}{n-2}=\dfrac{3.8875}{19}=0.2046.$$ $$s_{b_1}=\sqrt{\dfrac{MSE}{S_{xx}}}=\sqrt{\dfrac{0.2046}{80.0}}=0.05057,\qquad t_{0.025,19}=2.093.$$ $$b_1\pm t_{0.025,19}\,s_{b_1}=0.3375\pm2.093(0.05057)=0.3375\pm0.1059.$$ $$\boxed{0.232<\beta_1<0.443}$$

Final results — Question 8
Part	Result
(a)(i) Cov(X,Y)	1.35 (100 sq ft)·(1000 kWh)
(a)(ii) r	0.837
(b) 95% CI for ρ	(0.635, 0.932)
(c) b₀, b₁	−1.45, 0.3375
(d) SSE; 95% CI for β₁	3.8875; (0.232, 0.443)

Topic: Correlation, its confidence interval via the Fisher transform, and simple linear regression by least squares.

Key relations: $r=S_{xy}/\sqrt{S_{xx}S_{yy}}$; $z_r=\tfrac12\ln[(1+r)/(1-r)]\sim N(\cdot,\,1/\sqrt{n-3})$; $b_1=S_{xy}/S_{xx}$, $b_0=\bar Y-b_1\bar X$; $s_{b_1}=\sqrt{MSE/S_{xx}}$.

Why this works: the corrected sums of squares/cross-products are the sufficient statistics for both correlation and regression, since both quantities are ratios/ combinations of the same three numbers ($S_{xx}$, $S_{yy}$, $S_{xy}$); the Fisher z-transform is used for $\rho$'s CI because $r$'s own sampling distribution is skewed while $z_r$ is (nearly) normal.

Common pitfall: confusing $n-1$ (covariance denominator) with $n$ (used inside $S_{xx}$/$S_{yy}$/$S_{xy}$ themselves), or forgetting to transform the Fisher-z confidence limits back to the $\rho$-scale with $\tanh(\cdot)$ rather than reporting $z$ directly.

Source: Walpole et al., Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 11–12 (Correlation and simple linear regression).

Question 8 of 8: Correlation and Simple Linear Regression

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