Standard +0.3 This is a straightforward application of standard linear regression and correlation testing formulas. Students must calculate the regression line equation, make a prediction, then compute the product moment correlation coefficient and perform a hypothesis test. While it involves multiple steps and careful arithmetic with the given summations, all techniques are routine for Further Maths students with no novel problem-solving required—making it slightly easier than average.
A random sample of 8 students is chosen from those sitting examinations in both Mathematics and French. Their marks in Mathematics, \(x\), and in French, \(y\), are summarised as follows.
$$\Sigma x = 472 \qquad \Sigma x^2 = 29950 \qquad \Sigma y = 400 \qquad \Sigma y^2 = 21226 \qquad \Sigma xy = 24879$$
Another student scored 72 marks in the Mathematics examination but was unable to sit the French examination. Estimate the mark that this student would have obtained in the French examination. [5]
Test, at the 5% significance level, whether there is non-zero correlation between marks in Mathematics and marks in French. [6]
Calculate value of t using sX from above: t = (4⋅2 – 3⋅8) / (sX /√10) = 2⋅0 M1 A1
State or use correct tabular t value: t9, 0.95 = 1⋅833 [allow 1⋅83] B1*
(or can compare 0⋅4 with 0⋅367)
Correct conclusion (A.E.F., √ on t, dep *B1): t > 1⋅833, so claim is justified
or mean mass of Royals
Answer
Marks
> mean mass of Crowns B1
9
5
14
Question 9:
9 | Calculate gradient b 1 in y –y = b1(x –x) : Sxy = 24 879 – 472 × 400/8
= 1 279
2
Sxx = 29 950 – 472 /8 = 2 102
b1 = Sxy / Sxx = 0⋅6085 (3 s.f.) M1 A1
Find regression line of y on x: y = 400/8 + b1 (x – 472/8) M1 A1
= 50 + 0⋅6085 (x – 59)
= 0⋅6085x + 14⋅1
Find y when x = 72: = 57⋅9 or 58
Allow use of regression line of x on y
2
(since neither variable clearly independent): Syy = 21 226 – 400 /8 = 1 226
b2 = Sxy / Syy = 1⋅043 (M1 A1)
x = 472/8 + b2 (y – 400/8) (M1 A1)
= 1⋅043 y + 6⋅85
Y = 62⋅5 or 62 (A1) | 5
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Cambridge International A Level – October/November 2015 | 9231 | 21
Find product moment correlation coefficient r: r = 1 279 / √(2 102 × 1 226)
or √(0⋅6085 × 1⋅043) = 0⋅797 M1 A1*
State both hypotheses (B0 for r …): H0: ρ = 0, H1: ρ ≠ 0 B1
State or use correct tabular two-tail r-value: r8, 5% = 0⋅707 B1*
State or imply valid method for conclusion e.g.: Reject H0 if |r| > tab. value (AEF) M1
Correct conclusion (AEF, dep A1*, B1*): There is non-zero correlation A1 | 6 | 11
10A | 2 2
Find MI of lamina about Q: Ilamina = ⅓m{(3a) + (3a/2) }
+ m(9a/2) 2 M1 A1
2 2
[= (15/4 + 81/4) ma = 24 ma ]
State or find MI of rod about Q: Irod = (⅓ + 1) M (3a/2) 2 [= 3Ma 2 ] B1
2 2
Sum to find MI of object about Q: I 1 = 24 ma + 3 Ma
= 3 (8m + M) a 2 A.G. A1
2 2 2
Find MI of object about mid-point of PQ: I2 = (15/4 + 3 ) ma + ⅓ M (3a/2)
2 2
= (51/4) ma + ¾ Ma
= ¾ (17m + M) a 2 A.G. M1 A1
2 2
Use eqn of circular motion to find d θ/dt for axis l1:
[–]I1 d 2 θ/dt 2 = mg × (9a/2) sin θ
+ Mg × (3a/2) sin θ M1 A1
[ = (9m/2 + 3M/2) ga sin θ ]
2
[Approximate sin θ by θ and] find ω1 in SHM eqn:
ω1 2 = (3m + M)g / 2(8m + M) a M1
Find period T1 for axis l1 from 2π/ω1: (AEF) T1 = 2π√{2(8m + M) a / (3m + M)g} A1
2 2
Use eqn of circular motion to find d θ/dt for axis l2:
[–]I2 d 2 θ/dt 2 = mg × 3a sin θ M1
2
[Approximate sin θ by θ and] find ω2 in SHM eqn:
ω2 2 = 4mg / (17m + M) a M1
Find period T2 for axis l2 from 2π/ω2: (AEF) T2 = 2π√{(17m + M) a / 4mg} A1
Verify that T1 = T2 when m = M: (AEF) T1 = 2π√(18 a / 4g) = T2 B1
nd
[Taking m = M throughout 2 part can earn:
M1 A1 M1 A0 M1 M1 A0 B1 (max 6/8)] | 4
2
8 | 14
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Cambridge International A Level – October/November 2015 | 9231 | 21
10B | State hypotheses (B0 forx …), e.g.: H0: µX = µY , H1: µX ≠ µY B1
State assumption (AEF): Distributions have equal variances B1
Find sample means and estimate popln. variances: x = 4⋅2, y = 4⋅8
2 2
sX = (180 – 42 /10) / 9
(allow biased here: 0⋅36 or 0⋅6 2 ) = 0⋅4 or 0⋅6325 2
sY 2 = (281⋅5 – 57⋅6 2 /12) / 11
(allow biased here: 0⋅4183 or 0⋅6468 2 ) = 0⋅4564 or 251/550 or 0⋅6755 2 M1
2 2 2
Estimate (pooled) common variance: s = (9 sX + 11 sY ) / 20 (AEF)
(note sX 2 and sY 2 not needed explicitly) or (180 – 42 2 /10 + 281⋅5 – 57⋅6 2 /12) / 20
= 0⋅431 or 0⋅6565 2 M1 A1
Calculate value of t (to 3 s.f.): [-] t = (y –x) / s √(1/10 + 1/12)
= 2⋅13 M1 A1
State or use correct tabular t value: t20, 0.975 = 2⋅086 [allow 2⋅09] B1*
(or can comparey –x = 0⋅6 with 0⋅586)
Correct conclusion (AEF, √ on t, dep *B1): t > 2⋅09 so mean masses not same B1
S.R. Implicitly taking sX 2 , sY 2 as popln. variances: z = ( y – x) / √(sX 2 /10 + sY 2 /12)
(may also earn first B1 B1 M1) = 0⋅6 / √(0⋅078 = 2⋅15
Comparison with z0.975 and conclusion: 2⋅15 > 1⋅96
(can earn at most 5/9) so mean masses not same (B1)
State hypotheses (B0 forx …), e.g.: H0: µX = 3⋅8 , H1: µX > 3⋅8
or H0: µX = µZ, H1: µX > µZ B1
Calculate value of t using sX from above: t = (4⋅2 – 3⋅8) / (sX /√10) = 2⋅0 M1 A1
State or use correct tabular t value: t9, 0.95 = 1⋅833 [allow 1⋅83] B1*
(or can compare 0⋅4 with 0⋅367)
Correct conclusion (A.E.F., √ on t, dep *B1): t > 1⋅833, so claim is justified
or mean mass of Royals
> mean mass of Crowns B1 | 9
5 | 14
A random sample of 8 students is chosen from those sitting examinations in both Mathematics and French. Their marks in Mathematics, $x$, and in French, $y$, are summarised as follows.
$$\Sigma x = 472 \qquad \Sigma x^2 = 29950 \qquad \Sigma y = 400 \qquad \Sigma y^2 = 21226 \qquad \Sigma xy = 24879$$
Another student scored 72 marks in the Mathematics examination but was unable to sit the French examination. Estimate the mark that this student would have obtained in the French examination. [5]
Test, at the 5% significance level, whether there is non-zero correlation between marks in Mathematics and marks in French. [6]
\hfill \mbox{\textit{CAIE FP2 2015 Q9 [11]}}