Edexcel S1 2023 June — Question 2 13 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2023
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBivariate data
TypeEffect of data transformation on correlation
DifficultyModerate -0.3 This is a standard S1 bivariate data question testing routine calculations (Sxx, Sxy, correlation coefficient, regression line) and understanding of coding transformations. All parts follow textbook procedures with no novel problem-solving required, though the multi-part structure and coding section elevate it slightly above the most basic exercises.
Spec5.08a Pearson correlation: calculate pmcc5.08b Linear coding: effect on pmcc5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression5.09e Use regression: for estimation in context
Two students, Olive and Shan, collect data on the weight, \(w\) grams, and the tail length, \(t\) cm, of 15 mice. Olive summarised the data as follows \(S_tt = 5.3173\) \quad \(\sum w^2 = 6089.12\) \quad \(\sum tw = 2304.53\) \quad \(\sum w = 297.8\) \quad \(\sum t = 114.8\)
  1. Calculate the value of \(S_{ww}\) and the value of \(S_{tw}\) [3]
  2. Calculate the value of the product moment correlation coefficient between \(w\) and \(t\) [2]
  3. Show that the equation of the regression line of \(w\) on \(t\) can be written as $$w = -16.7 + 4.77t$$ [3]
  4. Give an interpretation of the gradient of the regression line. [1]
  5. Explain why it would not be appropriate to use the regression line in part (c) to estimate the weight of a mouse with a tail length of 2cm. [2]
Shan decided to code the data using \(x = t - 6\) and \(y = \frac{w}{2} - 5\)
  1. Write down the value of the product moment correlation coefficient between \(x\) and \(y\) [1]
  2. Write down an equation of the regression line of \(y\) on \(x\) You do not need to simplify your equation. [1]
Question 2:
AnswerMarks
2(a)297.8114.8 297.82
S =2304.53− or S =6089.12−
tw ww
AnswerMarks
15 15M1
S =25.367... awrt 25.4
AnswerMarks
twA1
S =176.797 awrt 177
AnswerMarks
wwA1
(3)
AnswerMarks
(b)"25.367"
r=
AnswerMarks
5.3173"176.797..."M1
= 0.82735…. awrt 0.827 or 0.828A1
(2)
AnswerMarks
(c)"25.367..."
b= =4.77065... 
AnswerMarks
5.3173M1
297.8 "25.367" 114.8
a= −  =−16.658.... 
AnswerMarks
15 5.3173 15M1
b = 4.771 or better or a = –16.66 or better seen andw=−16.7+ 4.77t*A1*cso
(3)
AnswerMarks Guidance
(d)[On average,] for each cm/1 cm of tail length/t the weight/w increases by 4.77 g/grams B1
(1)
AnswerMarks
(e)16.7
w=−16.7+4.772 =−7.16  or 4.772 =9.54  or t= =3.5  or sd = awrt 0.6
AnswerMarks
4.77M1
w=−7.16
or 9.54 < 16.7 or 2 < 3.5 which is negative/weight cannot be negative
AnswerMarks
or for sd extrapolation since a 2 cm tail is (approx 9 sd)/(more than 3 sd) from the meanA1
(2)
AnswerMarks Guidance
(f)0.827 B1ft
(1)
AnswerMarks
(g)2y+10=−16.7+4.77(x+6)
oeB1ft
(1)
AnswerMarks
NotesTotal 13
(a)M1 for a correct expression for S or S
tw ww
A1 awrt 25.4
A1 awrt 177
AnswerMarks
(b)M1 for a valid attempt at r with their S not equal to 2304.53 and S not equal to 6089.12
tw ww
A1 (M2 on epen) awrt 0.827 or awrt 0.828
AnswerMarks
(c)1st M1 for a correct method to find the value of b
2nd M1 ft their b. For a correct method to find a. Minimum shown
a=awrt 19.9−"theirb"awrt 7.65 =−16.658 
A1* Both method marks must be awarded, equation stated (no fractions) and sight of (4.771 or better)
or (−16.66 or better)
AnswerMarks
(d)B1 For a suitable contextual comment that implies that as length increases by 1 cm weight increases by
4.77g. Allow multiples eg each 10 cm increase in tail length weight increases by 47.7g Allow in terms
of t and w
AnswerMarks
(e)M1 for a correct method to calculate the value of w (condone if written as a fraction) or
4.772 =9.54  or correct method to find tail length when w = 0 or sd = awrt 0.6
A1 Method mark must be awarded. For –7.16 or 9.54 < 16.7 or 2 < 3.5 with a relevant explanation
stating that weight is negative. If sd = awrt 0.6 is given allow extrapolation since a 2 cm tail is (approx
9 sd)/(more than 3 sd) from the mean.
AnswerMarks
(f)B1ft follow through their answer to (b)
(g)−16.7+4.77(x+6)
B1 ISW no need to be simplified. Allow equivalent eg y= −5The correct
2
simplified equation is y=2.385x+0.96 allow awrt 2.39 and 0.96 – 0.98
AnswerMarks Guidance
QuScheme Marks 
Question 2:
--- 2(a) ---
2(a) | 297.8114.8 297.82
S =2304.53− or S =6089.12−
tw ww
15 15 | M1
S =25.367... awrt 25.4
tw | A1
S =176.797 awrt 177
ww | A1
(3)
(b) | "25.367"
r=
5.3173"176.797..." | M1
= 0.82735…. awrt 0.827 or 0.828 | A1
(2)
(c) | "25.367..."
b= =4.77065... 
5.3173 | M1
297.8 "25.367" 114.8
a= −  =−16.658.... 
15 5.3173 15 | M1
b = 4.771 or better or a = –16.66 or better seen andw=−16.7+ 4.77t* | A1*cso
(3)
(d) | [On average,] for each cm/1 cm of tail length/t the weight/w increases by 4.77 g/grams | B1
(1)
(e) | 16.7
w=−16.7+4.772 =−7.16  or 4.772 =9.54  or t= =3.5  or sd = awrt 0.6
4.77 | M1
w=−7.16
or 9.54 < 16.7 or 2 < 3.5 which is negative/weight cannot be negative
or for sd extrapolation since a 2 cm tail is (approx 9 sd)/(more than 3 sd) from the mean | A1
(2)
(f) | 0.827 | B1ft
(1)
(g) | 2y+10=−16.7+4.77(x+6)
oe | B1ft
(1)
Notes | Total 13
(a) | M1 for a correct expression for S or S
tw ww
A1 awrt 25.4
A1 awrt 177
(b) | M1 for a valid attempt at r with their S not equal to 2304.53 and S not equal to 6089.12
tw ww
A1 (M2 on epen) awrt 0.827 or awrt 0.828
(c) | 1st M1 for a correct method to find the value of b
2nd M1 ft their b. For a correct method to find a. Minimum shown
a=awrt 19.9−"theirb"awrt 7.65 =−16.658 
A1* Both method marks must be awarded, equation stated (no fractions) and sight of (4.771 or better)
or (−16.66 or better)
(d) | B1 For a suitable contextual comment that implies that as length increases by 1 cm weight increases by
4.77g. Allow multiples eg each 10 cm increase in tail length weight increases by 47.7g Allow in terms
of t and w
(e) | M1 for a correct method to calculate the value of w (condone if written as a fraction) or
4.772 =9.54  or correct method to find tail length when w = 0 or sd = awrt 0.6
A1 Method mark must be awarded. For –7.16 or 9.54 < 16.7 or 2 < 3.5 with a relevant explanation
stating that weight is negative. If sd = awrt 0.6 is given allow extrapolation since a 2 cm tail is (approx
9 sd)/(more than 3 sd) from the mean.
(f) | B1ft follow through their answer to (b)
(g) | −16.7+4.77(x+6)
B1 ISW no need to be simplified. Allow equivalent eg y= −5The correct
2
simplified equation is y=2.385x+0.96 allow awrt 2.39 and 0.96 – 0.98
Qu | Scheme | Marks
Two students, Olive and Shan, collect data on the weight, $w$ grams, and the tail length, $t$ cm, of 15 mice.

Olive summarised the data as follows

$S_tt = 5.3173$ \quad $\sum w^2 = 6089.12$ \quad $\sum tw = 2304.53$ \quad $\sum w = 297.8$ \quad $\sum t = 114.8$

\begin{enumerate}[label=(\alph*)]
\item Calculate the value of $S_{ww}$ and the value of $S_{tw}$ [3]
\item Calculate the value of the product moment correlation coefficient between $w$ and $t$ [2]
\item Show that the equation of the regression line of $w$ on $t$ can be written as
$$w = -16.7 + 4.77t$$ [3]
\item Give an interpretation of the gradient of the regression line. [1]
\item Explain why it would not be appropriate to use the regression line in part (c) to estimate the weight of a mouse with a tail length of 2cm. [2]
\end{enumerate}

Shan decided to code the data using $x = t - 6$ and $y = \frac{w}{2} - 5$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{5}
\item Write down the value of the product moment correlation coefficient between $x$ and $y$ [1]
\item Write down an equation of the regression line of $y$ on $x$
You do not need to simplify your equation. [1]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2023 Q2 [13]}}
This paper (7 questions)
View full paper
Q1 8 Q2 13 Q3 9 Q4 9 Q5 13 Q6 9 Q7 14