Question 3:
3 | 8 | 6.752 | 8.972443
> 4 | 5 | 2.221
Q6 (a)
(b) | µ~ mean length of upper shore limpets, µ ~ mean length of lower shore limpets
U L
H : µ = µ
0 u L
H : µ < µ both
1 u L
0.422 0.672
s.e. = +
120 150
= 0.0668
5.05−4.97 awrt + 1.20
z= =(±)1.1975
0.0668
Critical region is z ≥1.6449, or probability = awrt (0.115 or 0.116) z = + 1.6449
(1.1975 < 1.6449) therefore not in critical region / accept H /not significant
0
(or P(Z≥1.1975)=0.1151, 0.1151>0.05 or z not in critical region )
There is no evidence that the limpets on the upper shore are shorter than the limpets
on the lower shore.
Assume the populations or variables are independent
Standard deviation of sample = standard deviation of population
[Mention of Central Limit Theorem does NOT score the mark] | B1
M1
A1
dM1 A1
B1
M1
A1
(8)
B1
B1
(2)
[10]
(a)
(b) | 1st B1 If µ,µ used then it must be clear which refers to upper shore. Accept
1 2
sensible choice of letters such as u and l.
0.672 0.67 0.422
1st M1 Condone minor slips e.g. or + etc i.e. swapped n or one
120 150 120
0.67 0.42
sd and one variance but M0 for +
150 120
1st A1 can be scored for a fully correct expression. May be implied by awrt 1.20
2nd dM1 is dependent upon the 1st M1 but can ft their se value if this mark is scored.
2nd A1 for awrt (+) 1.20
3rd M1 for a correct statement based on their z value and their cv. No cv is M0A0
If using probability they must compare their p (<0.5) with 0.05 (o.e) so can
allow 0.884< 0.95 to score this 3rd M1 mark.
May be implied by their contextual statement and M1A0 is possible.
3rd A1 for a correct comment to accept null hypothesis that mentions length of
limpets on the two shores.
1st B1 for one correct statement. Accept ”samples are independent”
2nd B1 for both statements
Q7 (a)
(b) | 600.9
Estimate of Mean = = 120.18
5
600.92 0.148
Estimate of Variance = 1 { 72216.31 − } or = 0.037
4 5 4
P(-0.05 < µ−µˆ < 0.05) = 0.90 or P(−0.05< X −µ<0.05)=0.90 [ < is OK]
0.05
=1.6449
0.2
n
1.64492×0.22
n =
0.052
n = 43.29…
n = 44 | M1A1
M1
A1ft A1
(5)
B1
M1 A1
dM1
A1
A1
(6)
[11]
(a)
(b) | 1st M1 for an attempt at ∑x (accept 600 to 1sf)
600.9
1st A1 for =awrt 120 or awrt 120.2. No working give M1A1 for awrt 120.2
5
2nd M1 for the use of a correct formula including a reasonable attempt at
∑x2 (Accept 70 000 to 1sf) or ∑( x−x )2 =0.15(to 2 dp)
2nd A1ft for a correct expression with correct ∑x2 but can ft their mean (for
expression - no need to check values if it is incorrect)
3rd A1 for 0.037 Correct answer with no working scores 3/3 for variance
B1 for a correct probability statement or “width of 90% CI = 0.05×2=0.1”
0.05 0.2
1st M1 for = z value or 2× ×z =0.1
0.2 n
n
Condone 0.5 instead of 0.05 or missing 2 or 0.05 for 0.1 for M1
1st A1 for a correct equation including 1.6449
2nd dM1 Dependent upon 1st M1 for rearranging to get n = …Must see “squaring”
2nd A1 for n = awrt 43.3
3rd A1 for rounding up to get n = 44
Using e.g.1.645 instead of 1.6449 can score all the marks except the 1st A1 | 1st B1 may
be implied
by 1st A1
scored or
correct
equation.
Q8 (a)
(b)
(c) | E(4X-3Y)=4E(X) – 3E(Y)
= 4×30 – 3 ×20
= 60
Var(4X-3Y)= 16 Var (X) + 9 Var (Y) 16 or 9; adding
= 16 × 9 + 9 × 4
= 180
E(B) = 80
Var (B) = 16
E(B - A) = 20 E(B)-E(A)
Var (B - A) = 196 ft on 180 and 16
⎛ −20 ⎞
P (B - A >0) = P Z > =[P(Z >−1.428...)] stand. using their mean and var
⎜ ⎟
⎝ 196⎠
= 0.923 … awrt 0.923 – 0.924 | M1
A1
(2)
M1; M1
A1
(3)
B1
B1
M1
A1ft
dM1
A1 (6)
[11]
(a)
(b)
(c) | M1 for correct use of E(aX + bY) formula
1st M1 for 16Var(X) or 9Var(Y)
2nd M1 for adding variances
Key points are the 16, 9 and +. Allow slip e.g using Var(X)=4 etc to score Ms
1st M1 for attempting B - A and E(B - A) or A - B and E(A - B)
This mark may be implied by an attempt at a correct probability
⎛ 0−(80−60)⎞
e.g. P Z > . To be implied we must see the “0”
⎜ ⎟
⎝ 180+16 ⎠
1st A1ft for Var(B - A) can ft their Var(A) = 180 and their Var(B) = 16
2nd dM1 Dependent upon the 1st M1 in part (c).
for attempting a correct probability i.e. P(B-A>0) or P(A-B < 0) and
standardising with their mean and variance.
They must standardise properly with the 0 to score this mark
2nd A1 for awrt 0.923 ~ 0.924