| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Minimize sum of squared residuals |
| Difficulty | Standard +0.3 This is a structured Further Maths Statistics question that guides students through minimizing sum of squared residuals using a completed square form. While it involves Further Maths content, the question provides the algebraic form explicitly and asks for explanations rather than derivations. Parts (a)-(d) require understanding of regression concepts but minimal calculation. Part (e) requires applying transformations to find a new gradient, which is routine algebra. Overall, slightly easier than average due to the scaffolding provided. |
| Spec | 5.09a 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a)(i) | (Squares are 0 so the expression is minimised |
| by) making both squared brackets zero | B1 | |
| [1] | 2.1 | Needs “makes both brackets zero” oe, don’t need “squares” |
| Answer | Marks |
|---|---|
| (a)(ii) | This choice of a and b gives the minimum (sum |
| Answer | Marks |
|---|---|
| Dependent, response | B1 |
| Answer | Marks |
|---|---|
| [1] | 1.2 |
| 1.2 | OE. Needs “minimises” oe and “squares of residuals/ |
| Answer | Marks |
|---|---|
| (c) | 314 |
| Answer | Marks |
|---|---|
| 2 7 42 4 | B1 |
| Answer | Marks |
|---|---|
| M1 | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Exact or in range [14.9, 15.0] Allow 15 only if 3SF seen |
| Answer | Marks | Guidance |
|---|---|---|
| OR | Gradient of v on u = (gradient of y on x) –2 4 | M1A1 |
| Answer | Marks |
|---|---|
| OR | (x, y) = (0,−6),(36,0); (u, v) = (2, 6 8),(3 1 0,8) |
| Answer | Marks |
|---|---|
| 7 83 | M1 |
| M1 | Find any 2 points on (y on x) and convert to (u, v) |
| Answer | Marks | Guidance |
|---|---|---|
| OR | v = a + bu 8 – 2y = a + b(2 + 4x) | M1 |
| A1 | Substitute and compare coefficients of x |
| Answer | Marks | Guidance |
|---|---|---|
| 84 | A1 | |
| [3] | 3.2a | Obtain –0.988 or better. Ignore intercept constants. Don’t |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
7 | (a)(i) | (Squares are 0 so the expression is minimised
by) making both squared brackets zero | B1
[1] | 2.1 | Needs “makes both brackets zero” oe, don’t need “squares”
or “minimised” here
(a)(ii) | This choice of a and b gives the minimum (sum
of) squares of residues
Dependent, response | B1
[1]
B1
[1] | 1.2
1.2 | OE. Needs “minimises” oe and “squares of residuals/
differences/distances/errors” oe, but don’t need “sum of”
Both, no others
(b)
(c) | 314
or 15.0 (14.952)
21
(8) must be within the range of the given data, or
“must be interpolation” or “not extrapolation”.
x = 14 ( u − 2 ) , y = 12 ( 8 − v )
1(8−v)=−6+ 831(u−2)
2 7 42 4 | B1
[1]
B1
[1]
M1
M1 | 1.1
2.3
3.1a
1.1 | Exact or in range [14.9, 15.0] Allow 15 only if 3SF seen
correct in working.
Allow “within range of data”. Not “within range of y-values”,
not “must be one of the data values”. Ignore extra comments,
except that if anything definitely wrong seen: B0
Rearrange
Substitute into equation
(d)
(e)
OR | Gradient of v on u = (gradient of y on x) –2 4 | M1A1 | dv dy dv du
Use = or equivalent, e.g.
du dx dy dx
b = S / S = (S × (–2×4)) / (S 42)
uv uu xy xx
OR | (x, y) = (0,−6),(36,0); (u, v) = (2, 6 8),(3 1 0,8)
7 8 3
7 83
(68−8)(2−310)
New gradient is
7 83 | M1
M1 | Find any 2 points on (y on x) and convert to (u, v)
Find new gradient
OR | v = a + bu 8 – 2y = a + b(2 + 4x) | M1
A1 | Substitute and compare coefficients of x
4 and –2 correctly placed
Compare coefficients of x: 4 b ( − 2 ) = 84 32
−83
Gradient is (which is very close to –1, AG)
84 | A1
[3] | 3.2a | Obtain –0.988 or better. Ignore intercept constants. Don’t
need conclusion.
M1
M1
Find any 2 points on (y on x) and convert to (u, v)
Find new gradient
Question | Answer | Marks | AO | Guidance7 The coordinates of a set of 10 points are denoted by ( $\mathrm { x } _ { \mathrm { i } } , \mathrm { y } _ { \mathrm { i } }$ ) for $i = 1,2 , \ldots , 10$. For a particular set of values of ( $\mathrm { x } _ { \mathrm { i } } , \mathrm { y } _ { \mathrm { i } }$ ) and any constants $a$ and $b$ it can be shown that\\
$\Sigma \left( y _ { i } - a - b x _ { i } \right) ^ { 2 } = 10 ( 11 - a - 6 b ) ^ { 2 } + 126 \left( b - \frac { 83 } { 42 } \right) ^ { 2 } + \frac { 139 } { 14 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Explain why $\sum \left( \mathrm { y } _ { \mathrm { i } } - \mathrm { a } - \mathrm { bx } _ { \mathrm { i } } \right) ^ { 2 }$ is minimised by taking $b = \frac { 83 } { 42 }$ and $\mathrm { a } = 11 - 6 \mathrm {~b}$.
\item Hence explain why the equation of the regression line of $y$ on $x$ for these points is given by the corresponding values of $a$ and $b$ (so that the equation is $\mathrm { y } = \frac { 83 } { 42 } \mathrm { x } - \frac { 6 } { 7 }$ ).
\end{enumerate}\item State which of the following terms cannot apply to the variable $X$ if the regression line of $y$ on $x$ can be used for estimating values of $Y$.
Dependent Independent Controlled Response
\item Use the regression line to estimate the value of $y$ corresponding to $x = 8$.
\item State what must be true of the value $x = 8$ if the estimate in part (c) is to be reliable.
\item Variables $u$ and $v$ are related to $x$ and $y$ by the following relationships.\\
$u = 2 + 4 x \quad v = 8 - 2 y$
Show that the gradient of the regression line of $v$ on $u$ is very close to - 1 .
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2024 Q7 [8]}}