04-BS-2, December 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. $n=16$; $\sum X=384.0$, $\sum X^2=10{,}716.0$, $\sum Y=2{,}496.0$, $\sum Y^2=391{,}536.0$, $\sum XY=61{,}254.0$.

Find. (a) covariance and $r$; (b) 95% CI for $\rho$; (c) normal equations, $b_0$, $b_1$; (d) SSE and 95% CI for $\beta_1$.

Approach. Reduce every sum to the corrected sums of squares/cross-products $S_{xx},S_{yy},S_{xy}$; these feed the covariance, correlation, least-squares slope/intercept, error sum of squares, and (via the Fisher z-transform) the correlation confidence interval.

Corrected sums. $S_{xx}=\sum X^2-\dfrac{(\sum X)^2}{n}=10{,}716.0-\dfrac{384.0^2}{16}=10{,}716.0-9{,}216.0=1{,}500.0$. $S_{yy}=391{,}536.0-\dfrac{2{,}496.0^2}{16}=391{,}536.0-389{,}376.0=2{,}160.0$. $S_{xy}=61{,}254.0-\dfrac{384.0(2{,}496.0)}{16}=61{,}254.0-59{,}904.0=1{,}350.0$.
(a) Covariance and correlation. $\text{Cov}(X,Y)=\dfrac{S_{xy}}{n-1}=\dfrac{1{,}350.0}{15}=\boxed{90.0}$. $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{1{,}350.0}{\sqrt{1{,}500.0(2{,}160.0)}}=\dfrac{1{,}350.0}{1{,}800.0}=\boxed{0.75}$.
(b) 95% CI for ρ via Fisher's z-transform. $z_r=\dfrac12\ln\!\left(\dfrac{1+r}{1-r}\right)=\dfrac12\ln(7)=0.9730$, $\text{SE}(z_r)=\dfrac1{\sqrt{n-3}}=\dfrac1{\sqrt{13}}=0.2774$. $z_r\pm1.96(0.2774)=0.9730\pm0.5436\Rightarrow(0.4294,1.5166)$. Back-transforming $\rho=\dfrac{e^{2z}-1}{e^{2z}+1}$: $\boxed{0.405<\rho<0.908}$.
(c) Normal equations and least-squares estimates. The normal equations are $nb_0+b_1\sum X=\sum Y$ and $b_0\sum X+b_1\sum X^2=\sum XY$, i.e. $16b_0+384b_1=2{,}496.0$ and $384b_0+10{,}716b_1=61{,}254.0$. Slope $b_1=\dfrac{S_{xy}}{S_{xx}}=\dfrac{1{,}350.0}{1{,}500.0}=\boxed{0.900}$. Intercept $b_0=\bar Y-b_1\bar X=\dfrac{2{,}496.0}{16}-0.900\!\left(\dfrac{384.0}{16}\right)=156.0-0.900(24.0)=156.0-21.6=\boxed{134.4}$. Check: $16(134.4)+384(0.9)=2{,}150.4+345.6=2{,}496.0$ ✓.
(d) Error sum of squares and 95% CI for β₁. $\text{SSE}=S_{yy}-b_1S_{xy}=2{,}160.0-0.900(1{,}350.0)=2{,}160.0-1{,}215.0=\boxed{945.0}$. $\text{MSE}=\dfrac{\text{SSE}}{n-2}=\dfrac{945.0}{14}=67.5$. $\text{SE}(b_1)=\sqrt{\dfrac{\text{MSE}}{S_{xx}}}=\sqrt{\dfrac{67.5}{1{,}500.0}}=0.2121$. $b_1\pm t_{0.025,14}\,\text{SE}(b_1)=0.900\pm2.145(0.2121)=0.900\pm0.4550$, giving $\boxed{0.445<\beta_1<1.355}$ (thousands of $/year of experience).

Question 8 — final results
Part	Quantity	Value
(a)	Cov(X,Y); r	90.0; 0.75
(b)	95% CI for ρ	(0.405, 0.908)
(c)	b₀; b₁	134.4; 0.900
(d)	SSE; 95% CI for β₁	945.0; (0.445, 1.355)

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

Key relations: $r=S_{xy}/\sqrt{S_{xx}S_{yy}}$; $z_r=\tfrac12\ln\frac{1+r}{1-r}$, $\text{SE}(z_r)=1/\sqrt{n-3}$; normal equations $nb_0+b_1\sum X=\sum Y$, $b_0\sum X+b_1\sum X^2=\sum XY$; $b_1=S_{xy}/S_{xx}$; $\text{SSE}=S_{yy}-b_1S_{xy}$; $\text{SE}(b_1)=\sqrt{\text{MSE}/S_{xx}}$.

Why this works: Fisher's transform makes the sampling distribution of $r$ (which is skewed and bounded in $[-1,1]$) approximately normal in the $z$ scale, enabling a standard symmetric CI that is then mapped back through the inverse transform; the least-squares normal equations are exactly the calculus conditions ($\partial/\partial b_0=\partial/\partial b_1=0$) that minimize the sum of squared residuals.

Common pitfall: building the CI for $\rho$ directly on the raw $r$ scale (ignoring its skewed sampling distribution) instead of transforming to $z_r$ first; or reporting an intercept in the wrong units (thousands of dollars, per the problem's own scaling of $Y$).

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)