4769
Mark Scheme
June 2006
Q1
(i)
L=
1
σ 1 2π
e
−
ln L = const −
(W1 − μ ) 2
2σ 12
1
2σ
2
1
⋅
1
σ 2 2π
e
(W1 − μ ) 2 −
−
(W2 − μ ) 2
2σ 22
1
2σ
2
2
(W2 − μ ) 2
2
d ln L
2
=
(W2 − μ )
(W1 − μ ) +
dμ
2σ 22
2σ 12
= 0 ⇒ σ 22W1 − σ 22 μ + σ 12W2 − σ 12 μ = 0
σ 22W1 + σ 12W2
σ 12 + σ 22
Check this is a maximum.
⇒ μˆ =
E.g.
(ii)
(iii)
d ln L
1
1
=− 2 − 2 <0
2
σ1 σ 2
dμ
M1
A1
M1
A1
A1
Differentiate w.r.t. μ.
A1
BEWARE PRINTED ANSWER.
M1
11
A1
M1
σ 22 μ + σ 12 μ
=μ
σ 12 + σ 22
∴ unbiased.
E( μˆ ) =
2
A1
2
⎛
⎞
1
⎟ ⋅ (σ 24σ 12 + σ 14σ 22 )
Var(μˆ ) = ⎜⎜ 2
2 ⎟
+
σ
σ
1 ⎠
⎝ 1
B1
B1
σ 12σ 22
σ 12 + σ 22
First factor.
Second factor.
Simplification not required at this
point.
2
T = 12 (W1 + W2 )
B1
Var(T) = 14 (σ 12 + σ 22 )
Relative efficiency (y) =
(v)
Product form.
Two Normal terms.
Fully correct.
2
=
(iv)
M1
M1
A1
Var( μˆ )
Var(T )
=
σ 24σ 12 + σ 14σ 22
4
⋅ 2
2
2 2
(σ 1 + σ 2 )
σ 1 + σ 22
=
4σ 12σ 22
(σ 12 + σ 22 ) 2
E.g. consider σ 12 + σ 22 − 2σ 1σ 2 = (σ 1 − σ 2 ) 2 ≥ 0
∴ Denominator ≥ numerator, ∴ fraction ≤
1
[Both μ̂ and T are unbiased,] μ̂ has
smaller variance than T and is therefore
better.
M1
M1
Any attempt to compare
variances.
If correct.
A1
A1
BEWARE PRINTED ANSWER.
5
M1
E1
E1
E1
4
24
4769
Q2
Mark Scheme
f ( x) =
λk +1 x k e − λx
k!
Given:
(i)
∫
∞
0
[ x > 0 (λ > 0, k integer ≥ 0)]
,
u m e -u du = m!
M1
M X (θ ) = E[eθx ]
=∫
∞
λk +1
0
k!
x k e −( λ −θ ) x dx
M1
Put (λ – θ)x = u
λk +1
=
k!(λ − θ ) k +1
⎛ λ ⎞
=⎜
⎟
⎝ λ −θ ⎠
(ii)
June 2006
∫
∞
0
u k e − u du
M1
A1
A1
A1
A1
k +1
Y = X1 + X2 + … + Xn
By convolution theorem:- mgf of Y is
{MX(θ)}n
nk + n
⎛ λ ⎞
i.e. ⎜
⎟
⎝ λ −θ ⎠
μ = M ′(0)
B1
M ′(θ ) = λ nk + n (− nk − n)(λ − θ ) − nk − n −1 (−1)
M1
A1
∴μ =
nk + n
λ
σ = M ′′(0) − μ 2
A1
M ′′(θ ) = (nk + n)λnk + n (−nk − n − 1)(λ − θ ) − nk − n − 2 (−1)
M1
∴ M ′′(0) = (nk + n)(nk + n + 1) / λ 2
A1
For obtaining this expression
after substitution.
Take out constants. (Dep on
subst.)
Apply “given”: integral = k! (Dep
on subst.) BEWARE PRINTED
ANSWER.
7
2
∴σ 2 =
=
(iii)
(nk + n)(nk + n + 1)
λ
2
−
(nk + n)
2
λ2
nk + n
8
A1
λ2
[Note that MY(t) is of the same functional
form as MX(t) with k + 1 replaced by nk + n,
i.e. k replaced by nk + n –1. This must also
be true of the pdf.]
Pdf of Y is
λnk + n
(nk + n − 1)!
× y nk + n −1 × e −λy
[for y > 0]
(iv)
M1
λ = 1, k = 2, n = 5,
0·9165
Use of N(15, 15)
B1
B1
B1
One mark for each factor of the
expression. Mark for third factor
shown here depends on at least
one of the other two earned.
M1
M1
Mean. ft (ii).
Variance. ft (ii).
Exact P(Y > 10) =
3
4769
Mark Scheme
⎛
⎞
10 − 15
P(this > 10) = P⎜⎜ N(0, 1) >
= −1 ⋅ 291⎟⎟
⎝
15
⎠
= 0·9017
Reasonably good agreement – CLT working
for only small n.
June 2006
A1
c.a.o.
A1
E2
c.a.o.
(E1, E1)
[Or other sensible comments.]
6
24
4769
Mark Scheme
June 2006
Q3
(i)
x = 36.48
s = 9.6307
s 2 = 92.7507
y = 45.5
s = 14.8129
s 2 = 219.4218
Assumptions: Normality of both populations
equal variances
H0 : μA = μB
H1 : μA ≠ μB
Where μA, μB are the population means.
9 × 92.7507 + 11 × 219.4218
20
834.756 + 24136.64
=
= 162.4198
20
36 . 48 − 45 . 5
Test statistic is
1
1
162 . 4198
+
10 12
−9.02
=
= −1.653
5.4568
B1
B1
B1
B1
B1
If all correct. [No marks for use of
sn which are 9.1365 and 14.1823
respectively.]
Do NOT accept X = Y or similar.
Pooled s 2 =
(ii)
B1
M1
A1
Refer to t20.
Double tailed 5% point is 2·086.
Not significant.
No evidence that population mean times
differ.
M1
A1
A1
A1
Assumption: Normality of underlying
population of differences.
H0 : μD = 0
H1 : μD > 0
Where μD is the population mean of “before
– after” differences.
B1
Differences are
6.4, 4.4, 3.9, -1.0, 5.6, 8.8, -1.8,
12.1
( x = 4.8
= (12.7444)2
B1
B1
No ft from here if wrong.
No ft from here if wrong.
ft only c’s test statistic.
ft only c’s test statistic.
Do NOT accept D = 0 or similar.
The “direction” of D must be
CLEAR. Allow μA = μB etc.
M1
s = 4.6393)
[A1 can be awarded here if NOT
awarded in part (i)]. Use of sn
(=4.3396) is NOT acceptable,
even in a denominator of
Test statistic is
sn
n −1
4 .8 − 0
4 . 6393
8
=2.92(64)
(iii)
12
M1
A1
Refer to t7.
Single tailed 5% point is 1.895.
Significant.
Seems mean is lowered.
M1
A1
A1
A1
No ft from here if wrong.
No ft from here if wrong.
ft only c’s test statistic.
ft only c’s test statistic.
10
The paired comparison in part (ii) eliminates
the variability between workers.
E2
(E1, E1)
2
24
4769
Mark Scheme
June 2006
Q4
(i)
B1
Latin square.
Layout such as:
I
II
III
IV
V
Surf
-aces
(ii)
(iii)
Locations
2
3
4
B
C
D
C
D
E
D
E
A
E
A
B
A
B
C
1
A
B
C
D
E
5
E
A
B
C
D
B1
B1
Xij = μ + αi + eij
B1
μ = population
grand mean for whole
experiment.
B1
B1
αi = population
mean amount by which the ith
treatment differs from μ.
B1
eij are experimental errors
~ ind
B1
B1
B1
B1
N(0, σ2).
(letters = paints)
Correct rows and columns.
A correct arrangement of letters.
SC. For a description instead of
an example allow max 1 out of 2.
B1
Allow “uncorrelated”.
Mean.
Variance.
Totals are: 322, 351, 307, 355, 291
(each from sample of size 5)
Grand total: 1626
“Correction factor” CF =
1626 2
= 105755.04
25
Total SS = 106838 – CF = 1082.96
Between paints SS =
322 2
2912
+ ... +
− CF
5
5
= 106368 – CF =612.96
Residual SS (by subtraction) = 1082.96 –
612.96
= 470.00
Source of
variation
Between paints
SS
df
MS
612.96
4
Residual
Total
470.00
1082.96
20
24
153.2
4
23.5
MS ratio =
153.24
= 6.52
23.5
3
M1
M1
A1
For correct methods for any two
SS.
If each calculated SS is correct.
B1
B1
M1
Degrees of freedom “between
paints”.
Degrees of freedom “residual”.
MS column.
M1
A1
Independent of previous M1.
Dep only on this M1.
9
4769
Mark Scheme
June 2006
Refer to F4, 20
M1
Upper 5% point is 2.87
Significant.
A1
A1
Seems performances of paints are not all
the same.
A1
No ft if wrong. But allow ft of
wrong d.o.f. above.
No ft if wrong.
ft only c’s test statistic and
d.o.f.’s.
ft only c’s test statistic and
d.o.f.’s.
12
24
4769
1)
(i)
(ii)
Mark Scheme
f (x) =
1
θ
E[ X ] =
June 2007
0≤x ≤ θ
θ
B1
2
E[2 X ] = 2E[ X ] = 2E[ X ]
= θ
∴ unbiased
2 .3
∑ x = 2.3
∴x=
= 0.46 ∴ 2 x = 0.92
5
But we know θ ≥ 1
∴ estimator can give nonsense answers,
Write-down, or by symmetry, or by
integration.
M1
A1
E1
4
B1
E1
E2
(E1, E1)
i.e. essentially useless
(iii)
Y = max{Xi}, g(y) =
ny
4
n −1
0≤y≤ θ
θn
MSE (kY) = E[(kY − θ ) 2 ] =
E[k 2Y 2 − 2kθ Y + θ 2 ] =
k 2 E[Y 2 ] − 2kθ E[Y ] + θ 2
M1
1
M1
dMSE
=
dk
2kE[Y 2 ] − 2θ E[Y ] = 0
θ E[Y ]
for k =
E[Y 2 ]
M1
A1
d 2 MSE
= 2E[Y 2 ] > 0 ∴ this is a minimum
2
dk
θ
ny n
n θ n+1
nθ
dy
=
=
E[Y ] = ∫
n
n
θ n +1 n +1
θ
0
θ
E[Y 2 ] =
∫
0
ny n+1
θn
dy =
∴ minimising k = θ
BEWARE PRINTED ANSWER
M1
M1
A1
M1
A1
n θ n+ 2
nθ 2
=
θn n+2 n+2
nθ n + 2
n+2
=
2
n + 1 nθ
n +1
M1
A1
12
(iv) With this k, kY is always greater than the sample
maximum
So it does not suffer from the disadvantage in
part (ii)
89
E2
(E1 E1)
E2
(E1 E1)
4
4769
2(i)
(ii)
(iii)
(iv)
Mark Scheme
n
⎛n⎞
G (t ) = E[t X ] = ∑ ⎜⎜ ⎟⎟( pt ) x (1 − p ) n− x
x =0 ⎝ x ⎠
n
= [(1 − p ) + pt ]
M1
2
∴ M Z (θ ) = e
⎛ −
⎜ qe
⎜
⎝
(v)
pθ
npq
+ pe
M Z (θ ) = (q −
terms in n
−3
2
⎛
⎜ q + pe
⎜
⎝
1− p
θ
npq
⎞
⎟
⎟
⎠
1
1
1
M1
e
θ2
θ2
2n
6
1
B1
For BOTH
1
1
M1
1
θ
npq
n
⎞
⎟ =
⎟
⎠
1
1
n
1
BEWARE PRINTED ANSWER
5
qpθ
qp 2θ 2
+
+
npq 2npq
, n −2 , ………….. +
M1
For expansion of exponential terms
M1
For indication that these can be
neglected as n → ∞ . Use of result
given in question
pqθ
pq 2θ 2
p+
+
+ "") n =
npq 2npq
(1 +
4
1
M Z (θ ) = ebθ M X (aθ )
np
θ
q
Available as B2 for write-down or
as 1+1 for algebra
1
= (q + pt ) n
μ = G ′(1) G′(t ) = np(q + pt ) n−1
G ′(1) = np × 1 = np
σ 2 = G′′(1) + μ − μ 2
G′′(t ) = n(n − 1) p 2 (q + pt ) n−2
G' ' (1) = n(n − 1) p 2
∴σ 2 = n 2 p 2 − np 2 + np − n 2 p 2
= −np 2 + np = npq
X −μ
Z=
Mean 0, Variance 1
σ
M (θ ) = G (eθ ) = (q + peθ ) n
Z = aX + b with:
1
1
μ
np
a= =
and b = − = −
σ
σ
q
npq
−
June 2007
+ "") n →
1
2
1
90
4
4769
(vi)
Mark Scheme
N(0,1)
Because e
June 2007
1
θ2
is the mgf of N(0,1)
E1
and the relationship between distributions and
their mgfs is unique
E1
2
(vii) “Unstandardising”, N( μ , σ 2 ) ie N ( np , npq )
91
3
1
Parameters need to be given.
1
4769
3(i)
Mark Scheme
H 0 : μ A = μB
H1 : μ A ≠ μ B
1
Where μ A , μ B are the population means
1
June 2007
Do NOT allow X = Y or similar
Accept absence of “population” if
correct notation μ is used.
Hypotheses stated verbally must
include the word “population”.
Test statistic
26.4 − 25.38
=
2.45 1.40
+
7
5
1.02
= 1.285
0.63 = 0.7937
1.645 ×
No FT if wrong
No FT if wrong
10
B1
M1
(– 0.2856, 2.3256)
H 0 is accepted if –1.96< test statistic < 1.96
A1
cao
M1
i.e. if − 1.96 < x − y < 1.96
M1
0.7937
i.e. if − 1.556 < x − y < 1.556
A1
M1
In fact, X − Y ~ N(2,0.7937 )
So we want
P(−1.556 < N(2,0.7937 2 ) < 1.556) =
2
Zero out of 4 if not N(0,1)
4
SC1 Same wrong test can get
M1,M1,A0.
SC2 Use of 1.645 gets 2 out of 3.
BEWARE PRINTED ANSWER
M1
1.556 − 2 ⎞
⎛ − 1.556 − 2
P⎜
< N (0,1) <
⎟=
0.7937 ⎠
⎝ 0.7937
P(−4.48 < N(0,1) < −0.5594) = 0.2879
(iv)
Denominator two separate terms
correct
M1
0.7937 =
1.02 ± 1.3056 =
(iii)
M1
M1
1
1
1
1
CI ( for μ A – μ B ) is
1.02 ±
Numerator
A1
Refer to N(0,1)
Double-tailed 5% point is 1.96
Not significant
No evidence that the population means differ
(ii)
M1
M1
A1
cao
Wilcoxon would give protection if assumption
of Normality is wrong.
Wilcoxon could not really be applied if
underlying variances are indeed
different.
Wilcoxon would be less powerful (worse Type
II error behaviour) with such small samples
if Normality is correct.
92
Standardising
7
E1
E1
E1
3
4769
4 (i)
(ii)
(iii)
Mark Scheme
There might be some consistent source of plotto-plot variation that has inflated the residual
and which the design has failed to cater for.
Variation between the fertilisers should be
compared with experimental error.
E1
If the residual is inflated so that it measures
more than experimental error, the
comparison of between - fertilisers variation
with it is less likely to reach significance.
Randomised blocks
C
B
A
D
E
.
.
.
.
.
.
E2
E1 – Some reference to extra
variation.
E1 – Some indication of a reason.
2
E2
1
(E1, E1)
3
E1
Blocks (strips) clearly correctly
oriented w.r.t. fertiliser gradient.
E1
All fertilisers appear in a block.
E1
Different (random) arrangements in
the blocks.
2
(1, E1)
4
Totals are: 95.0 123.2 86.8 130.2 67.4
(each from sample of size 4)
Grand total 502.6
2
“Correction factor” CF = 502.6 = 12630.338
20
Total SS = 13610.22 – CF = 979.882
SPECIAL CASE: Latin Square
(iv)
June 2007
4
Between fertilisers SS =
M1
95.0 2
67.4 2 – CF =
+ ... +
4
4
M1
For correct method for any two
A1
If each calculated SS is correct
13308.07 – CF = 677.732
Residual SS (by subtraction) =
979.882 – 677.732=302.15
Source of variation
SS
df
Between fertiliser
Residual
677.732
302.15
4
15
Total
979.882
19
MS
169.433
20.143
MS Ratio
8.41
Refer to F4, 15
-upper 5% point is 3.06
Significant
- seems effects of fertilisers are not all the same
(vii) Independent N (0, σ2 [constant])
93
M1
M1
1,A1
1
1
1
1
1
1
1
1
No FT if wrong
No FT if wrong
12
3
4769
Mark Scheme
June 2008
4769 Statistics 4
Q1
(i)
L=
x
e −θ θ x1 e −θ θ xn ⎡ e −nθ θ ∑ i ⎤
…
⎢=
⎥
x1!
xn ! ⎣⎢ x1! x2 ! xn !⎦⎥
ln L = const − n θ +
∑x
i
ln θ
product form
A1
fully correct
M1
A1
d ln L
∑ xi = 0
= −n +
dθ
θ
x
∑ i (= x )
⇒ θˆ =
n
M1
A1
A1
Check this is a maximum
e.g.
M1
M1
d 2 ln L
∑ xi
=− 2 <0
2
θ
dθ
A1
(ii)
λ = P( X = 0) = e −θ
B1
(iii)
We have R ~ B(n, e −θ ) ,
M1
so E ( R) = ne
−θ
−θ
B1
R
n
M1
~
∴ E(λ ) = e −θ
A1
A1
~
9
1
B1
−θ
Var( R) = ne (1 − e )
λ=
CAO
i.e. unbiased
~ e −θ (1 − e −θ )
Var(λ ) =
n
A1
BEWARE PRINTED
ANSWER
7
77
4769
(iv)
Mark Scheme
June 2008
~
Relative efficiency of λ wrt ML est
Var(ML Est)
~
Var(λ )
=
=
θe
Eg:-
−2θ
n
⋅
M1
M1
n
e − θ (1 − e − θ )
Expression is
θ+
=
θ
θ2
2!
any attempt to compare
variances
if correct
θ
A1
e θ −1
BEWARE PRINTED
ANSWER
M1
+…
always < 1
E1
and this is ≈ 1 if θ is small
≈ 0 if θ is large
E1
E1
Allow statement that
θ
θ
e −1
→ 0 as θ → ∞
7
78
4769
Q2
(i)
Mark Scheme
B1
P( X = x) = q x −1 p
Pgf G (t ) = E (t X ) =
June 2008
∞
∑ pt
x
q x −1
M1
x =1
= pt (1 + qt + q 2 t 2 + …)
A1
= pt (1 − qt ) −1
A1
μ = G ' (1) σ 2 = G ' ' (1) + μ − μ 2
M1
G ' (t ) = pt (−1)(1 − qt ) −2 (−q ) + p (1 − qt ) −1
= pqt (1 − qt ) − 2 + p (1 − qt ) −1
∴ G ' (1) = pq (1 − q) −2 + p (1 − q) −1 =
FT into pgf only
BEWARE PRINTED
ANSWER
[consideration of |qt| < 1
not required]
for attempt to find G'(t)
and/or G"(t)
A1
q
1
+1 =
p
p
A1
BEWARE PRINTED
ANSWER
G ' ' (t ) = pqt (−2)(1 − qt ) −3 (−q ) + pq (1 − qt ) −2 +
p (−1)(1 − qt ) − 2 (− q )
A1
∴ G' ' (1) = 2 pq 2 (1 − q ) −3 + pq (1 − q ) −2 + pq(1 − q ) −2
=
2 q 2 2q
+
p2
p
A1
2q 2 2 q 1 1
2q 2 + 2 pq + p − 1
+
+ − 2 =
2
p p p
p
p2
q
q
= 2 (2q + 2 p − 1) = 2
p
p
∴σ 2 =
M1
For inserting their values
A1
BEWARE PRINTED
ANSWER
11
79
4769
(ii)
Mark Scheme
X1=number of trials to first success
X2= “ “ “ “ “ next “
.
.
.
.
Xn= “ “ “ “ “ nth “
∴ Y=X1+X2+…..Xn
= total no of trials
to the nth success
∴ pgf of Y = (pgf of X ) n = p nt n (1 − qt ) − n
(iii)
(iv)
μ Y = nμ X =
n
p
σ Y2 = nσ X2 =
nq
p2
1
N(candidate’s μ Y , candidate’s σ Y2 )
Y = no of tickets to be sold ~ random variable as
in (ii) with n = 140 and p = 0.8
~ Approx N ( 140 = 175, 140 × 0.2 = 43.75 )
(0.8)2
1
P(Y ≥ 160) ≈ P(N(175,43.75) > 159 )
2
5
1
E1
1
Do not award if cty corr
absent or wrong, but FT if
160 used →
-2.268, 0.9884
A1
A1
For any sensible discussion in context (eg groups
of passengers ⇒ not indep.)
(ii)
1
1
M1
= P(N(0,1)>-2.343)
= 0.9905
(i)
E1
E1
1
0.8
Q3
June 2008
CAO
E1
E1
7
X = amount of salt ~ N( μ [750],σ 2 [20 2 ] )
Sample of n=9
Type I error: rejecting null hypothesis …
… when it is true.
B1
B1
Allow B1 for
P(rej H0 when true)
Type II error: accepting null hypothesis …
… when it is false.
B1
B1
Allow B1 for
P(acc H0 when false)
OC: P (accepting null hypothesis …
… as a function of the parameter under
investigation)
Reject if x < 735 or x > 765
20 2
α = P ( X < 735 or X > 765 | X ~ N(750,
))
9
(735 − 750)3
= P( Z <
= −2.25
20
(765 - 750)3
or Z >
= 2.25)
20
B1
B1
[ P(type II error | the true
value of the parameter)
scores B1+B1]
M1
Might be implicit
A1
A1
= 2(1-0.9878) = 2 × 0.0122 = 0.0244
A1
This is the probability of rejecting good output
and unnecessarily re-calibrating the machine –
seems small
[but not very small?]
E1
E1
80
6
CAO
Accept any sensible
comments
6
4769
(iii)
(iv)
Mark Scheme
June 2008
M1
Accept if 735 < x < 765, and now μ = 725 .
2
β = P(735 < X < 765 | X ~ N(725, 20 9 ))
= P(1.5
< Z< 6)
= 1 – 0.9332 = 0.0668
A1
A1
A1
This is the probability of accepting output and
carrying on when in fact μ has slipped to 725 –
small[-ish?]
E1
E1
OC = P( 735 < X < 765 | X ~ N( μ , 20
M1
=Ф
⎛ ( 765 − μ ) 3 ⎞ –
⎟
⎜
20
⎠
⎝
Ф
2
9
))
might be implicit
CAO
If upper limit 765 not
considered, maximum 2 of
these 4 marks. If Ф(6) not
considered, maximum 3
out of 4.
accept sensible comments
6
⎛ ( 735 − μ ) 3 ⎞
⎜
⎟
20
⎝
⎠
″ Ф − Ф″
μ=720:Ф(6.75) – Ф(2.25)=1– 0.9878 =0.0122
730: 5.25
0.75 =1– 0.7734 =0.2266
740: 3.75
−0.75 =1– (1–0.7734)= 0.7734
M1
A1
both correct
1
if any two correct
750: similarly or by write-down from part (ii)
[ FT ] : 0.9756
1
760, 770, 780 by symmetry
[FT]:
0.7734, 0.2266, 0.0122
1
6
Q4
(i)
xij = μ + α i + eij
μ = population …
.. grand mean for whole experiment
α i = population …
μ
.. mean by which i th treatment differs from
1
3
eij are experimental errors…
~ ind N (0, σ 2 )
(ii)
Totals are 240, 246, 254, 264, 196
each from sample of size 5
Grand total 936
“Correction factor” CF =
1
1
1
1
1
936 2
= 43804.8
20
Total SS = 44544 - CF = 739.2
81
Allow “uncorrelated”
1 for ind N; 1 for 0; 1 for
σ2.
9
4769
Mark Scheme
June 2008
M1
M1
For correct methods for
any two,
A1
if each calculated SS is
correct.
Between contractors SS =
240 2
196 2
+…+
− CF = 44209.6 – CF = 404.8
5
5
Residual SS ( by subtraction) = 739.2 – 404.8 = 334.4
M1
Source of
Variation
SS
df
MS
MS ratio
M1
1
Between
Contractors
404.8
3
Residual
334.4
16
134 .93
6.456
A1
20.9
CAO
Total
739.2
1
19
1
NO FT IF WRONG
Upper 5% point is 3.24
1
NO FT IF WRONG
Significant
1
Seems performances of contractors are not all
the same
1
Refer to F3,16
12
(iii)
Randomised blocks
B1
Description
E1
E1
82
Take the subject areas as
“blocks”, ensure each
contractor is used at least
once in each block
3
4769
Mark Scheme
June 2009
4769 Statistics 4 June 2009
Q1
(i)
Follow-through all intermediate results in this
question, unless obvious nonsense.
P(X ≥ 2) = 1 – θ – θ (1 – θ) = (1 – θ)2 [o.e.]
L = [θ]
=θ
(ii)
n0
n0 + n1
[θ(1 – θ)] n 1 [(1-θ)2]
(1 – θ)
n − n0 − n1
2 n − 2 n0 − n1
ln L = (n0 + n1 ) ln θ + (2n − 2n0 − n1 ) ln (1 – θ)
d ln L
dθ
n0 + n1
=
θ
=0
M1
A1
M1
A1
A1
Product form
Fully correct
BEWARE PRINTED
ANSWER
5
M1
A1
M1
–
2n − 2n0 − n1
1−θ
A1
M1
⇒ (1 - θˆ ) (n 0 + n 1 ) = θˆ (2n − 2n0 − n1 )
⇒ θˆ =
(iii)
∞
E(X) =
A1
n0 + n1
2n − n0
∑ xθ (1 − θ )
6
M1
x
x=0
= θ {0 + (1 – θ) + 2(1 – θ) 2 + 3(1 − θ ) 3 + …}
=
A2
1−θ
θ
Divisible, for algebra; e.g.
by “GP of GPs”
BEWARE PRINTED
ANSWER
So could sensibly use (method of moments)
~
1−θ
θ given by ~ = X
~
1
⇒θ =
1+ X
~
M1
θ
A1
To use this, we need to know the exact
numbers of faults for components with “two or
more”.
(iv)
BEWARE PRINTED
ANSWER
E1
6
B1
14
= 0 ⋅ 14
100
1
~
= 0·8772
θ =
1 + 0.14
x=
B1
Also, from expression given in question,
~
0 ⋅ 8772 2 (1 − 0 ⋅ 8772)
Var(θ ) =
100
B1
= 0·000945
CI is given by 0·8772 ± 1·96 x
(0·817, 0·937)
0 ⋅ 000945 =
80
M1
B1
M1
A1
For 0.8772
For 1.96
For 0 ⋅ 000945
7
4769
Mark Scheme
June 2009
Q2
(i)
Mgf of Z = E (e tZ ) =
∞
1
−∞
2π
∫
Complete the square
e
tz −
z2
2
M1
dz
M1
A1
A1
1
1
z2
tz −
= − (z − t) 2 + t 2
2
2
2
= e
t2
2
∞
1
−∞
2π
∫
e
−
( z −t ) 2
2
dt = e
t2
2
M1
M1
M1
Pdf of N(t,1)
∴∫=1
(ii)
A1
Y has mgf M Y (t )
M1
1
1
1
Mgf of aY + b is E[e t ( aY +b ) ]
= e bt E[e ( at )Y ] = e bt M Y (at )
(iii)
Z=
∴M
(iv)
X −μ
σ
X
, so X = σZ + μ
( t ) = e μ t .e
(σ t )
2
2
=e
μt +
μ+
σ
σ t
2
2
2
For final answer e
For factor e bt
For factor E[e ( at )Y ]
For final answer
M1
1
For factor e μt
1
1
For factor e 2
For final answer
M1
A1
For E[(e X ) k ]
A1
For M X (k )
t2
2
8
4
(σt ) 2
4
For E (e kX )
M1 A1
2
E (W 2 ) = M X (2) = e 2 μ + 2σ
For taking out factor e
For use of pdf of N(t,1)
For ∫ pdf = 1
2 2
W = eX
E (W k ) = E[(e X ) k ] = E (e kX ) = M X (k )
∴ E (W ) = M X (1) = e
t2
2
2
2
2
2
∴ Var (W ) = e 2 μ + 2σ − e 2 μ +σ [ = e 2 μ +σ (e σ − 1)]
81
M1 A1
A1
8
4769
Mark Scheme
June 2009
Q3
(i)
x = 126 ⋅ 2 s = 8·7002 s 2 = 75 ⋅ 693
y = 133 ⋅ 9 s = 10·4760 s 2 = 109 ⋅ 746
A1
H0 : μ A = μB
1
H0 : μ A ≠ μB
Where μ A , μ B are the population means.
A1 if all correct. [No
mark for use of s n , which
are 8·2537 and 9·6989
respectively.]
Do not accept X = Y or
similar.
1
Pooled s 2
9 × 75 ⋅ 693 + 6 × 109 ⋅ 476 681 ⋅ 24 + 658 ⋅ 48
=
15
15
= 89 ⋅ 3146
=
B1
[√ = 9·4506]
Test statistic is
126 ⋅ 2 − 133 ⋅ 9
89 ⋅ 3146
(ii)
1 1
+
10 7
=−
7⋅7
= − 1⋅ 653
4 ⋅ 6573
M1
A1
Refer to t15
1
No FT if wrong
Double-tailed 10% point is 1·753
Not significant
No evidence that population mean
concentrations differ.
There may be consistent differences between
days (days of week, types of rubbish, ambient
conditions,…) which should be allowed for.
1
1
1
No FT if wrong
Assumption: Normality of population of
differences.
Differences are 7·4 -1·2 11·1 5·5 6·2 3·7 −0·3
1·8 3·6
[ d = 4·2, s = 3·862 ( s 2 =14·915)]
Use of s n (= 3 ⋅ 641) is not acceptable, even in a
n −1 ]
4⋅2 − 0
10
E1
E1
1
M1
A1 Can be awarded here
if NOT awarded in part (i)
denominator of s n /
Test statistic is
3 ⋅ 862 / 9
M1
A1
= 3·26
1
1
1
1
Refer to t 8
Double-tailed 5% point is 2·306
Significant
Seems population means differ
No FT if wrong
No FT if wrong
10
82
4769
(iii)
Mark Scheme
June 2009
B1
B1
1
1
Wilcoxon rank sum test
Wilcoxon signed rank test
H0: medianA = medianB
H1: medianA ≠ medianB
[Or more formal
statements]
4
Q4
(i)
Description must be in context. If no context
given, mark according to scheme and then give
half-marks, rounded down.
Clear description of “rows”.
And “columns”
(ii)
As extraneous factors to be taken account of in
the design, with “treatments” to be compared.
Need same numbers of each
Clear contrast with situations for completely
randomised design and randomised trends.
eij ~ ind N (0, σ2)
α i is population mean effect by which ith
treatment differs from overall mean
(iii)
Source of
Variation
E1
E1
E1
E1
E1
E1
E1
E1
E1
1
1
1
1
1
9
Allow uncorrelated
For 0
For σ2
5
SS
df
MS
MS
ratio
Between
Treatments
92·30
4
23·075
5·034
Residual
68·76
15
4·584
Total
161·06
19
1
1
1
1
1
1
1
Refer to F4,15
1
No FT if wrong
Upper 1% point is 4·89
Significant, seems treatments are not all the
same
1
1
No FT if wrong
83
10
4769
Mark Scheme
June 2010
Question 1
f ( x) =
(i)
xe − x / λ
E( X ) =
=
=
1
λ2
1
λ2
{
( x > 0)
λ2
1
λ

2
∞
x 2 e − x / λ dx
0
 −λ x 2 e − x / λ  +  λ.2 xe− x / λ dx
0
0
{[0 − 0]} + 2λ.1 =
2λ .
(
1
2
E( X ) = E( X )
(ii)
}
∞
∞
M1 for integral for E(X)
M1 for attempt to
integrate by parts
∴ E λˆ [ =
For second term: M1 for
use of integral of pdf or
for integr'g by parts again
A1
X
)
] =λ
∴ λˆ is unbiased.
E1
[7]
M1
E( X 2 ) =
M1 for use of E(X2)
By parts M1
( )
=
1
λ2

∞
0
x3e− x / λ dx
 −λ x e
λ {
1
3 −x/λ
2
1
λ2
∞
}
∞
 +  3λ x 2 e − x / λ dx
0
0
{[0 − 0]} + 3λ E ( X )
= 6λ 2 .
M1 for use of E(X)
A1 for 6λ2
∴ Var ( X ) = E ( X 2 ) − {E ( X )} = 6λ 2 − 4λ 2 = 2λ 2 .
2
λ
∴ Var λˆ =
.
2n
Variance of λ̂ becomes very small as n increases.
( )
A1
2
It is unbiased and so becomes increasingly concentrated at the
correct value λ.
(iv)
A1
1
1 Var ( X )
Var λˆ = Var ( X ) =
4
4
n
=
(iii)
M1
( )
E λ = ( 18 + 14 + 18 ) 2λ = λ .
∴ λ is unbiased.
( )
A1
E1
E1
( )
M1 A1
λ2 / 6
8

∴ relative efficiency of λ to λ̂ is
= .
2
9
3λ /16
M1 any comparison of
variances
M1 correct comparison
A1 for 8/9
So λ̂ is preferred.
[Note. This M1M1A1 is allowable in full
as FT if everything is plausible.]
E1 (FT from above)
1
[2]
E λ : B1; "unbiased": E1
Var λ = ( 641 + 161 + 641 ) 2λ 2 = 163 λ 2 .
 ) / Var( λ̂ ), award 1 out of 2
Special case. If done as Var( λ
for the second M1 and the A1 in the scheme.
[7]
[8]
4769
Mark Scheme
June 2010
Question 2
(i)
e− λ ( λt )
G (t ) = E (t ) = 
x!
x =0
∞
X
= e − λ eλt = eλ (t −1) [A1]
(ii)
x


λ 2t 2
+ ...  [A1]
[M1] = e 1 + λ t +
2!


−λ
[Allow omission of previous A1 step and write-down of
this for A2 provided opening M1 has been earned (NB
answer is given)]
G ' ( t ) = λ eλ ( t −1)
Mean = G'(1)
(iii)
(iv)
Z=
X −μ
σ
:
[A1]
G'' ( t ) = λ 2 eλ (t −1) [M1]
Variance = G''(1) + mean – mean2
∴ variance = λ 2 + λ − λ 2 = λ
G'(1) = λ
[M1]
[3]
G''(1) = λ2 [A1]
[A1]
[5]
mean 0 [B1]
variance 1 [B1]
[2]
( )
Mgf of X is M (θ ) = G eθ = e
(
)
λ eθ −1
[B1]
Linear transformation result is M aX +b (θ ) = ebθ M X ( aθ )
[B2 if fully correct, any equivalent form. Allow B1 if either factor correct.]
Use with a =
M Z (θ ) = e
1
σ
− λθ
e
[A1]
(v)
Consider
=
θ2
2
1
=

λ  eθ /

and b = −
λ
λ  eθ /
λ

−1

= e
λ  eθ /

[A1]
λ
−
μ
=− λ
σ
λ
[M1]
− θ −1
λ

[A1]
[NB answer is given]


θ
θ
θ2
θ3
θ

− 1 = λ  1 +
+
+
+ ... −
− 1
3/ 2
λ 
λ 
λ 2!λ 3!λ

+ terms in λ −1/ 2 , λ −1 , λ −3/ 2 ,... [A1]
→
θ2
2
[7]
[M1]
as λ → ∞ [M1]
[some explanation required]
∴ M Z (θ ) → eθ
2
/2
as λ → ∞
[A1] [answer given]
[4]
(vi)
θ2 /2
e
is the mgf of N(0, 1)
[E1] ,
and the relationship between distributions and their mgfs is unique
[E1] .
"Unstandardising", X tends to N(μ, σ 2) i.e. N(λ, λ) [B1, parameters must be given].
2
[3]
4769
Mark Scheme
June 2010
Question 3
(i)
H 0 is accepted if –1.96 < value of test statistic < 1.96
i.e. if
−1.96 <
( x1 − x2 ) − ( 0 )
2
2
1.2 1.4
+
8
10
< 1.96
i.e. if
−1.96 × 0.6132 < x1 − x2 < 1.96 × 0.6132
i.e. if
−1.20(18) < x1 − x2 < 1.20(18)
Note. Use of μ 1 – μ 2 instead of x1 − x2 can score M1 B1 M0 M1 A0 A0.
(ii)
M1 double inequality
B1 1.96
M1
num of test statistic
M1
den of test statistic
A1
Special case. Allow 1 out of 2 of the A1 marks
if 1.645 used provided all 3 M marks have
been earned.
[6]
B1
M1
FT if wrong
[FT can's acceptance region if
reasonable]
E1
so H 0 is rejected.
CI for μ 1 – μ 2 :
r
A1
x1 − x2 = 1.4
which is outside the acceptance region
r
1.4 ± (2.576 × 0.6132),
i.e. 1.4 ± 1.5796,
i.e. ( –0.18 [–0.1796] , 2.97[96] )
M1 for 1.4
B1 for 2.576
M1 for 0.6132
A1 cao for interval
[7]
(iii)
Wilcoxon rank sum test (or Mann-Whitney form of test)
Ranks are:
First
Second
14 13 10 8 6 11
2 12 3 1 4 7 5 9
M1
M1
A1
W = 14 + 13 + 10 + 8 + 6 + 11 = 62
[or 8 + 8 + 7 + 7 + 6 + 5 = 41 if M-W used]
Refer to W 6,8 [or MW 6,8 ] tables.
B1
M1
Lower 2½% critical point is 29 [or 8 if M-W used].
Combined ranking
Correct [allow up to 2 errors;
FT provided M1 earned]
No FT if wrong
A1
Special case 1. If M1 for W 6,8 has not
been awarded (likely to be because rank
sum 43 has been used, which should be
referred to W 8.6 ), the next two M1 marks
can be earned but nothing beyond them.
Consideration of upper 2½% point is also needed.
Eg: by using symmetry about mean of
(
1
2
× 6 × 8 ) + ( 12 × 6 × 7 )
= 45, critical point is 61.
[For M-W: mean is 12 × 6 × 8 = 24, hence critical point is 40.]
Result is significant.
Seems (population) medians may not be assumed equal.
M1
M1 for any correct method
A1 if 61 correct
E1,
E1
Special case 2 (does not apply if
Special Case 1 has been invoked).
These 2 marks may be earned even if
only 1 or 2 of the preceding 3 have been
earned.
[11]
3
4769
Mark Scheme
June 2010
Question 4
(i)
Randomised blocks
B1
Eg:WEST
D
A
C
B
C
B
A
D
D
C
A
B
EAST
Plots in strips (blocks)
correctly aligned w.r.t. fertility trend.
Each letter occurs at least once in each block
in a random arrangement.
M1
E1
M1
E1
[5]
(ii)
μ = population [B1] grand mean for whole experiment [B1]
α i = population [B1] mean amount by which the ith treatment differs
from μ [B1]
e ij ~ ind N [B1, accept "uncorrelated"] (0 [B1] , σ 2 [B1] )
(ii)
Totals are
62.7 65.6 69.0 67.8
Grand total 265.1
4 marks, as
shown
3 marks, as
shown
[7]
all from samples of size 5
"Correction factor" CF = 265.12/20 = 3513.9005
M1 for attempt to
Total SS = 3524.31 – CF = 10.4095
2
2
2
2
form three sums
of squares.
62.7 65.6 69.0 67.8
Between varieties SS =
– CF
+
+
+
5
5
5
5
M1 for correct
= 3518.498 – CF = 4.5975
method for any
two.
Residual SS (by subtraction) = 10.4095 – 4.5975 = 5.8120
Source of variation
Between varieties
Residual
Total
SS
df
4.5975 3 [B1]
5.8120 16 [B1]
10.4095 19
MS [M1]
1.5325
0.36325
MS ratio [M1]
4.22 [A1 cao]
A1 if each
calculated SS is
correct.
5 marks within
the table, as
shown
Refer MS ratio to F 3,16 .
M1 No FT if
wrong
Upper 5% point is 3.24.
Significant.
Seems the mean yields of the varieties are not all the same.
A1
E1
E1
No FT if wrong
[12]
4
4769
Mark Scheme
June 2011
4769 June 2011 Qu 1
f ( x) =
(i)
L=
2
1
e − x / 2θ
2πθ
[N(0, θ )]
2
2
2
1
1
1
e − x1 / 2θ .
e − x2 / 2θ .....
e − xn / 2θ
2πθ
2πθ
2πθ
 = ( 2πθ )

n
1
ln L = − ln ( 2πθ ) −
2
2θ
−n / 2
e
x
−Σxi 2 / 2θ
d ln L
= 0
dθ
n
1
= 2
ˆ
2θ 2θˆ
x
n
M1
A1
x
M1
A1
2
i
2
A1
i
Check this is a maximum. Eg:
M1
d 2 ln L
n 1
1
= . 2 − 3  xi 2
2
dθ
2 θ
θ
A1
which, for θ = θˆ , is
for ln L
fully correct
M1
for differentiating
A1, A1 for each term
2
i
gives
1
i.e. θˆ =


2
x
product form
fully correct
Note. This A1 mark and the
next five A1 marks depend
on all preceding M marks
having been earned.
i
d ln L
n 1
1
= − . + 2
dθ
2 θ 2θ
M1
A1
n
n
n
− 2 = − 2 < 0.
2
ˆ
ˆ
2θ
2θˆ
θ
A1
for expression
involving θˆ
A1
for showing < 0
[14]
(ii)
First consider E(X 2) = Var(X) + {E(X)}2 = θ + 0
M1
A1
1
∴ E θˆ = (θ + θ + ... + θ ) = θ
n
A1
i.e. θˆ is unbiased.
A1
()
[4]
(iii)
()
Here θˆ = 10 and Est Var θˆ = 2 × 102/100 = 2
Approximate confidence interval is given by
10 ± 1.96 2 = 10 ± 2.77 , i.e. it is (7.23, 12.77).
1
B1, B1
M1
centred at 10
B1
1.96
M1
Use of √2
A1 c.a.o. Final interval
[6]
4769
Mark Scheme
June 2011
4769 June 2011 Qu 2
(i)
f ( x ) = 12 e − x / 2
n=2
M (θ ) = E ( eθ X ) = 
− x 1 −θ
1  e (2 ) 
=  1

2  − ( 2 − θ ) 
∞
1
0 2
e
− x ( 12 −θ )
∞
[A1] =
0
dx
1
2
1
2
−θ
A1
[A1] = (1 − 2θ )
−1
[A1]
Any equivalent form
A1, A1, A1 for each
expression, as shown,
beware printed answer
f ( x ) = 14 xe − x / 2
n=4
M (θ ) = 
∞
xe
1
0 4
− x ( 12 −θ )
M1 for attempt to integrate
this by parts
dx
∞
− x ( 1 −θ )
− x ( 12 −θ )


∞ e
1   xe 2 

=  1
d
x
[A1]
 [A1] − 

1
0 −
4   − ( 2 − θ ) 
( 2 −θ )

0

=

1
1
−1
.2 (1 − 2θ ) [A1] 
[ 0 − 0] [A1] + 1
4

2 −θ
=
1
2
A1, A1 for each component,
as shown
A1, A1 for each component,
as shown
1
−1
−2
1
2
1
2
−
θ
=
−
θ
(
)
(
)
1
2 (1 − 2θ )
A1 for final answer,
beware printed answer
[10]
(ii)
Mean = M'(0)
M ' (θ ) = −2 ( − n2 ) (1 − 2θ )
− n2 −1
= n (1 − 2θ )
− n2 −1
∴ mean = n
M1 A1
A1
Variance = M''(0) – {M'(0)}2
M '' (θ ) = n ( − n2 − 1) ( −2 )(1 − 2θ )
− n2 − 2
= n ( n + 2 )(1 − 2θ )
− n2 − 2
M1 A1
∴ M''(0) = n(n + 2)
A1
∴ variance = n(n + 2) – n2 = 2n
A1
[Note. This part of the question may also be done by expanding the
mgf.]
Solution continued on next page
2
[7]
4769
Mark Scheme
June 2011
4769 June 2011 Qu 2 continued
(iii)
By convolution theorem,
{
M W (θ ) = (1 − 2θ )
M1
− 12
}
k
= (1 − 2θ )
−k / 2
.
B1
This is the mgf of χ 2k ,
M1
so (by uniqueness of mgfs)
W ~ χ 2k .
B1
[4]
(iv)
W ~ χ 2100 has mean 100, variance 200. Can regard W as
the sum of a large "random sample" of χ 21 variates.
118.5 − 100


∴ P ( χ 2100 < 118.5 ) ≈ P  N(0,1) <
= 1.308 
200


= 0.9045.
M1
for use of N(0,1)
A1 c.a.o. for 1.308
A1 c.a.o.
[3]
3
4769
Mark Scheme
June 2011
4769 June 2011 Qu 3
8 separate B1 marks
for components of
answer, as shown
(i)
Type I error: rejecting null hypothesis [B1] when it is true [B1]
Allow B1 out of 2 for P(...)
Type II error: accepting null hypothesis [B1] when it is false [B1]
Allow B1 out of 2 for P(...)
OC: P(accepting null hypothesis [B1] as a function of the
parameter under investigation [B1])
P(Type II error | the true value of
the parameter) scores B1+B1
Power: P(rejecting null hypothesis [B1] as a function of the
parameter under investigation [B1])
P(Type I error | the true value of
the parameter) scores B1+B1.
"1 – OC" as definition scores zero.
[8]
(ii)
X ~ N(μ, 25)
H 0 : μ = 94
(
H 1 : μ > 94
(
)
We require 0.02 = P reject H 0 μ = 94 = P X > c μ = 94
c − 94 

= P ( N ( 94, 25 / n ) > c ) = P  N ( 0,1) >

5/ n 

∴
c − 94
= 2.054
5/ n
(
M1 for first expression
M1 for standardising
)
c − 97 

= P ( N ( 97, 25 / n ) > c ) = P  N ( 0,1) >

5/ n 

c − 97
= − 1.645
5/ n
∴ we have c = 94 +
M1
B1 for 2.054
We also require 0.95 = P reject H 0 μ = 97
∴
)
M1 for first expression
M1 for standardising
B1 for –1.645
10.27
8.225
and c = 97 −
n
n
Attempt to solve;
c = 95.666 [allow 95.7 or awrt]
√n = 6.165, n = 38.01
Take n as "next integer up" from candidate's value
M1 two equations
A1 both correct
(FT any previous
errors)
M1
A1 c.a.o.
A1 c.a.o.
A1
[13]
(iii)
Power function:
step function from 0
with step marked at 94
to height marked as 1
G1
G1
G1
Zero out of 3 if step is wrong way
round.
[3]
4
4769
Mark Scheme
June 2011
4769 June 2011 Qu 4
(a)
Each E2 in this part is available as E2, E1, E0.
(i)
Description of situation where randomised blocks would be suitable, ie
one extraneous factor (eg stream down one side of a field).
E2
Explanation of why RB is suitable (the design allows the extraneous
factor to be "taken out "separately).
E2
Explanation of why LS is not appropriate (eg: there is only one
extraneous factor; LS would be unnecessarily complicated; not
enough degrees of freedom would remain for a sensible estimate of
experimental error).
E2
Description of situation where Latin square would be suitable, ie two
extraneous factors (and all with same number of levels) (eg streams
down two sides of a field).
E2
Explanation of why LS is suitable (the design allows the extraneous
factors to be "taken out "separately).
E2
Explanation of why RB is not appropriate (RB cannot cope with two
extraneous factors).
E2
(ii)
[12]
(b)
Totals are 56.5 57.4 60.6 82.3 from samples of sizes 4 3 5 4
Grand total 256.8
"Correction factor" CF = 256.82/16 = 4121.64
M1 for attempt to
Total SS = 4471.92 – CF = 350.28
2
Between treatments SS =
2
2
2
56.5 57.4 60.6 82.3
– CF
+
+
+
4
3
5
4
= 4324.1103 – CF = 202.47
Residual SS (by subtraction) = 350.28 – 202.47 = 147.81
Source of variation
Between treatments
Residual
Total
SS
202.47
147.81
350.28
df
3 [B1]
12 [B1]
15
MS [M1]
67.49
12.3175
MS ratio [M1]
5.47(92) [A1 cao]
Refer MS ratio to F 3,12 .
Upper 5% point is 3.49.
Significant.
Seems the effects of the treatments are not all the same.
form three sums of
squares.
M1 for correct
method for any two.
A1 if each
calculated SS is
correct.
5 marks within
the table, as
shown
M1
A1
E1
E1
No FT if wrong
No FT if wrong
[12]
5
4769
Question
(i)
1
Mark Scheme
Marks
M1
use of cdfs
Answer
P  X  x   FB ( x).  FG ( x).
1
2
ie cdf of X is F  x  
1
2
FB ( x)  FG ( x)
ie (by differentiating) pdf of X is f  x   12 f B ( x)  f G ( x)
1
1
(ii)
(iii)
E X

1
2
2
B
G
1
2
   x f ( x)dx
2
2
B
2
 B2
G

Answer given
[3]
M1
[answer given; needs some indication of method]
M1
( x)dx
M1
 2  G 2 

A1
[1]
M1
 x f ( x)dx   x f

 B  G
1
2
2
Use of "E  X 2    2   2 "

1
2
 Var( X )  E( X 2 )  E( X )
A1
M1
2
  2  12  B 2  12 G 2  14  B 2  14 G 2  12  B G
=  2  14   B  G 
A1
A1
2
Answer given
[7]
1
(iv)
[Central Limit Theorem] Approx dist of X is
N

1
2
 B  12 G ,
1
2n
B1
B1
B1
1
(v)

2
 14   B  G 
2

B1
 X  X  Var  X   
E X      
2
and
Var  X         2n
X st 
1
2
B
st
G
1
2
either
B
st
B4
[4]
M1M1
2
n
B1
G
1
4
2
n
Guidance
A1
1
2
  xf ( x)dx   xf ( x)dx 
E X  
1
2
June 2012
B1
2
n
[4]
5
4 marks as shown
4769
Mark Scheme
Question
(vi)
1
Marks
E1
Answer
1
E ( X )  E ( X st )  (  B  G )  E ( X )
2
ie they are unbiased.
 
June 2012
E1
M1
 
Clearly Var X  Var X st ,
 X st is the more efficient.
Guidance
for any attempt to compare variances
Candidates are not required to note that the variances are equal in
the case μB  μG .
 
for deduction that Var X  Var X st [FT c's variances]
E1
More efficient
[5]
2
(i)
Mean of X = 3.5 (immediate by symmetry)
E X 2  16 1  4  ...  36   91
6
B1
M1
 
A1
2
91  7  35
 Var  X      
6  2  12
Answer given
[3]
2
(ii)
G  t   E  t X    t1. 16    t 2 . 16   ...   t 6 . 16 

1
6
t  t
2
 ...  t
6
M1
t 1  t 6 
A1
  6 1  t 
Answer given
[2]
2
(iii)
[P(N = 0) = ½, P(N = 1) = (½)(½),] P(N = r) = (½)r.(½)
B1
[1]
2
(iv)
H  t   E  t N    t 0 . 12    t1. 14    t 2 . 81   ...
M1

A1
1
1

 2  t 
t
1 2 2  t
1
2
6
 
M1
answer given; must be convincing
Answer given
4769
Question
(v)
2
Mark Scheme
Marks
M1
for use of 1st derivative
A1
Answer
Mean = H'(1), variance = H''(1) + mean – mean2.
H '  t    1 2  t 
 1   2  t 
3
3
H ''  t    2  2  t   1  2  2  t 
2
2
 mean = 1
 variance = 2 + 1 – 1 = 2
2
(vi)
K(t) = H{G(t)} = {2 – G(t)}–1

 12 1  t   t 1  t  1  t  t 2  ...  t 5  
t 1  t 6  
 

 2



6 1  t  
6 1  t 




1
June 2012
1
1
 12  t  t 2  t 3  ...  t 6 
2
6 1

  6 12  t  t  ...  t 
6


M1
for use of 2nd derivative
A1
[4]
M1
M1
M1
For variance
A1
Answer given
Guidance
inserting G(t)
use of hint given
[4]
2
(vii)
K '  t   6 12  t  t 2  ...  t
 1  2t  3t
6 2
 4t 3  5t 4  6t 5 
M1
reasonable attempt to differentiate K(t)
M1
reasonable attempt at 2nd derivative
M1
for use of derivatives
mean = K '(1)  6(12 – 6)–2(21) = 21/6 = 7/2
A1
Substitution shown
K''(1) = (12  6–3  212) + (6  6–2  70) = (49/2) + (70/6)
variance =
A1
49 70 7 49
294  140  42  147
329
  


2
6 2 4
12
12
A1
K''  t   12 12  t  t 2  ...  t 6 
 6 12  t  t 2  ...  t 6 
2
3
2
1  2t  3t
 2  6t  12t
2
2
 4t 3  5t 4  6t 5 
 20t 3  30t 4 
2
[6]
7
Ft c’s K’(1) and/or K’’(1) provided variance positive
Exact.
4769
Mark Scheme
Question
2 (viii)
June 2012
Marks
Answer
Guidance
We have:
X  7 / 2
M1
for correct use of candidate's values for means and variances
A1
answer honestly obtained (common denominator shown).
A0 if different from (vii)
 X 2  35 /12
N  1
N2  2
 Q 2  329 /12
Inserting in the quoted formula gives
  7  2   35 
294  35
329

 2      1  
12
12
  2    12 
3
(i)
as required.
H0: population medians are equal
[2]
B1
H1: population median for A < population median for B
B1
Wilcoxon rank sum test (or Mann-Whitney form of test)
Ranks are: A
1 2 4 5 9 11
B
3 6 7 8 10 12 13 14
M1
A1
W = 1 + 2 + 4 + 5 + 9 + 11 = 32
[or 0 + 0 + 1 + 1 + 4 + 5 = 11 if M-W used]
Refer to W6,8 [or MW6,8] tables
Lower 5% critical point is 31 [or 10 if M-W used]
Result is not significant
[Note:
"population" must be explicit]
1) Explicit statement re shapes of distributions.
(eg that they are the same shape) is not required.
2) More formal statements of hypotheses gain
both marks [eg cdfs are F(x) and F(x – Δ), H0 is
Δ = 0 etc).]
Combined ranking
Correct [allow up to 2 errors; FT provided M1
earned]
B1
M1
A1
A1
Seems median yields may be assumed equal
A1
[9]
8
No FT if wrong
No FT if wrong
4769
Mark Scheme
Question
(ii)
3
June 2012
Marks
Guidance
B1
B1
"population" must be explicit, either in words or
by notation
B1
For all. Use of sn scores B0
Answer
H0: population means are equal
H1: population mean for A < population mean for B
For A: x  11.4, sn 12  1.912 [ sn 1 1.38275]
y  12.575, sn 12  1.051[ sn 1 1.025]
(5 1.912)  (7 1.051) 16.915
Pooled s2 =

 1.4096
12
12
For B:
M1
for any reasonable attempt at pooling (but not if sn2 used)
A1
If correct
M1
A1
Ft if incorrect
Refer to t12
Lower single-tailed 5% critical point is –1.782
M1
A1
No FT if wrong
No FT if wrong
must compare –1.83 with –1.782 unless it is clear and explicit that
absolute values are being used
Significant
Seems mean yield for A is less than that for B
A1
A1
[11]
E1
E1
E1
E1
[4]
B1
Test statistic =
11.4  12.575
1.175

  1.83(25)
0.6412
1.4096 16  18
3
(iii)
t test is "more sensitive" if Normality is correct.
Non-rejection of Normality supports t.
But Wilcoxon is more reliable if not Normal –
and we do not have proof of Normality.
4
(i)
Latin square
4  4 layout,
with rows clearly representing casting techniques
and columns clearly representing chemical compositions [or vice
versa]
Labels each appearing exactly once in each row and in each
column
representing manufacturers.
9
B2,1,0
B2 if completely correct, B1 if one point omitted
B1
for correct structure
B1
[5]
within the square
4769
Question
(ii)
4
Mark Scheme
Answer
Totals are 440.6 453.5 458.2 459.4 all from samples of size 4
Grand total 1811.7 "Correction factor" CF = 1811.72/16
= 205141.06
Total SS = 205202.57 – CF = 61.5144
Between manufacturers SS =
June 2012
Marks
Guidance
M1
for attempt to form three sums of squares.
= 205196.55 – CF = 55.4969
Residual SS (by subtraction) = 61.5144 – 55.4969 = 6.0175
M1
A1
for correct method for any two
if each calculated SS is correct.
Source of variation
SS
df
Between treatments 55.4969 3
Residual
6.0175 12
Total
61.5144
15
B1
B1
M1
M1
A1
Between treatments df
Residual df
Method for calculating either mean square
MS ratio
cao at least 3sf, condone up to 6 sf
M1
A1
No FT if wrong
2
2
2
2
440.6 453.5 458.2 459.4



– CF
4
4
4
4
MS
18.4989
0.5014(6)
MS ratio
36.89
Refer MS ratio to F3,12.
quotation of an extreme point (not just the 5%point); eg 1% point,
5.95, or 0.1% point, 10.80.
Must be some reference to very highly significant
and very strong/overwhelming evidence that "the manufacturers
are not all the same”.
4
(iii)
A1
A1
[12]
B2
yij     i  eij
 = population …
…grand mean for whole exp't
i = population mean amount by which the ith treatment differs
from 
B1
B1
B1
B1 if any two RHS terms correct
“Population”; award here or for i
eij is experimental error – does not need to be stated explicitly
here, is subsumed in error assumptions below.
eij ~ ind N [*] ( accept "uncorrelated") (0 [*] ,  2 [*] )
B2
[7]
10
if all three * components correct, B1 if any two correct.
4769
Mark Scheme
Question
1
Answer
(i)
P ( X= x=
)
L=
Marks
−θ
e θ
x!
e − θ θ x1
e − θ θ xn
e − nθ θ Σxi
=
. ... .
x1 !
xn !
x1 ! x2 !... xn !
M1 A1
M1 for general product form. A1 (a.e.f.) for answer.
ln L = – nθ + Σx i ln θ – Σln x i !
M1 A1
M1 is for taking logs (base e).
Allow (±) constant instead of last term.
Σx
d ln L
=− n + i
dθ
θ
= 0 for ML Est θ̂ .
M1 A1
M1 for differentiating, A1 for answer.
M1
A1
[10]
M1
P ( X= 0=
) e .
ˆ
(iii)
M1
−θ
ML Est of e − θ = e − θ ,
1
Any or all of the four M1 marks down to here can be awarded in
part (iv) if not awarded here.
A1
Confirmation that this is a maximum:
Σx
d 2 ln L
=
− 2i < 0 .
2
θ
dθ
(ii)
Guidance
x
i.e. θˆ x .
=
∴ nθˆ Σxi , =
1
June 2013
M1 A1
i.e. the estimate is e − x .
We have estimate of P( X= 0)= e = 0.0067 ,
[3]
M1
so we might reasonably expect around 1000e −5 ≈ 6.7 cases
of zero in a sample of size 1000
E1
– finding no such cases seems suspicious.
E1
−5
[3]
6
M1 for "invariance property", not necessarily named.
=
=
and x 5.
Sensible
use of n 1000
4769
Mark Scheme
Question
1
(iv)
Answer
Marks
X has Poisson distribution "scaled up" so that ,
therefore ,
∞
 ∞ eθ− θ x

eθ− θ x
=k ∑
− e − θ  =k {1 − e − θ } .
where 1 =k ∑
x!
=
x 1=
 x 0 x!

1
∴ k = −θ .
1− e
(e
[for x = 1, 2, ... ].
− 1) x1 ! x2 !... xn !
(
1− )Σ−ln x !
θ
M1
M1 for sum from 0 to ∞ minus value for x = 0.
Beware printed answer.
A1
i
neθ
d ln L Σxi
,
= − θ
θ e −1
dθ
and on setting this equal to zero we get that θ̂ satisfies
Σxi
θeθ
= =
x.
θ
e −1 n
∴
(i)
M1 for this idea, however expressed.
A1
n
∴ ln
=
LΣ lnxθ
n−ln e
i
2
M1
A1
θ Σxi
θ
Guidance
M1
1 e−θ θ x
θx
∴P( X =
=θ
x ) = −θ .
1− e
x!
( e − 1) x!
∴L=
June 2013
A1
A1
Beware printed answer.
[8]
B1
B1
B1
(A) μ = 0.
(B) E(X2) = 8/3.
(C) Var(X) = 8/3.
[3]
2
(ii)
Mgf of X is M X (θ) = E(eθX)
1  1 
1

=  e −2θ .  +  e0 .  +  e 2θ . 
3  3 
3

1
= (1 + e 2θ + e −2θ ) .
3
M1
A1
[2]
7
Any equivalent form.
4769
2
Mark Scheme
Question
Answer
Marks
(iii)
General results:
[Convolution theorem] Mgf of sum of independent
random variables = product of their mgfs.
[Linear transformation result] M aX + b ( θ ) = ebθ M X ( aθ ) .
B1
M1
M1
1

M ΣX ( θ ) =  (1 + e 2θ + e −2θ ) 
3

Guidance
B1 for explicit mention of "independent".
Note M1 mark required for f.t. to A marks below.
Allow implicit b = 0.
Note M1 mark required for f.t. to A marks below.
n
θ
θ
1 

2
−2  

M X ( θ )=   1 + e n + e n  


3 

Z=
June 2013
A1
n
A1
X μn n
nX 3
−
=
σ
σ
2 2
B1
2 3n
2 3n  
 1 
−
θ
θ 
M Z ( θ ) = 1 + e n 2 2 + e n 2 2  

 3 

Might be implicit in what follows.
n
A1
n
θ 3
θ 3 
 1 
−

=  1 + e 2 n + e 2 n   .


3
 
 
A1
[8]
8
Beware printed answer.
4769
Mark Scheme
Question
2
(iv)
Answer
Marks
1
(1
3
θ 3 3θ 2
+1+
+
+ terms in n −3/ 2 , n −2 , ...
2n 2n.2!
MZ (θ ) = {
θ 3 3θ 2
+
+ terms in n −3/ 2 , n −2 , ...) }n
2
.2!
n
2n
cancel
neglect
+1−
 1  3θ 2  
≈   3+

2n  
 3 
June 2013
Guidance
M1
M1 for reasonable attempt to expand exponentials.
A1
A1 for this line.
A1
A1 for this line.
M1
M1
M1 for cancelling first order terms, may be implicit.
M1 for neglecting higher order terms in n −1 , MUST be explicit.
n
A1
n
 θ2 
= 1+  .
 2n 
A1
Beware printed answer.
[7]
2
(v)
Limit is e
θ2 / 2
. This is mgf of N(0, 1).
M1 M1
B1
A1
[4]
B1
Mgfs are unique (even in this limiting process).
So (approximately) distribution of Z is N(0, 1).
3
(i)
Type I error:
Type II error:
OC:
Power:
rejecting null hypothesis
when it is true
B1
accepting null hypothesis
B1
when it is false
B1
P(accepting null hypothesis
B1
as a function of the parameter under investigation)
B1
P(rejecting null hypothesis
B1
as a function of the parameter under investigation)
B1
[8]
9
M1 for limit, M1 for recognising mgf.
Bracketed phrase is not needed to earn the mark.
Bracketed word is not needed to earn the mark.
Allow B1 out of 2 for P(...).
Allow B1 out of 2 for P(...).
P(Type II error | the true value of the parameter) scores B1+B1.
P(Type I error | the true value of the parameter) scores B1+B1.
4769
Mark Scheme
Question
Answer
Marks
3
(ii)
Correct "spike" shape, or point shown.
Location of spike or point at θ 0 correct and correctly labelled.
Height of spike correct and correctly labelled as 1.
3
(iii)
X ~ N(μ, 9). n = 25. H 0 : μ = 0. H 1 : μ ≠ 0.
Accept H 0 if –1 < x < 1.
G1
G1
G1
[3]

P(Type I error) = P  X > 1

 9 
X ~ N  0,  
 25  
M1
M1
1− 0 

=P  N(0,1) >
 = 2 × 0.0478 = 0.0956.
3/ 5 

P(Type II error when μ = 0.5)

9 

= P  −1 < X < 1 X ~ N  0.5,  
25  


0.5 
 −1.5
< N(0,1) <
P ( 2.5 < N(0,1) < 0.8333)
= P
 =−
3/ 5 
 3/ 5
= 0.7976 – 0.0062 = 0.7914.
3
(iv)
Either or both marks might be implicit in what follows.
M1
M1
As for the two M1 marks for the Type I error.
M1
Standardising with correct end-points.
[9]
G1
G1
G1
G1
[4]
10
M1 for use of X > 1 , M1 for distribution of X .
Accept a.w.r.t. 0.096.
E1
E1
Correct shape – must be symmetrical, and reasonable
approximation to Normal pdf.
"Centre" at 0 and clearly labelled.
Height at 0 distinctly less than 1 by about 10% (ft candidate's
P(Type I error)).
Some indication that height at 0.5 is about 0.8 (ft candidate's
P(Type II error)).
Guidance
A1
A1
This is high.
We are trying to detect only a small departure from H 0 , the
"error" (σ2) being comparatively large.
June 2013
4769
4
Mark Scheme
June 2013
Question
Answer
Marks
Guidance
(i)
Randomisation is mainly to guard against possible sources of
bias, which may be due to subjective allocation of treatments
to units or unsuspected.
B1 B1 B1
Award up to B3 marks. These include B1 for idea of "sources of
bias" and another B1 for either idea of " unsuspected" or
“subjective”.
Replication enables an estimate of experimental error to be
made.
B1 B1 B1
Award up to B3 marks. These include B1 for some concept of
replication addressing experimental error and another B1 for the
explicit idea that it enables this error to be estimated. The third B1
should be awarded for the general quality of the response,
especially the lack of extraneous material in it
Alternative suggestions offered by candidates may be rewarded
provided they are statistically sound. Suggestions that are limited
to particular designs (eg for the advantages of randomisation within
a randomised blocks design) may be rewarded if statistically sound,
but for at most 2 of the B3 marks in each set.
4
(ii)
μ is the population mean for the entire experiment.
[6]
B1 B1
α i is the population amount by which the mean for the ith
treatment differs from μ.
B1 B1
e ij ~ ind N (0, σ2)
B1
B1
B1
[7]
11
B1 for explicit mention of "population", B1 for idea of mean for
whole experiment.
B1 for explicit mention of "population", B1 for idea of difference
of means.
For "ind N"; allow "uncorrelated".
For mean 0.
For variance σ2 [i.e. that the variance is constant].
4769
Mark Scheme
Question
4
(iii)
Answer
Marks
Guidance
B1
B1
No need for definition of k as the number of treatments, and accept
simply α 1 = α 2 =... with no explicit upper end to the sequence.
Accept hypotheses stated verbally provided it is clear that
population parameters (means) are being referred to.
B1 for each d.f. (4 and 15).
Null hypothesis: α 1 = α 2 = ... = α k ( = 0 )
Alternative hypothesis: not all α i are equal
Note that alternative hypothesis is NOT that all the α i are
different – B0 for alternative hypothesis if this is stated.
B1
Source of
variation
Between
fertilisers
Residual
Total
Sum of
squares
219.2
d.f.
304.5
523.7
15
19
4
Mean
squares
54.8
June 2013
B1
MS ratio
M1
For method of mean squares.
2.7
M1
For method of mean square ratio.
A1
A1 c.a.o. for 2.7 (2.6695).
20.3
f.t. from here provided all M marks earned.
Refer 2.7 to F 4,15 .
M1
Upper 5% point is 3.06.
Not significant.
Seems mean effects of fertilisers are all the same.
A1
A1
E1
[11]
12
No f.t. if wrong but allow M1 for F with candidate’s df if both
positive and totalling 19.
cao. No f.t. if wrong (or if not quoted).
Verbal conclusion in context, and not "too assertive".