04-BS-2, May 2018, Q8 | Nivaar Exam Prep

Question 8

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. $n=39$ urban centres with summary statistics on $X$ (population, hundred-thousands) and $Y$ (deaths/day).

Given data
Symbol	Value
$n$	39
$\sum X$	3,861.0
$\sum X^2$	497,189.0
$\sum Y$	4,641.0
$\sum Y^2$	924,717.0
$\sum XY$	583,605.0

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

Approach. Reduce every sum to the corrected sums of squares/cross-products $S_{xx}, S_{yy}, S_{xy}$, from which $r$, the least-squares slope/intercept, $SSE$ and $se(b_1)$ all follow directly; the CI for $\rho$ needs Fisher's $z$-transform since $r$'s own sampling distribution is skewed.

Corrected sums of squares/cross-products. $$S_{xx}=\sum X^2-\frac{(\sum X)^2}{n}=497{,}189.0-\frac{(3{,}861.0)^2}{39}=114{,}950.0,$$ $$S_{yy}=\sum Y^2-\frac{(\sum Y)^2}{n}=924{,}717.0-\frac{(4{,}641.0)^2}{39}=372{,}438.0,$$ $$S_{xy}=\sum XY-\frac{(\sum X)(\sum Y)}{n}=583{,}605.0-\frac{(3{,}861.0)(4{,}641.0)}{39}=124{,}146.0.$$
a) Correlation coefficient. $$r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{124{,}146.0}{\sqrt{114{,}950.0\times 372{,}438.0}}=\boxed{0.600}.$$
b) 95% CI for ρ (Fisher's z-transform). $$z_r=\tfrac12\ln\!\frac{1+r}{1-r}=\tfrac12\ln\!\frac{1.600}{0.400}=0.6931,\qquad \sigma_{z}=\frac{1}{\sqrt{n-3}}=\frac{1}{\sqrt{36}}=0.1667.$$ With $z_{0.025}=1.960$: $z_r\pm 1.960(0.1667)=0.6931\pm 0.3268\Rightarrow(0.3665,\ 1.0198)$. Back-transforming $r=\tanh(z)$: $$\boxed{0.351 < \rho < 0.770}.$$
c) Normal equations and least-squares estimates. The least-squares normal equations for $Y=\beta_0+\beta_1 X+\varepsilon$ are $$\sum Y = n b_0 + b_1\sum X,\qquad \sum XY = b_0\sum X + b_1\sum X^2,$$ i.e. $4{,}641.0=39b_0+3{,}861.0\,b_1$ and $583{,}605.0=3{,}861.0\,b_0+497{,}189.0\,b_1$. Solving (equivalently $b_1=S_{xy}/S_{xx}$, $b_0=\bar Y-b_1\bar X$): $$b_1=\frac{124{,}146.0}{114{,}950.0}=\boxed{1.080},\qquad \bar X=99.0,\ \bar Y=119.0,\qquad b_0=119.0-1.080(99.0)=\boxed{12.08}.$$ So $\hat Y = 12.08+1.080X$.
d) Error sum of squares and 95% CI for β₁. $$SSE=S_{yy}-b_1S_{xy}=372{,}438.0-1.080(124{,}146.0)=\boxed{238{,}360.3}.$$ $$MSE=\frac{SSE}{n-2}=\frac{238{,}360.3}{37}=6{,}442.2,\qquad se(b_1)=\sqrt{\frac{MSE}{S_{xx}}}=\sqrt{\frac{6{,}442.2}{114{,}950.0}}=0.2367.$$ With $t_{0.025,37}=2.026$: $$b_1\pm t_{0.025,37}\,se(b_1)=1.080\pm 2.026(0.2367)=1.080\pm 0.4797,$$ $$\boxed{0.600 < \beta_1 < 1.560}.$$

Question 8 — final results
Part	Result
a) $r$	0.600
b) 95% CI for $\rho$	(0.351, 0.770)
c) $b_0,\ b_1$	12.08, 1.080
d) $SSE$	238,360.3
d) 95% CI for $\beta_1$	(0.600, 1.560)

Topic: Simple linear regression and inference for the correlation coefficient and the slope.
Key relations: $r=S_{xy}/\sqrt{S_{xx}S_{yy}}$; $b_1=S_{xy}/S_{xx}$, $b_0=\bar Y-b_1\bar X$; $SSE=S_{yy}-b_1S_{xy}$; $se(b_1)=\sqrt{MSE/S_{xx}}$; Fisher $z=\tfrac12\ln\frac{1+r}{1-r}\sim N(0,\tfrac{1}{n-3})$.
Why this works: The corrected sums of squares/cross-products are algebraically equivalent to (but far cheaper than) working from centred data; Fisher's transform is used for $\rho$ because $r$'s own sampling distribution is bounded and skewed near $\pm1$, while $z$ is (approximately) normal for any $\rho$.
Common pitfall: Using $S_{yy}$ (total sum of squares of $Y$) in place of $SSE$ (residual sum of squares) when computing $MSE$ — only the residual variation after removing the fitted line estimates $\sigma^2$.
Source: Walpole, Myers, Myers & Ye, Ch. 11 – Simple Linear Regression and Correlation.

Question 8 of 8

Question 8