Solution. Use the computational form, which avoids working out every deviation: \[r=\frac {n\sum XY-\sum X\sum Y} {\sqrt {n\sum X^2-\left (\sum X\right )^2}\ \sqrt {n\sum Y^2-\left (\sum Y\right )^2}}.\] From the table, with \(n=12\): \[\sum X=18,\quad \sum Y=12,\quad \sum XY=837,\quad \sum X^2=1550,\quad \sum Y^2=592.\] Then \[n\sum XY-\sum X\sum Y = 12(837)-(18)(12)=10044-216=9828,\] \[n\sum X^2-\left (\sum X\right )^2=12(1550)-324=18276,\] \[n\sum Y^2-\left (\sum Y\right )^2=12(592)-144=6960,\] so \[r=\frac {9828}{\sqrt {18276}\ \sqrt {6960}}=\frac {9828}{135.19\times 83.43}=0.871,\] as required.

A word on why the scores being “quoted relative to the mean” does not matter. Correlation is unaffected by adding a constant to either variable, so subtracting each test’s mean beforehand leaves \(r\) exactly as it was. It only makes the arithmetic pleasanter.

An \(r\) of \(0.871\) is a strong positive association: pupils who did well on one test tended to do well on the other, which is what one would hope of two tests intended to measure the same thing.

Solution. (a). With \(n=10\), \[\sum X=75,\quad \sum Y=110,\quad \sum XY=707,\quad \sum X^2=677.\] \[\hat {b}=\frac {n\sum XY-\sum X\sum Y}{n\sum X^2-\left (\sum X\right )^2} =\frac {10(707)-75(110)}{10(677)-75^2}=\frac {7070-8250}{6770-5625} =\frac {-1180}{1145}=-1.031.\] \[\hat {a}=\overline {Y}-\hat {b}\,\overline {X}=11-(-1.031)(7.5)=18.729.\] So the regression line is \[Y=18.729-1.031X.\] The negative slope says \(Y\) falls by about \(1.03\) units for each unit increase in \(X\). As a check, the line passes through \((\overline {X},\overline {Y})=(7.5,11)\), as it must.

(b). \(r=-0.822\) and \(R^2=0.676\).

The sign of \(r\) matches the sign of the slope, as it always does. \(R^2=0.676\) says the line accounts for about \(68\%\) of the variation in \(Y\); the remaining \(32\%\) is scatter about the line that \(X\) does not explain.

(c). The fitted values give \[SS_T=180.0,\qquad SS_R=121.61,\qquad SS_E=58.39,\] and \(SS_R/SS_T=0.676=R^2\), which is a useful arithmetic check.

Testing \(H_0:\beta =0\), the critical value is \(F_{1,8,\,0.05}=5.32\) and \(16.66\) is well beyond it. The \(P\)-value is \(0.0035\), so we reject \(H_0\): there is strong evidence of a linear relationship.

The equivalent \(t\)-test gives the same verdict. With \(s=\sqrt {MS_E}=2.70\) and \(\sum (X-\overline {X})^2=114.5\), \[SE(\hat {b})=\frac {2.70}{\sqrt {114.5}}=0.2525, \qquad t=\frac {-1.031}{0.2525}=-4.08\] on \(8\) degrees of freedom, against a critical \(2.306\) – and the same \(P\)-value of \(0.0035\). Note that \[t^2=(-4.08)^2=16.66=F,\] exactly as it must be: in simple linear regression the two tests are one test written two ways.

A \(95\%\) confidence interval for the slope is \[-1.031\pm 2.306(0.2525)=-1.031\pm 0.582=(-1.613,\ -0.449).\] It excludes zero, which is the same conclusion again – but it also says how large the effect plausibly is, which neither test does on its own.

Solution. Both use \[r=\frac {S_{XY}}{\sqrt {S_{XX}S_{YY}}},\qquad S_{XY}=\sum XY-\frac {\sum X\sum Y}{n},\quad S_{XX}=\sum X^2-\frac {(\sum X)^2}{n}.\]

First set (\(n=9\)): \(\sum X=81\), \(\sum Y=18.9\), giving \[S_{XX}=170.0,\qquad S_{YY}=9.64,\qquad S_{XY}=-36.7\] \[r=\frac {-36.7}{\sqrt {170.0\times 9.64}}=\frac {-36.7}{40.48}=-0.907.\]

Second set (\(n=8\)): \(\sum X=76\), \(\sum Y=20.9\), giving \[S_{XX}=134.0,\qquad S_{YY}=10.05,\qquad S_{XY}=-35.95\] \[r=\frac {-35.95}{\sqrt {134.0\times 10.05}}=\frac {-35.95}{36.70}=-0.980.\]

Both are strong negative correlations: more chemical, fewer blooms. The second set is the tighter of the two, \(-0.980\) against \(-0.907\).

Two cautions. The chemical is described as vital as well as toxic, so the true relationship is unlikely to be a straight line over the whole range — too little would harm the plants as well. These data only cover the descending part of that curve. And correlation this strong still does not establish that the chemical causes the reduction; the plots inside and outside glasshouses may differ in other ways too.

Solution. A rank test is the right tool here, and not merely an alternative to the product-moment coefficient. The two judges are using quite different parts of their scales — X’s marks run from \(46\) to \(55\), Y’s from \(11\) to \(29\) — so what matters is whether they put the competitors in the same order, not whether their numbers agree.

(a) Rank each judge’s marks, sharing ranks where there are ties:

With \(\sum d^2=20\) and \(n=10\), \[r_s=1-\frac {6\sum d^2}{n(n^2-1)}=1-\frac {6(20)}{10(99)}=1-0.121=0.879.\] The critical value for \(n=10\) at the \(5\%\) level (two-tailed) is \(0.648\). Since \(0.879>0.648\), the judges’ opinions are consistent.

(b) Since the judges broadly agree, the sensible order of merit adds their ranks: C (\(10+10=20\)) first, then I (\(9+8=17\)), E (\(7.5+9=16.5\)), D (\(7.5+6.5=14\)), B (\(5.5+5=10.5\)), G (\(3.5+6.5=10\)), J (\(5.5+3=8.5\)), A (\(3.5+4=7.5\)), F (\(2+2=4\)), H (\(1+1=2\)).

Adding the raw marks would be wrong: Judge X’s range is only \(9\) marks wide while Judge Y’s is \(18\), so Y’s opinion would silently count for twice as much.

Solution. Before anything else. The \(2000\) figure of \(3.5\) sits among values of \(35\) to \(62\) and is followed by \(37.1\). It is almost certainly \(33.5\) with a digit dropped. The working below uses the data as given, and then shows what difference the anomaly makes — altering someone’s data without saying so is not an option.

(a), (b) Code the years as \(x=\text {year}-2000\), so \(x=0,1,\ldots ,9\). Then \(n=10\), \(\sum x=45\), \(\sum Y=408.0\), and \[S_{XX}=82.5,\qquad S_{XY}=361.2\] \[\widehat {b}=\frac {S_{XY}}{S_{XX}}=\frac {361.2}{82.5}=4.378,\qquad \widehat {a}=\overline {Y}-\widehat {b}\,\overline {x}=40.8-4.378(4.5)=21.10\] \[\implies \quad \widehat {Y}=21.10+4.378x.\]

(c) For \(2010\), \(x=10\): \[\widehat {Y}=21.10+4.378(10)=64.9\ \text {million litres}.\]

(d) The actual figure was \(78.3\), so the model is short by \(13.4\) million litres — about \(17\%\). Three things are going on.

First, the prediction is an extrapolation: \(x=10\) lies outside the range of the data, where a fitted line carries no guarantee at all.

Second, the anomalous first value drags the line down. Reading it as \(33.5\) instead gives \(\widehat {Y}=31.46+2.742x\), a \(2010\) prediction of \(58.9\) — further from the truth, but with \(r\) improving from \(0.860\) to \(0.935\). The single bad point was flattering the fit’s slope while spoiling its correlation.

Third, and most importantly, the residuals are not random: consumption is rising faster than linearly towards the end of the series, and \(2010\) continues that. A straight line is the wrong shape. A model allowing for growth — a quadratic, or a line fitted to \(\log Y\) — would suit these data better.

Solution. Which line. Mass on age. The two regression lines answer different questions and are not interchangeable. Least squares minimises the squared deviations in the dependent variable only, so the choice depends on which variable is which. Here age is fixed by the experimenter — the animal was weighed at chosen weekly intervals, and those ages carry no measurement error — while mass is the response that varies. Age is therefore the independent variable and mass the dependent one.

Fitting it. With \(n=6\), \(\overline {x}=13.5\) and \(\overline {y}=403.5\): \[S_{XX}=17.5,\qquad S_{XY}=277.5\] \[\widehat {b}=\frac {277.5}{17.5}=15.86,\qquad \widehat {a}=403.5-15.86(13.5)=189.43\] \[\implies \quad \widehat {\text {mass}}=189.43+15.86\,(\text {age}).\]

At \(17\) weeks, \[\widehat {y}=189.43+15.86(17)=459.0\ \text {kg}.\]

Two reservations worth recording. The reading at week \(15\), \(404\) kg, is lower than week \(14\)’s \(423\) and identical to week \(13\)’s — an animal losing weight then gaining \(47\) kg in a week is odd, and it may be a recording error. And \(17\) weeks is beyond the data, so the estimate assumes growth continues at the same steady rate, which for a growing animal it eventually will not.

Solution. The question as printed gives \(\sum (y-\overline {y})=62\). That sum is identically zero for any data whatever, since deviations about their own mean always cancel. The intended quantity is \(\sum (y-\overline {y})^2=62\), and that is what is used here.

(a) \(\overline {x}=\frac {104}{26}=4\), and \[\widehat {b}=\frac {S_{XY}}{S_{XX}}=\frac {-56}{56}=-1.\] The value of \(\overline {y}\) is not given, so the intercept cannot be found as a number. The line is still fully determined in deviation form: \[y-\overline {y}=-1(x-4)\qquad \text {that is}\qquad \widehat {y}=\overline {y}-x+4.\] Longer training reduces performance time by one unit per unit of training.

(b) At \(x=12\), \[\widehat {y}=\overline {y}-12+4=\overline {y}-8,\] so the predicted performance time is \(8\) units below the mean. Without \(\overline {y}\) that is as far as the arithmetic can go, and it is a complete answer to what was asked.

(c) The residual sum of squares is \[SSE=S_{YY}-\widehat {b}\,S_{XY}=62-(-1)(-56)=6\] \[\implies \quad \widehat {\sigma }^2_{y/x}=\frac {SSE}{n-2}=\frac {6}{24}=0.25.\]

(d) The standard error of the slope is \[\text {se}(\widehat {b})=\sqrt {\frac {\widehat {\sigma }^2_{y/x}}{S_{XX}}}=\sqrt {\frac {0.25}{56}}=0.0668\] and with \(t_{24,\,0.025}=2.064\), \[-1\pm 2.064(0.0668)=-1\pm 0.138\quad \implies \quad (-1.138,\ -0.862).\] The interval is comfortably clear of zero, so the effect of training is significant. The fit is very tight — of the total \(62\) in \(Y\), all but \(6\) is explained by \(x\), giving \(r=-0.950\).