Question 1 - A-Level Maths

Edexcel S1 2018 Specimen — Question 1 12 marks

Exam Board	Edexcel
Module	S1 (Statistics 1)
Year	2018
Session	Specimen
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear regression
Type	Calculate from raw data
Difficulty	Moderate -0.8 This is a standard S1 linear regression question requiring straightforward application of formulas for Sxx, Sxy, correlation coefficient, and regression line. All necessary summary statistics are provided, making it a routine calculation exercise with no problem-solving or conceptual challenges—easier than average A-level.
Spec	5.08a Pearson correlation: calculate pmcc 5.09b Least squares regression: concepts 5.09c Calculate regression line 5.09d Linear coding: effect on regression 5.09e Use regression: for estimation in context

The percentage oil content, $p$, and the weight, $w$ milligrams, of each of 10 randomly selected sunflower seeds were recorded. These data are summarised below.

$$\sum w ^ { 2 } = 41252 \quad \sum w p = 27557.8 \quad \sum w = 640 \quad \sum p = 431 \quad \mathrm {~S} _ { p p } = 2.72$$

Find the value of $\mathrm { S } _ { w w }$ and the value of $\mathrm { S } _ { w p }$
Calculate the product moment correlation coefficient between $p$ and $w$
Give an interpretation of your product moment correlation coefficient. The equation of the regression line of $p$ on $w$ is given in the form $p = a + b w$
Find the equation of the regression line of $p$ on $w$
Hence estimate the percentage oil content of a sunflower seed which weighs 60 milligrams. $\_\_\_\_$ VAYV SIHI NI JIIIM ION OC
VJYV SIHI NI JIIIM ION OC
VJYV SIHI NI JLIYM ION OC

Show mark scheme Show mark scheme source

Question 1 Mark Scheme

1(a)

Answer	Marks
$S_{ww} = \frac{6402}{10} = 292$	M1A1
$S_{wp} = \frac{27557.8}{10} = -26.2$	A1

(3 marks)

1(b)

Answer	Marks
$r = \frac{-26.2}{\sqrt{292 \times 2.72}}$	M1
$= -0.9297$ awrt $-0.930$	A1

(2 marks)

1(c)

Answer	Marks
As weight increases the percentage of oil content decreases o.e.	B1

(1 mark)

1(d)

Answer	Marks
$b = \frac{-26.2}{292} = -0.0897...$ awrt $-0.09$	M1 A1
$a = \frac{431}{10} - (-0.0897) \times \frac{640}{10} = 48.842...$	M1
$p = 48.8 - 0.0897w$	A1

(4 marks)

1(e)

Answer	Marks
$p = 48.8 - 0.0897 \times 60$	M1
$= 43.4/43.5$ awrt $43.4/43.5$	A1

(2 marks)

Total: 12 marks

Notes

(a)

- M1: for a correct expression for $S_{ww}$ or $S_{wp}$ (may be implied by one correct answer)

- 1st A1: for either $S_{ww} = 292$ or $S_{wp} = -26.2$

- 2nd A1: for both $S_{ww} = 292$ and $S_{wp} = -26.2$

(b)

- M1: for a correct expression (Allow ft of their $S_{ww}$ or $S_{wp}$ provided $S_{ww} \neq 41252$ and $S_{wp} \neq 27557.8$). Condone missing "$-$"

- A1: for awrt $-0.930$ (Condone $-0.93$ for M1A1 if correct expression is seen). (Answer only awrt $-0.930$ scores 2/2 but answer only $-0.93$ is M1A0)

(c)

- B1: For a correct contextual description of negative correlation which must include weight and oil (but w increases as p decreases is not sufficient)

(d)

- 1st M1: for a correct expression for $b$ (Allow ft)

- 1st A1: for awrt $-0.09$

- 2nd M1: for a correct method for $a$ ft their value of $b$ (Allow $a = 43.1 - b \times 64$)

- 2nd A1: for a correct equation for $p$ and $w$ with $a = \text{awrt } 48.8$ and $b = \text{awrt } -0.0897$. No fractions. Equation in $x$ and $y$ is A0

(e)

- M1: substituting $w = 60$ into their equation

- A1: awrt $43.4$ or $43.5$ (Answer only scores 2/2)

# Question 1 Mark Scheme

## 1(a)
$S_{ww} = \frac{6402}{10} = 292$ | M1A1

$S_{wp} = \frac{27557.8}{10} = -26.2$ | A1

(3 marks)

## 1(b)
$r = \frac{-26.2}{\sqrt{292 \times 2.72}}$ | M1

$= -0.9297$ awrt $-0.930$ | A1

(2 marks)

## 1(c)
As weight increases the percentage of oil content decreases o.e. | B1

(1 mark)

## 1(d)
$b = \frac{-26.2}{292} = -0.0897...$ awrt $-0.09$ | M1 A1

$a = \frac{431}{10} - (-0.0897) \times \frac{640}{10} = 48.842...$ | M1

$p = 48.8 - 0.0897w$ | A1

(4 marks)

## 1(e)
$p = 48.8 - 0.0897 \times 60$ | M1

$= 43.4/43.5$ awrt $43.4/43.5$ | A1

(2 marks)

**Total: 12 marks**

---

## Notes

**(a)**
- M1: for a correct expression for $S_{ww}$ or $S_{wp}$ (may be implied by one correct answer)
- 1st A1: for either $S_{ww} = 292$ or $S_{wp} = -26.2$
- 2nd A1: for both $S_{ww} = 292$ and $S_{wp} = -26.2$

**(b)**
- M1: for a correct expression (Allow ft of their $S_{ww}$ or $S_{wp}$ provided $S_{ww} \neq 41252$ and $S_{wp} \neq 27557.8$). Condone missing "$-$"
- A1: for awrt $-0.930$ (Condone $-0.93$ for M1A1 if correct expression is seen). (Answer only awrt $-0.930$ scores 2/2 but answer only $-0.93$ is M1A0)

**(c)**
- B1: For a correct contextual description of negative correlation which must include weight and oil (but w increases as p decreases is not sufficient)

**(d)**
- 1st M1: for a correct expression for $b$ (Allow ft)
- 1st A1: for awrt $-0.09$
- 2nd M1: for a correct method for $a$ ft their value of $b$ (Allow $a = 43.1 - b \times 64$)
- 2nd A1: for a correct equation for $p$ and $w$ with $a = \text{awrt } 48.8$ and $b = \text{awrt } -0.0897$. No fractions. Equation in $x$ and $y$ is A0

**(e)**
- M1: substituting $w = 60$ into their equation
- A1: awrt $43.4$ or $43.5$ (Answer only scores 2/2)

Show LaTeX source

\begin{enumerate}
  \item The percentage oil content, $p$, and the weight, $w$ milligrams, of each of 10 randomly selected sunflower seeds were recorded. These data are summarised below.
\end{enumerate}

$$\sum w ^ { 2 } = 41252 \quad \sum w p = 27557.8 \quad \sum w = 640 \quad \sum p = 431 \quad \mathrm {~S} _ { p p } = 2.72$$

(a) Find the value of $\mathrm { S } _ { w w }$ and the value of $\mathrm { S } _ { w p }$\\
(b) Calculate the product moment correlation coefficient between $p$ and $w$\\
(c) Give an interpretation of your product moment correlation coefficient.

The equation of the regression line of $p$ on $w$ is given in the form $p = a + b w$\\
(d) Find the equation of the regression line of $p$ on $w$\\
(e) Hence estimate the percentage oil content of a sunflower seed which weighs 60 milligrams.\\

$\_\_\_\_$ VAYV SIHI NI JIIIM ION OC\\
VJYV SIHI NI JIIIM ION OC\\
VJYV SIHI NI JLIYM ION OC

\begin{center}

\end{center}

\hfill \mbox{\textit{Edexcel S1 2018 Q1 [12]}}

This paper (7 questions)

View full paper

Q1 12 Q2 3 Q3 14 Q4 12 Q5 8 Q6 11 Q7 15

Edexcel S1 2018 Specimen — Question 1 12 marks

Question 1 Mark Scheme

1(a)

(3 marks)

1(b)

(2 marks)

1(c)

(1 mark)

1(d)

(4 marks)

1(e)

(2 marks)

Total: 12 marks

Notes

(a)

- M1: for a correct expression for \(S_{ww}\) or \(S_{wp}\) (may be implied by one correct answer)

- 1st A1: for either \(S_{ww} = 292\) or \(S_{wp} = -26.2\)

- 2nd A1: for both \(S_{ww} = 292\) and \(S_{wp} = -26.2\)

(b)

- M1: for a correct expression (Allow ft of their \(S_{ww}\) or \(S_{wp}\) provided \(S_{ww} \neq 41252\) and \(S_{wp} \neq 27557.8\)). Condone missing "\(-\)"

- A1: for awrt \(-0.930\) (Condone \(-0.93\) for M1A1 if correct expression is seen). (Answer only awrt \(-0.930\) scores 2/2 but answer only \(-0.93\) is M1A0)

(c)

- B1: For a correct contextual description of negative correlation which must include weight and oil (but w increases as p decreases is not sufficient)

(d)

- 1st M1: for a correct expression for \(b\) (Allow ft)

- 1st A1: for awrt \(-0.09\)

- 2nd M1: for a correct method for \(a\) ft their value of \(b\) (Allow \(a = 43.1 - b \times 64\))

- 2nd A1: for a correct equation for \(p\) and \(w\) with \(a = \text{awrt } 48.8\) and \(b = \text{awrt } -0.0897\). No fractions. Equation in \(x\) and \(y\) is A0

(e)

- M1: substituting \(w = 60\) into their equation

- A1: awrt \(43.4\) or \(43.5\) (Answer only scores 2/2)

This paper (7 questions)

Answer	Marks
\(S_{ww} = \frac{6402}{10} = 292\)	M1A1
\(S_{wp} = \frac{27557.8}{10} = -26.2\)	A1

Answer	Marks
\(r = \frac{-26.2}{\sqrt{292 \times 2.72}}\)	M1
\(= -0.9297\) awrt \(-0.930\)	A1

Answer	Marks
\(b = \frac{-26.2}{292} = -0.0897...\) awrt \(-0.09\)	M1 A1
\(a = \frac{431}{10} - (-0.0897) \times \frac{640}{10} = 48.842...\)	M1
\(p = 48.8 - 0.0897w\)	A1

Answer	Marks
\(p = 48.8 - 0.0897 \times 60\)	M1
\(= 43.4/43.5\) awrt \(43.4/43.5\)	A1