| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of Pearson’s product-moment correlation coefficient |
| Type | One-tailed test for negative correlation |
| Difficulty | Standard +0.3 This is a straightforward application of standard correlation and regression formulas with all summary statistics provided. Parts (i)-(iii) are routine calculations, part (iv) requires computing r using the formula, and part (v) is a standard hypothesis test comparing r to a critical value. While it's a multi-part question worth 13 marks total, each step follows textbook procedures with no conceptual challenges or novel insights required—slightly easier than average due to its mechanical nature. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09a Dependent/independent variables5.09c Calculate regression line |
| Answer | Marks |
|---|---|
| 10 | (i) Find mean values to 3 s.f.: x = 2⋅024, y = 3⋅817 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Valid method for reaching conclusion: Reject H0 if | r | > tabular value M1 |
| Answer | Marks |
|---|---|
| Correct conclusion (needs values correct): There is negative corrln. (A.E.F.) A1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | [13] | |
| Page 6 | Mark Scheme: Teachers’ version | Syllabus |
| GCE A LEVEL – October/November 2010 | 9231 | 02 |
| Answer | Marks |
|---|---|
| EITHER | Equate tensions at O (A1 for each): 60 (1 + MO)/2 = 20 (2 – MO)/1 M1A1A1 |
| Answer | Marks |
|---|---|
| = 0⋅0937 [s] A1 | 4 |
| Answer | Marks |
|---|---|
| 3 | [14] |
| Answer | Marks |
|---|---|
| OR | Integrate f(t) to find F(t): F(t) = ∫ 0 t λe -λx dx = [– e -λx ] 0 t |
| Answer | Marks |
|---|---|
| Correct conclusion (A.E.F., requires correct values): 3⋅58 < 14⋅07 so suitable model A1 | 2 |
| Answer | Marks |
|---|---|
| 7 | [14] |
Question 10:
10 | (i) Find mean values to 3 s.f.: x = 2⋅024, y = 3⋅817 B1
(ii) Calculate gradient b in y –y = b(x –x):
b = (88⋅415 – 24⋅29 × 45⋅8/12) / (50⋅146 – 24⋅29 2 /12) M1
= – 4⋅292 / 0⋅979
or – 0⋅358 / 0⋅0816 = – 4⋅38[4] A1
(iii) Find regression line: y – 3⋅817 = – 4⋅384 (x – 2⋅024) M1
y = 12⋅7 – 4⋅38x A1
(iv) Find correlation coefficient r:
r = (88⋅415 – 24⋅29 × 45⋅8/12) / √{(50⋅146 – 24⋅29 2 /12) (211⋅16 – 45⋅8 2 /12)} M1
= – 4⋅292 / √(0⋅979 × 36⋅36) A1
or – 0⋅358 / √(0⋅0816 × 3⋅03)
= – 0⋅719 A1
State valid comment in context (A.E.F.): [Moderate] negative correlation
between rainfall and sunshine A1
(v) State both hypotheses: H 0: ρ = 0, H1: ρ < 0 B1
Use correct tabular r value: r12, 1% = 0⋅658 B1
Valid method for reaching conclusion: Reject H0 if |r| > tabular value M1
S.R. Calculate t-value: t = r√10 / √(1 – r 2 ) = – 3⋅27 (B1)
Use correct tabular t value: t10, 0.99 = 2⋅76[4] (B1)
Correct conclusion (needs values correct): There is negative corrln. (A.E.F.) A1 | 1
2
2
4
4 | [13]
Page 6 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2010 | 9231 | 02
11
EITHER | Equate tensions at O (A1 for each): 60 (1 + MO)/2 = 20 (2 – MO)/1 M1A1A1
Simplify to evaluate MO: MO = 10/50 = 0⋅2 A.G. A1
Apply Newton’s law at general point: 0⋅1 d 2 y/dt 2 =
(lose A1 for each incorrect term) 20 (1⋅8 – y) – 60 (1⋅2 + y)/2 M1 A2
Simplify: d 2 y/dt 2 = – 500y A.G. A1
State period (A.E.F.): T = 2π/√500 or π /5√5
or 0⋅281 [s] B1
(i) Find speed v when first at 0⋅3 from M: v 2 = 500 (0⋅2 2 – 0⋅1 2 ) M1
v = √15 or 3⋅87 [m s -1 ] A1
(ii) Find time t to reach this point: t = (1/ω) cos -1 (-0⋅1/0⋅2)
(A.E.F.) or ¼ T + (1/ω) sin -1 (0⋅1/0⋅2) M1
= (2π/3) /ω or (π/2 + π/6) /ω A1
= 2⋅094/√500
[or = 0⋅07025 + 0⋅02342]
= 0⋅0937 [s] A1 | 4
5
2
3 | [14]
11
OR | Integrate f(t) to find F(t): F(t) = ∫ 0 t λe -λx dx = [– e -λx ] 0 t
= 1 – e -λt A.G. M1 A1
EITHER: Deduce λ directly from mean: λ = 1/20 or 0⋅05
OR: Deduce λ from a tabular value, e.g.: 1 – e -40λ = 0⋅8647, λ = 0⋅05 M1
Substitute for λ and put t = 15 to give F(15) to 4 dp: 1 – e -15/20 = 0⋅5276 [or 0⋅5277] A1
Calculate expected values to 2 dp (5 values earn A1):
22⋅12 17⋅23 13⋅41 10.45 8.14 6.34 4.93 3⋅85 13⋅53 M1 A2
-t/20
State (at least) null hypothesis: H0: 1 – e fits data (A.E.F.) B1
Combine two adjacent cells with exp. value < 5: O: . . . 8 6 17
E: . . . 6⋅34 8⋅78 13⋅53 M1
Calculate value of χ2 (to 2 dp): χ2 = 3⋅58 M1 A1
(Cells not combined gives 4⋅81 earning M1 A0, max 4/7)
Compare with consistent tabular value (to 2 dp): χ 7, 0.95 2 = 14⋅07 (cells combined)
χ 8, 0.95 2 = 15⋅51 (not combined) B1
χ2
Valid method for reaching conclusion: Reject H0 if > tabular value M1
Correct conclusion (A.E.F., requires correct values): 3⋅58 < 14⋅07 so suitable model A1 | 2
2
3
7 | [14]For each month of a certain year, a weather station recorded the average rainfall per day, $x$ mm, and the average amount of sunshine per day, $y$ hours. The results are summarised below.
$n = 12$, $\Sigma x = 24.29$, $\Sigma x^2 = 50.146$, $\Sigma y = 45.8$, $\Sigma y^2 = 211.16$, $\Sigma xy = 88.415$.
\begin{enumerate}[label=(\roman*)]
\item Find the mean values, $\bar{x}$ and $\bar{y}$. [1]
\item Calculate the gradient of the line of regression of $y$ on $x$. [2]
\item Use the answers to parts (i) and (ii) to obtain the equation of the line of regression of $y$ on $x$. [2]
\item Find the product moment correlation coefficient and comment, in context, on its value. [4]
\item Stating your hypotheses, test at the 1% level of significance whether there is negative correlation between average rainfall per day and average amount of sunshine per day. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2010 Q10 [13]}}