04-BS-2, May 2016, Q8 | Nivaar Exam Prep

Question 8 (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=16)
ΣX	911.0
ΣX²	52,245.0
ΣY	1,313.0
ΣY²	128,283.0
ΣXY	77,196.0

Find. (a) covariance and $r$; (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 $S_{xx},S_{yy},S_{xy}$, from which the covariance, correlation, and regression coefficients all follow directly; use Fisher's $z$-transform for the correlation interval and the standard regression-inference formulas for $\beta_1$.

Corrected sums and means. $\bar x=911/16=56.9375$, $\bar y=1313/16=82.0625$. $S_{xx}=52{,}245-911^2/16=374.94$. $S_{yy}=128{,}283-1313^2/16=20{,}534.94$. $S_{xy}=77{,}196-(911)(1313)/16=2{,}437.06$.
(a) Covariance and correlation. $\widehat{\text{Cov}}(X,Y)=\dfrac{S_{xy}}{n-1}=\dfrac{2437.06}{15}=162.47$. $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{2437.06}{\sqrt{374.94(20534.94)}}=\boxed{0.8783}$.
(b) 95% CI for ρ via Fisher's z-transform. $z_r=\tfrac12\ln\!\left(\dfrac{1+r}{1-r}\right)=1.3683$, $\sigma_z=\dfrac{1}{\sqrt{n-3}}=0.2774$. $z_r\pm1.96\sigma_z=(0.8247,\,1.9119)$. Back-transforming $\rho=\dfrac{e^{2z}-1}{e^{2z}+1}$ gives $\boxed{0.678<\rho<0.957}$.
(c) Normal equations and least-squares estimates. The normal equations are $\Sigma y_i=nb_0+b_1\Sigma x_i$ and $\Sigma x_iy_i=b_0\Sigma x_i+b_1\Sigma x_i^2$. Solving via the corrected sums: $b_1=\dfrac{S_{xy}}{S_{xx}}=\dfrac{2437.06}{374.94}=6.500$, $b_0=\bar y-b_1\bar x=82.0625-6.500(56.9375)=\boxed{-288.03}$, so $\hat y=-288.03+6.500x$.
(d) Error sum of squares and CI for β1. $SSE=S_{yy}-b_1S_{xy}=20534.94-6.500(2437.06)=4{,}694.23$. $MSE=\dfrac{SSE}{n-2}=\dfrac{4694.23}{14}=335.30$, $s=18.311$. $SE(b_1)=\dfrac{s}{\sqrt{S_{xx}}}=\dfrac{18.311}{\sqrt{374.94}}=0.9457$. $t_{0.025,14}=2.145$. $b_1\pm t\cdot SE(b_1)=6.500\pm2.028$, so $\boxed{4.47<\beta_1<8.53}$.

Question 8 — final results
Part	Quantity	Value
(a)	Cov(X,Y); r	162.47; 0.8783
(b)	95% CI for ρ	(0.678, 0.957)
(c)	b₀, b₁	−288.03, 6.500
(d)	SSE; 95% CI for β₁	4,694.23; (4.47, 8.53)

Topic: Correlation, Fisher's z-transform confidence interval, and simple linear regression by least squares.

Key relations: $r=S_{xy}/\sqrt{S_{xx}S_{yy}}$; $z_r=\tfrac12\ln\!\left(\frac{1+r}{1-r}\right)\sim N(0,\,1/(n-3))$; least-squares normal equations $\Sigma y=nb_0+b_1\Sigma x$, $\Sigma xy=b_0\Sigma x+b_1\Sigma x^2$; $b_1\pm t_{\alpha/2,n-2}\,s/\sqrt{S_{xx}}$.

Why this works: minimizing $\Sigma(y_i-b_0-b_1x_i)^2$ by calculus produces exactly the two normal equations above, whose solution reduces to the ratio of corrected sums $S_{xy}/S_{xx}$ for the slope; Fisher's transform is used (rather than a direct interval on $r$) because $r$'s own sampling distribution is skewed near ±1, while $z_r$ is very nearly normal for any $\rho$.

Common pitfall: confusing the correlation interval (uses Fisher's z) with the slope interval (uses the t-distribution and $SE(b_1)$ from the residual variance) — they use different machinery even though both describe the same X–Y relationship.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 11–12 (Correlation and Simple Linear Regression).

Question 8 of 8

Question 8 (20 marks)