Question 8 of 8

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. $n=16$; $\sum X=90$, $\sum X^2=600$, $\sum Y=56$, $\sum Y^2=296$, $\sum XY=231$.

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

Approach. Reduce to the corrected sums of squares/cross-products $S_{xx}, S_{yy}, S_{xy}$; these drive the correlation coefficient, the least-squares slope/intercept, and (via a residual-based standard error) the slope's confidence interval. The correlation's own CI uses Fisher's $z$-transform since $r$'s sampling distribution is not Normal.

1. Corrected sums. $$\bar X=\frac{90}{16}=5.625,\quad \bar Y=\frac{56}{16}=3.5$$ $$S_{xx}=\sum X^2-\frac{(\sum X)^2}{n}=600-\frac{90^2}{16}=\boxed{93.75}$$ $$S_{yy}=\sum Y^2-\frac{(\sum Y)^2}{n}=296-\frac{56^2}{16}=\boxed{100.0}$$ $$S_{xy}=\sum XY-\frac{(\sum X)(\sum Y)}{n}=231-\frac{90(56)}{16}=\boxed{-84.0}$$
2. (a) Sample correlation coefficient. $$r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{-84.0}{\sqrt{93.75(100.0)}}=\boxed{-0.8675}$$ The strong negative correlation indicates that employees with more days of absence tend to have fewer years of service.
3. (b) 95% CI for $\rho$ via Fisher's z-transform. $$z'=\frac12\ln\frac{1+r}{1-r}=\boxed{-1.3231}\qquad SE_{z'}=\frac{1}{\sqrt{n-3}}=\frac{1}{\sqrt{13}}=0.2774$$ $$z'\pm1.96\,SE_{z'}=-1.3231\pm0.5437 \implies (-1.8668,\,-0.7795)$$ Converting back via $\rho=(e^{2z'}-1)/(e^{2z'}+1)$: $$\boxed{-0.953 < \rho < -0.652}$$
4. (c) Normal equations and least-squares estimates. The normal equations of $\hat Y=b_0+b_1X$ are $$\sum Y=nb_0+b_1\sum X, \qquad \sum XY=b_0\sum X+b_1\sum X^2$$ Solving (equivalently, $b_1=S_{xy}/S_{xx}$ and $b_0=\bar Y-b_1\bar X$): $$b_1=\frac{-84.0}{93.75}=\boxed{-0.896}\qquad b_0=3.5-(-0.896)(5.625)=\boxed{8.54}$$ So $\hat Y = 8.54 - 0.896\,X$: each additional day of absence is associated with about 0.9 fewer years of service, on average.
5. (d) 95% CI for the slope $\beta_1$. Residual standard error: $$s_{Y|X}^2=\frac{S_{yy}-b_1S_{xy}}{n-2}=\frac{100.0-(-0.896)(-84.0)}{14}=\boxed{1.767}\qquad s_{Y|X}=1.329$$ $$SE_{b_1}=\frac{s_{Y|X}}{\sqrt{S_{xx}}}=\frac{1.329}{\sqrt{93.75}}=0.1373$$ $$b_1\pm t_{0.025,14}\,SE_{b_1}=-0.896\pm2.145(0.1373)=-0.896\pm0.2945$$ $$\boxed{-1.190 < \beta_1 < -0.602}$$