04-BS-2, May 2017, 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. $n=28$ candidates; $\sum X=2{,}240.0$, $\sum X^2=190{,}000.0$, $\sum Y=1{,}848.0$, $\sum Y^2=124{,}155.0$, $\sum XY=151{,}971.0$.

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

Approach. Reduce to the corrected sums of squares/cross-products $S_{xx}, S_{yy}, S_{xy}$; these feed the covariance, $r$, the least-squares slope/intercept, and (via the residual sum of squares) the standard error needed for the CIs. $\rho$'s CI uses the Fisher $z$-transformation; $\beta_1$'s CI uses the $t$-distribution.

Reduce to corrected sums. $\bar x=\dfrac{2240}{28}=80.0$, $\bar y=\dfrac{1848}{28}=66.0$. $S_{xx}=\sum X^2-\dfrac{(\sum X)^2}{n}=190{,}000-\dfrac{2240^2}{28}=10{,}800.0$. $S_{yy}=124{,}155-\dfrac{1848^2}{28}=2{,}187.0$. $S_{xy}=151{,}971-\dfrac{(2240)(1848)}{28}=4{,}131.0$.
(a) Covariance and correlation. $\widehat{\text{Cov}}(X,Y)=\dfrac{S_{xy}}{n-1}=\dfrac{4131}{27}=153.0$. $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{4131}{\sqrt{10800(2187)}}=\dfrac{4131}{4860.0}=0.850$. So $\boxed{\widehat{\text{Cov}}(X,Y)=153.0,\ r=0.850}$ — a strong positive linear association between study hours and mark.
(b) 95% CI for ρ (Fisher z). $z'=\dfrac12\ln\!\left(\dfrac{1+r}{1-r}\right)=\dfrac12\ln\!\left(\dfrac{1.850}{0.150}\right)=1.2562$, with $\text{SE}=\dfrac{1}{\sqrt{n-3}}=\dfrac{1}{\sqrt{25}}=0.200$. $z'\pm1.96(0.200)=(0.8642,\,1.6482)$. Back-transforming $\rho=\dfrac{e^{2z'}-1}{e^{2z'}+1}$ at each limit gives $\boxed{0.698<\rho<0.929}$.
(c) Normal equations and least-squares fit. The normal equations are $\sum Y=nb_0+b_1\sum X$ and $\sum XY=b_0\sum X+b_1\sum X^2$. Solving via the corrected sums, $b_1=\dfrac{S_{xy}}{S_{xx}}=\dfrac{4131}{10800}=0.3825$, $b_0=\bar y-b_1\bar x=66-0.3825(80)=35.40$. So $\boxed{\hat Y=35.40+0.3825X}$: each extra hour of study is associated with about $0.38$ extra marks.
(d) Error sum of squares and CI for β₁. $\text{SSE}=S_{yy}-b_1S_{xy}=2187-0.3825(4131)=606.89$. $\text{MSE}=\dfrac{\text{SSE}}{n-2}=\dfrac{606.89}{26}=23.34$, so $s=4.831$. $\text{SE}(b_1)=\dfrac{s}{\sqrt{S_{xx}}}=\dfrac{4.831}{\sqrt{10800}}=0.04649$. With $t_{0.025,26}=2.056$: margin $=2.056(0.04649)=0.0956$. So $\boxed{0.287<\beta_1<0.478}$.

Question 8 — final results
Part	Quantity	Value
(a)	Cov(X,Y); r	153.0; 0.850
(b)	95% CI for ρ	(0.698, 0.929)
(c)	b₀; b₁	35.40; 0.3825
(d)	SSE; 95% CI for β₁	606.89; (0.287, 0.478)

Topic: Simple linear correlation and least-squares regression, with interval estimates for ρ and β₁.

Key relations: $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$; Fisher's $z'=\tfrac12\ln\frac{1+r}{1-r}\sim N\!\left(\tfrac12\ln\frac{1+\rho}{1-\rho},\ \tfrac{1}{n-3}\right)$; least-squares $b_1=S_{xy}/S_{xx}$, $b_0=\bar y-b_1\bar x$; $\dfrac{b_1-\beta_1}{s/\sqrt{S_{xx}}}\sim t_{n-2}$ with $s^2=\text{SSE}/(n-2)$.

Why this works: the least-squares normal equations are the calculus first-order conditions for minimizing $\sum(y_i-b_0-b_1x_i)^2$; Fisher's transform is used for $\rho$ (rather than a direct normal CI) because $r$'s own sampling distribution is skewed except very close to zero, while $z'$ is (to a good approximation) normal for any $\rho$.

Common pitfall: confusing the sample covariance (divide by $n-1$) with the corrected cross-product $S_{xy}$ (no division) when computing $r$ or $b_1$, and forgetting the $n-2$ (not $n-1$ or $n$) degrees of freedom used for SSE/MSE and the $t$-critical value in a simple regression.

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

Question 8 of 8

Question 8 (20 marks)