| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Calculate r from raw bivariate data |
| Difficulty | Moderate -0.3 This is a straightforward application of the product moment correlation coefficient formula with all summary statistics provided. Part (i)(a) is pure substitution into a standard formula, (i)(b) tests basic understanding of correlation vs causation, and part (ii) involves simple regression line interpretation and mean calculations. While multi-part, each component is routine S1 material requiring no problem-solving insight. |
| Spec | 2.02e Correlation does not imply causation5.08a Pearson correlation: calculate pmcc5.08b Linear coding: effect on pmcc5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line |
| 21 | 23 | 19 | 15 | 14 | 5 | 2 | 10 | 9 | 20 | 18 | 23 | ||
| Number of absences, \(y\) | 23 | 25 | 18 | 18 | 12 | 10 | 4 | 9 | 11 | 15 | 20 | 26 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{xx} = 3215 - \frac{179^2}{12} = 544.9167\) or \(\frac{6539}{12}\) | ||
| \(S_{yy} = 3565 - \frac{191^2}{12} = 524.9167\) or \(\frac{6299}{12}\) | ||
| \(S_{xy} = 3343 - \frac{179 \times 191}{12} = 493.9167\) or \(\frac{5927}{12}\) | M1 | Correct sub in a correct \(S\) formula |
| \(r = \frac{\text{"493.9167"}}{\sqrt{\text{"544.9167"} \times \text{"524.9167"}}}\) | M1 | Correct sub in 3 correct \(S\) formulae and a correct \(r\) formula; Correct ans (\(\geq 3\) sf) no wking, M1M1A1 |
| \(= 0.924\) (3 sf) or \(0.9235...\) | A1 [3] | Must see \(\geq 3\) sf; Ignore any comparison with 0.92 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correlation does not imply causation; or It depends on the time of year; or Both depend on a third variable; or There could be other factors | B1 [1] | Any answer which implies or is equivalent to any of these; Allow without context; Ignore incorrect comments |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 'Increased' and positive gradient or positive coeff of \(n\) or 'Output goes up by 0.6 each month'; Both needed | B1 [1] | 'Increased' and values of \(z\) shown: at least 6 values or 1st and last values or 1st, 2nd/3rd/4th and 9th/10th/11th/12th ie 17.6 or 18.2 or 18.8 or 19.4 and 22.4 or 23 or 23.6 or 24.2; NOT: 'Increased' and 'Value of \(z\) incr as \(n\) incr' / '\(z\) incr as no. of mths incr' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{n} = 6.5\) or \(\frac{78}{12}\) oe seen | B1 | |
| \(\bar{z} = 0.6 \times \text{'6.5'} + 17\) alone, eg not \(\div 12\); or \(17 = \bar{z} - 0.6 \times \text{'6.5'}\) oe | M1 | or \((0.6\times1+17+0.6\times2+17+\ldots0.6\times12+17)\div12\) oe M1; or '250.8'\(\div 12\) M1; ft their '6.5' only if comes from \(\ldots\div12\); Long method, all correct terms seen and \(\div12\) M1; NB ans 20.9 may not score the B1 |
| \(\bar{z} = 20.9\) | A1 [3] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total output \(= \text{"20.9"} \times 12\) | M1 | or \(0.6\times1+17+0.6\times2+17+\ldots0.6\times12+17\) oe; or eg \(\frac{88}{5}+\frac{91}{5}+\ldots+\frac{121}{5}\) oe; Long method, all correct terms seen |
| \(251\) (3 sf) | A1f [2] | ft their (ii)(b); Not ISW, eg 25100 scores A0, even if 251 seen |
# Question 2:
## Part (i)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{xx} = 3215 - \frac{179^2}{12} = 544.9167$ or $\frac{6539}{12}$ | | |
| $S_{yy} = 3565 - \frac{191^2}{12} = 524.9167$ or $\frac{6299}{12}$ | | |
| $S_{xy} = 3343 - \frac{179 \times 191}{12} = 493.9167$ or $\frac{5927}{12}$ | M1 | Correct sub in a correct $S$ formula |
| $r = \frac{\text{"493.9167"}}{\sqrt{\text{"544.9167"} \times \text{"524.9167"}}}$ | M1 | Correct sub in 3 correct $S$ formulae and a correct $r$ formula; Correct ans ($\geq 3$ sf) no wking, M1M1A1 |
| $= 0.924$ (3 sf) or $0.9235...$ | A1 **[3]** | Must see $\geq 3$ sf; Ignore any comparison with 0.92 |
## Part (i)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correlation does not imply causation; or It depends on the time of year; or Both depend on a third variable; or There could be other factors | B1 **[1]** | Any answer which implies or is equivalent to any of these; Allow without context; Ignore incorrect comments |
## Part (ii)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 'Increased' and positive gradient or positive coeff of $n$ or 'Output goes up by 0.6 each month'; Both needed | B1 **[1]** | 'Increased' and values of $z$ shown: at least 6 values or 1st and last values or 1st, 2nd/3rd/4th **and** 9th/10th/11th/12th ie 17.6 or 18.2 or 18.8 or 19.4 **and** 22.4 or 23 or 23.6 or 24.2; NOT: 'Increased' and 'Value of $z$ incr as $n$ incr' / '$z$ incr as no. of mths incr' |
## Part (ii)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{n} = 6.5$ or $\frac{78}{12}$ oe seen | B1 | |
| $\bar{z} = 0.6 \times \text{'6.5'} + 17$ alone, eg not $\div 12$; or $17 = \bar{z} - 0.6 \times \text{'6.5'}$ oe | M1 | or $(0.6\times1+17+0.6\times2+17+\ldots0.6\times12+17)\div12$ oe M1; or '250.8'$\div 12$ M1; ft their '6.5' only if comes from $\ldots\div12$; Long method, all correct terms seen and $\div12$ M1; NB ans 20.9 may not score the B1 |
| $\bar{z} = 20.9$ | A1 **[3]** | cao |
## Part (ii)(c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total output $= \text{"20.9"} \times 12$ | M1 | or $0.6\times1+17+0.6\times2+17+\ldots0.6\times12+17$ oe; or eg $\frac{88}{5}+\frac{91}{5}+\ldots+\frac{121}{5}$ oe; Long method, all correct terms seen |
| $251$ (3 sf) | A1f **[2]** | ft their (ii)(b); Not ISW, eg 25100 scores A0, even if 251 seen |
---2 (i) The table shows the amount, $x$, in hundreds of pounds, spent on heating and the number of absences, $y$, at a factory during each month in 2014.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | l | l | l | l | l | }
\hline
\begin{tabular}{ l }
Amount, $x$, spent on \\
heating (£ hundreds) \\
\end{tabular} & 21 & 23 & 19 & 15 & 14 & 5 & 2 & 10 & 9 & 20 & 18 & 23 \\
\hline
Number of absences, $y$ & 23 & 25 & 18 & 18 & 12 & 10 & 4 & 9 & 11 & 15 & 20 & 26 \\
\hline
\end{tabular}
\end{center}
$n = 12 \quad \Sigma x = 179 \quad \Sigma x ^ { 2 } = 3215 \quad \Sigma y = 191 \quad \Sigma y ^ { 2 } = 3565 \quad \Sigma x y = 3343$
\begin{enumerate}[label=(\alph*)]
\item Calculate $r$, the product moment correlation coefficient, showing that $r > 0.92$.
\item A manager says, 'The value of $r$ shows that spending more money on heating causes more absences, so we should spend less on heating.' Comment on this claim.\\
(ii) The months in 2014 were numbered $1,2,3 , \ldots , 12$. The output, $z$, in suitable units was recorded along with the month number, $n$, for each month in 2014. The equation of the regression line of $z$ on $n$ was found to be $z = 0.6 n + 17$.\\
(a) Use this equation to explain whether output generally increased or decreased over these months.\\
(b) Find the mean of $n$ and use the equation of the regression line to calculate the mean of $z$.
\item Hence calculate the total output in 2014.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2016 Q2 [10]}}