Exam SRM — Linear Regression Practice Flashcards

**Answer: (B).** The slope is $\hat\beta_1 = \dfrac{S_{xy}}{S_{xx}}$ using corrected (centered) sums. $\bar x = \dfrac{40}{8}=5$, $\bar y = \dfrac{120}{8}=15$. $S_{xx} = \sum x^{2} - n\bar x^{2} = 240 - 8(25) = 40$. $S_{xy} = \sum xy - n\bar x\bar y = 680 - 8(5)(15) = 680 - 600 = 80$. $\hat\beta_1 = \dfrac{80}{40} = 2.00$. Distractor (E) $17.00 = \dfrac{\sum xy}{\sum x} = \dfrac{680}{40}$ divides the raw cross-product by the raw $x$-sum; (C) $2.83 = \dfrac{\sum xy}{\sum x^{2}} = \dfrac{680}{240}$ uses the raw (uncorrected) sums instead of the centered $S_{xy}/S_{xx}$; (A) and (D) come from other miscentering slips. Always use the corrected sums $S_{xy}/S_{xx}$, not the raw moments.

**Answer: (C).** The least-squares line passes through the centroid $(\bar x,\bar y)$, so $\hat\beta_0 = \bar y - \hat\beta_1\bar x = 15 - 2(5) = 15 - 10 = 5.00$. The fitted line is $\hat y = 5 + 2x$. (Forgetting to subtract and reporting $\bar y = 15$ gives distractor (E); using $\hat\beta_0 = \bar y - \hat\beta_1\bar y$ or other centroid slips give the others.)

**Answer: (B).** $n=5$, $\sum x = 15$, $\sum y = 30$, so $\bar x = 3$, $\bar y = 6$. $\sum x^{2} = 1+4+9+16+25 = 55 \Rightarrow S_{xx} = 55 - 5(9) = 10$. $\sum xy = 3+10+12+32+50 = 107 \Rightarrow S_{xy} = 107 - 5(3)(6) = 107 - 90 = 17$. $\hat\beta_1 = \dfrac{S_{xy}}{S_{xx}} = \dfrac{17}{10} = 1.70$. (The intercept is $\hat\beta_0 = 6 - 1.7(3) = 0.90$, which appears as distractor (A) — the intercept, not the slope. Using the uncorrected $\sum xy/\sum x^2$-type ratios produces the others.)

**Answer: (C).** The sample correlation uses both corrected sums of squares in the denominator: $r = \dfrac{S_{xy}}{\sqrt{S_{xx}\,S_{yy}}} = \dfrac{180}{\sqrt{80 \cdot 500}} = \dfrac{180}{\sqrt{40000}} = \dfrac{180}{200} = 0.90$. Distractor (B) $0.81$ is $r^{2}=R^{2}$, not $r$. Distractor (D) $2.25 = \dfrac{S_{xy}}{S_{xx}}=\hat\beta_1$, the slope (a common slip: computing the slope when correlation is asked). Distractor (A) $0.36$ squares the wrong quantity.

**Answer: (C).** In simple linear regression $R^{2}=r^{2}$, the squared correlation. $r = \dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} = \dfrac{144}{\sqrt{64 \cdot 400}} = \dfrac{144}{\sqrt{25600}} = \dfrac{144}{160} = 0.90$. $R^{2} = r^{2} = 0.90^{2} = 0.81$. Equivalently $R^{2} = \dfrac{\text{SSR}}{\text{SST}} = \dfrac{S_{xy}^{2}/S_{xx}}{S_{yy}} = \dfrac{144^{2}/64}{400} = \dfrac{324}{400} = 0.81$. (Reporting $r = 0.90$ instead of $r^{2}$ is distractor (D).)

**Answer: (A).** The regression (explained) sum of squares in SLR is $\text{SSR} = \dfrac{S_{xy}^{2}}{S_{xx}} = \dfrac{144^{2}}{64} = \dfrac{20736}{64} = 324$. By the ANOVA decomposition $\text{SST}=\text{SSR}+\text{SSE}$ with $\text{SST}=S_{yy}=400$: $\text{SSE} = \text{SST} - \text{SSR} = 400 - 324 = 76$. Distractor (D) $324$ is the SSR (explained), not the SSE (unexplained) — swapping the two is the most common error here.

**Answer: (B).** The residual standard error is $s = \sqrt{\dfrac{\text{SSE}}{n-2}}$, with $n-2$ degrees of freedom because two parameters ($\hat\beta_0,\hat\beta_1$) are estimated. $s = \sqrt{\dfrac{40}{10-2}} = \sqrt{\dfrac{40}{8}} = \sqrt{5} \approx 2.236$. Distractor (D) $5.00$ is $s^{2}=\text{MSE}$ (the variance, not the standard error). Distractor (A) divides by $n$ instead of $n-2$: $\sqrt{40/10}=2.0$.

**Answer: (C).** The slope standard error is $SE(\hat\beta_1) = \dfrac{s}{\sqrt{S_{xx}}} = \dfrac{2.608}{\sqrt{100}} = \dfrac{2.608}{10} = 0.2608$. The test statistic is $t = \dfrac{\hat\beta_1}{SE(\hat\beta_1)} = \dfrac{0.8}{0.2608} \approx 3.07$, compared to $t_{n-2}=t_{10}$. Since $|t|=3.07 > t_{0.025,10}\approx 2.228$, reject $H_0$. (Distractor (A) $0.31$ is the standard error scaled; (D) $8.00$ divides by $s/S_{xx}$ instead of $s/\sqrt{S_{xx}}$.)