Edexcel S1 Specimen — Question 6 14 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
SessionSpecimen
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear regression
TypeCalculate y on x from raw data table
DifficultyModerate -0.8 This is a standard S1 linear regression question with all summary statistics provided. Students follow a routine algorithm: calculate Sdd and Sfd using given sums, find b and a using formulas, then interpret. The final part requires simple algebraic manipulation. No conceptual difficulty or novel insight required—purely procedural application of memorized formulas.
Spec5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context
  1. A travel agent sells flights to different destinations from Beerow airport. The distance \(d\), measured in 100 km , of the destination from the airport and the fare \(\pounds f\) are recorded for a random sample of 6 destinations.
Destination\(A\)\(B\)\(C\)\(D\)\(E\)\(F\)
\(d\)2.24.06.02.58.05.0
\(f\)182025233228
$$\text { [You may use } \sum d ^ { 2 } = 152.09 \quad \sum f ^ { 2 } = 3686 \quad \sum f d = 723.1 \text { ] }$$
  1. Using the axes below, complete a scatter diagram to illustrate this information.
  2. Explain why a linear regression model may be appropriate to describe the relationship between \(f\) and \(d\).
  3. Calculate \(S _ { d d }\) and \(S _ { f d }\)
  4. Calculate the equation of the regression line of \(f\) on \(d\) giving your answer in the form \(f = a + b d\).
  5. Give an interpretation of the value of \(b\). Jane is planning her holiday and wishes to fly from Beerow airport to a destination \(t \mathrm {~km}\) away. A rival travel agent charges 5 p per km.
  6. Find the range of values of \(t\) for which the first travel agent is cheaper than the rival. \includegraphics[max width=\textwidth, alt={}, center]{61983561-79f7-4883-8ae7-ab1f4955d444-20_967_1630_1722_164}
Question 6:
Part (a)
AnswerMarks
Points plotted correctlyB1 B1
Part (b)
AnswerMarks
The points lie reasonably close to a straight lineB1
Part (c)
AnswerMarks
\(\bar{d} = 27.7\), \(\bar{f} = 146\) (both, may be implied)B1
\(S_{dd} = 152.09 - \frac{(27.7)^2}{6} = 24.208\ldots\) awrt 24.2M1 A1
\(S_{fd} = 723.1 - \frac{27.7 \times 146}{6} = 49.06\ldots\) awrt 49.1A1
Part (d)
AnswerMarks
\(b = \frac{S_{fd}}{S_{dd}} = 2.026\ldots\) awrt 2.03M1 A1
\(a = \frac{146}{6} - b \times \frac{27.7}{6} = 14.97\ldots\) so \(f = \mathbf{15.0 + 2.03d}\)M1 A1
Part (e)
AnswerMarks
A flight costs £2.03 (or about £2) for every extra 100km or about 2p per kmB1ft
Part (f)
AnswerMarks
\(15.0 + 2.03d < 5d\) so \(d > \frac{15.0}{(5-2.03)} = 5.00 \sim 5.05\)M1
So \(t > 500\sim505\)A1
WST01/01: Statistics S1 Mark Scheme
Question 6 (Regression/Correlation):
Part (a)
AnswerMarks Guidance
AnswerMark Guidance
At least 4 points correctB1 Allow \(\pm\) one 2mm square
All points correctB1 Allow \(\pm\) one 2mm square
Part (b)
AnswerMarks Guidance
AnswerMark Guidance
Line drawn through correct pointsB1 Ignore extra points and lines; require reference to points and line
Part (c)
AnswerMarks Guidance
AnswerMark Guidance
Correct method for either \(S_{dd}\) or \(S_{fd}\)M1 Correct expression seen
\(S_{dd}\) awrt 24.2A1
\(S_{fd}\) awrt 49.1A1
Part (d)
AnswerMarks Guidance
AnswerMark Guidance
Correct expression for \(b\)M1 Can follow through from (c)
Correct method to find \(a\) using \(b\) and meansM1 Follow through
\(f = \ldots\) in terms of \(d\), all values awrtA1 Accept 15 as rounding from correct answer only
Part (e)
AnswerMarks Guidance
AnswerMark Guidance
Contextual interpretation of cost and distance— Follow through their value of \(b\)
Part (f)
AnswerMarks Guidance
AnswerMark Guidance
Attempt intersection of 2 linesM1 \(t\) in range 500 to 505 seen
\(d\) in range 5 to 5.05A1 Accept \(t\) greater than 500 to 505 inclusive for M1A1 
# Question 6:

## Part (a)
| Points plotted correctly | B1 B1 | |

## Part (b)
| The points lie reasonably close to a straight **line** | B1 | |

## Part (c)
| $\bar{d} = 27.7$, $\bar{f} = 146$ (both, may be implied) | B1 | |
|---|---|---|
| $S_{dd} = 152.09 - \frac{(27.7)^2}{6} = 24.208\ldots$ awrt **24.2** | M1 A1 | |
| $S_{fd} = 723.1 - \frac{27.7 \times 146}{6} = 49.06\ldots$ awrt **49.1** | A1 | |

## Part (d)
| $b = \frac{S_{fd}}{S_{dd}} = 2.026\ldots$ awrt **2.03** | M1 A1 | |
|---|---|---|
| $a = \frac{146}{6} - b \times \frac{27.7}{6} = 14.97\ldots$ so $f = \mathbf{15.0 + 2.03d}$ | M1 A1 | |

## Part (e)
| A flight costs **£2.03 (or about £2)** for every extra **100km** or about 2p per km | B1ft | |

## Part (f)
| $15.0 + 2.03d < 5d$ so $d > \frac{15.0}{(5-2.03)} = 5.00 \sim 5.05$ | M1 | |
|---|---|---|
| So $t > 500\sim505$ | A1 | |

# WST01/01: Statistics S1 Mark Scheme

---

## Question 6 (Regression/Correlation):

**Part (a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| At least 4 points correct | B1 | Allow $\pm$ one 2mm square |
| All points correct | B1 | Allow $\pm$ one 2mm square |

**Part (b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Line drawn through correct points | B1 | Ignore extra points and lines; require reference to points and line |

**Part (c)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct method for either $S_{dd}$ or $S_{fd}$ | M1 | Correct expression seen |
| $S_{dd}$ awrt 24.2 | A1 | |
| $S_{fd}$ awrt 49.1 | A1 | |

**Part (d)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct expression for $b$ | M1 | Can follow through from (c) |
| Correct method to find $a$ using $b$ and means | M1 | Follow through |
| $f = \ldots$ in terms of $d$, all values awrt | A1 | Accept 15 as rounding from correct answer only |

**Part (e)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Contextual interpretation of cost and distance | — | Follow through their value of $b$ |

**Part (f)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt intersection of 2 lines | M1 | $t$ in range 500 to 505 seen |
| $d$ in range 5 to 5.05 | A1 | Accept $t$ greater than 500 to 505 inclusive for M1A1 |

---
\begin{enumerate}
  \item A travel agent sells flights to different destinations from Beerow airport. The distance $d$, measured in 100 km , of the destination from the airport and the fare $\pounds f$ are recorded for a random sample of 6 destinations.
\end{enumerate}

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
Destination & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ \\
\hline
$d$ & 2.2 & 4.0 & 6.0 & 2.5 & 8.0 & 5.0 \\
\hline
$f$ & 18 & 20 & 25 & 23 & 32 & 28 \\
\hline
\end{tabular}
\end{center}

$$\text { [You may use } \sum d ^ { 2 } = 152.09 \quad \sum f ^ { 2 } = 3686 \quad \sum f d = 723.1 \text { ] }$$

(a) Using the axes below, complete a scatter diagram to illustrate this information.\\
(b) Explain why a linear regression model may be appropriate to describe the relationship between $f$ and $d$.\\
(c) Calculate $S _ { d d }$ and $S _ { f d }$\\
(d) Calculate the equation of the regression line of $f$ on $d$ giving your answer in the form $f = a + b d$.\\
(e) Give an interpretation of the value of $b$.

Jane is planning her holiday and wishes to fly from Beerow airport to a destination $t \mathrm {~km}$ away. A rival travel agent charges 5 p per km.\\
(f) Find the range of values of $t$ for which the first travel agent is cheaper than the rival.\\
\includegraphics[max width=\textwidth, alt={}, center]{61983561-79f7-4883-8ae7-ab1f4955d444-20_967_1630_1722_164}

\hfill \mbox{\textit{Edexcel S1  Q6 [14]}}
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