Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, December 2017 — 04-BS-2 Probability and Statistics, 2 hours, closed book except one hand-written information sheet and an approved calculator. Eight questions of equal value; the exam instructs "any 5 questions constitute a complete paper," but all 8 are answered here as a full study resource.
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists (9th ed.) — used throughout for distribution theory, estimation, and hypothesis-testing procedures.
Find. (a) Covariance and r. (b) 95% CI for ρ. (c) Least-squares normal equations and $b_0, b_1$. (d) SSE and the 95% CI for β₁.
Approach. Compute the corrected sums of squares/products $S_{xx}, S_{yy}, S_{xy}$ first; they feed the covariance, correlation, regression coefficients, and (via SSE) the standard error of $b_1$. Use the Fisher $z$-transformation for the CI on ρ (since $r$'s own sampling distribution is skewed).
Preliminary: means and corrected sums.
$$\bar{x}=\frac{800}{25}=32, \qquad \bar{y}=\frac{125}{25}=5.$$
$$S_{xx}=\sum X^2-n\bar{x}^2=25{,}984-25(1024)=384, \qquad S_{yy}=\sum Y^2-n\bar{y}^2=2161-25(25)=1536,$$
$$S_{xy}=\sum XY-n\bar{x}\bar{y}=4624-25(32)(5)=624.$$
(a)(i) Covariance and (ii) correlation.
$$\text{Cov}(X,Y)=\frac{S_{xy}}{n-1}=\frac{624}{24}=\boxed{26.0}, \qquad r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{624}{\sqrt{384(1536)}}=\frac{624}{768}=\boxed{0.8125}.$$
(b) 95% CI for ρ, Fisher z.
$$z_r=\tfrac12\ln\!\frac{1+r}{1-r}=1.1300, \qquad SE=\frac{1}{\sqrt{n-3}}=\frac{1}{\sqrt{22}}=0.2132.$$
$$z_r\pm1.96(0.2132) = 1.1300\pm0.4178 = (0.7122,\ 1.5478).$$
Back-transforming with $\rho=(e^{2z}-1)/(e^{2z}+1)$ gives $\boxed{(0.6147,\ 0.9141)}$.
(c) Normal equations and least-squares estimates. The normal equations are
$$\sum Y = nb_0+b_1\sum X, \qquad \sum XY = b_0\sum X+b_1\sum X^2.$$
Solving (equivalently, $b_1=S_{xy}/S_{xx}$, $b_0=\bar{y}-b_1\bar{x}$):
$$b_1=\frac{624}{384}=\boxed{1.625}, \qquad b_0=5-1.625(32)=\boxed{-47.0}.$$
So the fitted line is $\hat{Y}=-47.0+1.625X$.
(d) Error sum of squares and CI for β₁.
$$SSE = S_{yy}-b_1S_{xy} = 1536-1.625(624) = \boxed{522.0}, \qquad MSE=\frac{SSE}{n-2}=\frac{522}{23}=22.70,\quad s=4.764.$$
$$SE(b_1)=\frac{s}{\sqrt{S_{xx}}}=\frac{4.764}{\sqrt{384}}=0.2431, \qquad t_{0.025,23}=2.069.$$
$$b_1\pm t_{0.025,23}\,SE(b_1) = 1.625\pm 0.503, \qquad \boxed{(1.122,\ 2.128)}.$$
