IB SL – 2
IB Packet – Correlation and Regression
1.
Markscheme
(a)
(A1) for correct scales and labels
(A3) for all ten points plotted correctly
(A2) for eight or nine points plotted correctly
(A1) for six or seven points plotted correctly
Note: Award at most (A0)(A3) if axes reversed.
(A4)
(b)
(c)
(i)
x = 42
(A1)
(ii)
y = 64
(A1)
( x, y) plotted on graph and labelled, M
(A1)(ft)(A1)
Note: Award (A1)(ft) for position, (A1) for label.
(d)
–0.998
(e)
line on graph
(A1)(ft)(A1)
Notes: Award (A1)(ft) for line through their M, (A1) for approximately
correct intercept (allow between 83 and 85). It is not necessary that the line
is seen to intersect the y-axis. The line must be straight for any mark to be
awarded.
(f)
y = –0.470(25) + 83.7
(M1)
Note: Award (Ml) for substitution into formula or some indication of method
on their graph. y = –0.470(0.25) + 83.7 is incorrect.
(G2)
Note: Award (G1) for correct sign, (G1) for correct absolute value.
= 72.0 (accept 71.95 and 72)
(A1)(ft)(G2)
Note: Follow through from graph only if they show working on their graph.
Accept 72 ±0.5.
(g)
Yes since 25 % lies within the data set and r is close to –1
Note: Accept Yes, since r is close to –1
Note: Do not award (R0)(A1).
(R1)(A1)
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2.
(a)
(A1)(A3)
Notes: Award (A1) for scales and labels (accept x/y).
Award (A3) for all points correct.
Award (A2) for 7 or 8 points correct.
Award (A1) for 5 or 6 points correct.
Award at most (A1)(A2) if points are joined up.
If axes are reversed award at most (A0)(A3)(ft).
(b)
Negative
(A1)
(c)
(i)
17
(G1)
(ii)
23
(G1)
(d)
Point correctly placed and labelled M
Note: Accept an error of ±0.5.
(A1)(ft)(A1)
(e)
y = –0.708x + 35.0
(G1)(G1)
Note: Award at most (G1)(G0) if y = not seen. Accept 35.
(f)
Regression line drawn that passes through M and (0, 35)
(A1)(ft)(A1)(ft)
Note: Award (A1) for straight line that passes through M, (A1) for line
(extrapolated if necessary) that passes through (0, 35) (accept error of ±1).
If ruler not used, award a maximum of (A1)(A0).
(g)
y = –0.708(30) + 35.0
= 14 (Accept 13)
(M1)
(A1)(ft)(G2)
OR
Using graph: (M1) for some indication on graph of point, (A1)(ft)
(M1)
for answers. Final answer must be consistent with their graph. (A1)(ft)(G2)
Note: The final answer must be an integer.
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3.
(a)
(A1)(A2)
Notes: Award (A1) for axes labelled with d and t and correct scale, (A2) for
6 or 7 points correctly plotted, (A1) for 4 or 5 points, (A0) for 3 or less points
correctly plotted. Award at most (A1)(A1) if points are joined up.
If axes are reversed award at most (A0)(A2).
(b)
(i)
(ii)
t =4
(G1)
 568 
d  81.1
(G1)

 7 
Note: If answers are the wrong way around award in (i) (G0) and in (ii)
(G1)(ft).
(c)
Point marked and labelled with M or (t , d ) on their graph
(A1)(ft)(A1)(ft)
(d)
Line of best fit drawn that passes through their M and (0, 48) (A1)(ft)(A1)(ft)
Notes: Award (A1)(ft) for straight line that passes through their M, (A1) for
line (extrapolated if necessary) that passes through (0, 48).
Accept error of ±3. If ruler not used award a maximum of (A1)(ft)(A0).
(e)
4.5h (their answer ±0.2)
(M1)(A1)(ft)(G2)
Note: Follow through from their graph. If method shown by some indication
on graph of point but answer is incorrect, award (M1)(A0).
(f)
d = 8.25t + 48.1
Notes: Award (G1) for 8.25, (G1) for 48.1.
Award at most (G1)(G0) if d = (or y = ) is not seen.
Accept d – 81.1 = 8.25(t – 4) or equivalent.
(g)
(i)
d = 8.25 × 10.3 + 48.1
d = 133 km
(ii)
No
Outside the set of values of t or equivalent.
Note: Do not award (A1)(R0).
(G1)(G1)
(M1)
(A1)(ft)(G2)
(A1)
(R1)
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4.
(a)
(A1)(A1)(A1)
Notes: (A1) for label and scales, (A2) for all
points correct, (A1) for 5 or 6 correct.
Award a maximum of (A2) if points are joined.
(b)
r = −0.141
(G2)
Note: If negative sign is missing award (G1)(G0).
(c)
“The coefficient of correlation is too low, (very) weak
(linear) relationship”.
Not a sensible thing to do (accept “no”).
Note: Do not award (R0)(A1)
The correlation coefficient has to be mentioned in
their reasoning.
(R1)
(A1)
[7]
5.
(a)
0.965
(A1) (C1)
(b)
y = 1.15x + 0.976
(A1) for 1.15x (A1) for +0.976
(A1)(A1) (C2)
(c)
y = 1.15 (7) + 0.976
(M1)
Chemistry = 9.03
(accept 9)
(A1)(ft) (C2)
Note: Follow through from candidate’s answer to (b) even if no working is
seen. Award (A2)(ft).
(d)
the correlation coefficient is close to 1
OR strongly correlated variables
OR 7 lies within the range of physics marks.
(R1) (C1)
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6.
(a)
r=
S xy
SxSy
4.16
(8.96)(0.610)
= 0.76
=
(M1)
(A1)
2
(b)
There is a fairly strong positive correlation between high school
grades and university grades.
(A1) (A1)
2
Note: Award (A1) for strong (or fairly strong) or high, (A1) for positive.
(c)
y– y
_
S xy
S x2
_
( x  x)
4.16
(x – 83.5)
8.96 2
y = 0.052x – 1.29 (3 s.f.)
Note: Award (C2) for correct answer (from calculator).
y – 3.04 =
(M1)
(A1)
2
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7.
(a)
For applying r =
S xy
S x S y 
or any correct formulae.
(M1)
For: Sx = 4.0034568…  4
(M2)
Sy = 13.992456…  14
(M2)
r = 0.6399706…  0.64 (2 d.p.)
(A1)
6
Note: Follow through with candidate’s Sx + Sy.
Accept solutions that use the unbiased estimates for the population standard
deviations.
(b)
This indicates that there is a degree of positive correlation between scores in Mathematics and scores
in English.
(R1)
Note: Follow through using candidate’s v,2
and  52% (v) from table.
Therefore those who do well in Mathematics are likely to do well in English also. (Or equivalent
statements.)
(R1)
2
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8.
(a)
II
(A1) (C1)
(b)
V
(A1) (C1)
(c)
III
(A1) (C1)
(d)
I
(A1) (C1)
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