Question 8 of 8: Study Hours vs. Exam Mark — Correlation & Regression

Question 8: Study Hours vs. Exam Mark — Correlation & Regression (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=21; ∑X=2,940.0, ∑X²=436,100.0, ∑Y=1,470.0, ∑Y²=110,900.0, ∑XY=219,100.0.

Find. (a) Cov(X,Y), r; (b) test H₀: ρ=0.9; (c) the normal equations and b₀, b₁; (d) SSE and the 95% CI for β₁.

Approach. Reduce every sum to the corrected sums of squares/cross-products $S_{xx}$, $S_{yy}$, $S_{xy}$ first; every later quantity (r, b₁, SSE) is built from these three numbers. Test ρ=0.9 (a nonzero null) with Fisher's z-transform, since the simple $t=r\sqrt{n-2}/\sqrt{1-r^2}$ statistic is only valid for testing ρ=0.

1. Corrected sums and sample means. $$\bar x=\dfrac{2940}{21}=140.0\ \text{hrs},\qquad \bar y=\dfrac{1470}{21}=70.0\ \text{marks}.$$ $$S_{xx}=\sum x^2-\dfrac{(\sum x)^2}{n}=436100-\dfrac{2940^2}{21}=\boxed{24{,}500},$$ $$S_{yy}=\sum y^2-\dfrac{(\sum y)^2}{n}=110900-\dfrac{1470^2}{21}=\boxed{8{,}000},$$ $$S_{xy}=\sum xy-\dfrac{(\sum x)(\sum y)}{n}=219100-\dfrac{2940(1470)}{21}=\boxed{13{,}300}.$$
2. (a) Sample covariance and correlation. $$\mathrm{Cov}(X,Y)=\dfrac{S_{xy}}{n-1}=\dfrac{13300}{20}=\boxed{665.0},\qquad r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{13300}{\sqrt{24500(8000)}}=\boxed{0.95}.$$
3. (b) Test H₀: ρ=0.9 via Fisher's z-transform. Since 0.9≠0, transform both r and ρ₀: $$z_r=\tfrac12\ln\dfrac{1+0.95}{1-0.95}=1.8318,\qquad z_{\rho_0}=\tfrac12\ln\dfrac{1+0.9}{1-0.9}=1.4722,$$ $$Z=\dfrac{z_r-z_{\rho_0}}{1/\sqrt{n-3}}=\dfrac{1.8318-1.4722}{1/\sqrt{18}}=\boxed{1.525}.$$ The two-sided critical value is $z_{0.025}=1.960$. Since $1.525<1.960$, fail to reject H₀: the true correlation is not significantly different from 0.9.
4. (c) Normal equations and least-squares estimates. The normal equations for $\hat y=b_0+b_1x$ 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 $b_0=\bar y-b_1\bar x$): $$b_1=\dfrac{13300}{24500}=\boxed{0.5429}\ \text{marks/hr},\qquad b_0=70.0-0.5429(140.0)=\boxed{-6.00}.$$ So $\hat y=-6.00+0.5429x$: each extra hour of study is associated with roughly 0.54 more exam marks.
5. (d) Error sum of squares and 95% CI for β₁. $$\mathrm{SSE}=S_{yy}-b_1S_{xy}=8000-0.5429(13300)=\boxed{780.0},\qquad \mathrm{MSE}=\dfrac{\mathrm{SSE}}{n-2}=\dfrac{780.0}{19}=41.05.$$ $$\mathrm{SE}(b_1)=\sqrt{\dfrac{\mathrm{MSE}}{S_{xx}}}=\sqrt{\dfrac{41.05}{24500}}=0.04093,\qquad t_{0.025,19}=2.093.$$ $$b_1\pm t\,\mathrm{SE}(b_1)=0.5429\pm2.093(0.04093)=0.5429\pm0.0857\ \Rightarrow\ \boxed{0.457<\beta_1<0.629}.$$