S3 June 2009 ms

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June 2009
6691 Statistics S3
Mark Scheme

Question
Number
Scheme Marks



Q1 (a) Randomly select a number between 00 and 499 (001 and 500)
B1

select every 500th person
B1


(2)
(bi) Quota




Advantage:


Representative sample can be achieved (with small sample size)
B1

Cheap (costs kept to a minimum) not

"quick"



Administration relatively easy






Disadvantage


Not possible to estimate sampling errors (due to lack of randomness)
B1

Not a random process


Judgment of interviewer can affect choice of sample - bias


Non-response not recorded


Difficulties of defining controls e.g. social class







(2)
(bii) Systematic


B1

Advantage:


Simple or easy to use not "quick" or "cheap" or "efficient"


It is suitable for large samples (not populations)
B1 (2)

Disadvantage


Only random if the ordered list is (truly) random

Requires a list of the population or must assign a number to each member of the pop.
[6]




(a) 1st B1 for idea of using random numbers to select the first from1 - 500 (o.e.)

2nd B1 for selecting every 500th (name on the list)

If they are clearly trying to carry out stratified sample then score B0B0



(b) Score B1 for any one line

(i) 1st B1 for Quota advantage
2nd B1 for Quota disadvantage

(ii) 3rd B1 for Systematic Advantage
4th B1 for Systematic Disadvantage



8371_9374 GCE Mathematics January 2009
137

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Question
Number
Scheme Marks



Q2 (a) Limits are 20.1 1.96 x 0.5
M1 B1





A1cso

(19.1, 21.1)
(3)



(b) 98 % confidence limits are



0.5
M1
20.1 2.3263 x


B1
10



(19.7, 20.5)
A1A1

(4)
(c)

The growers claim is not correct
B1
Since 19.5 does not lie in the interval (19.7, 20.5)
dB1

(2)

[9]









(a) M1 for 20.1 zx0.5. Need 20.1 and 0.5 in correct places with no 10

B1 for z = 1.96 (or better)

A1 for awrt 19.1 and awrt 21.1 but must have scored both M1 and B1
[ Correct answer only scores 3/3]

(b)
0.5
M1 for 20.1 z x
, need to see 20.1, 0.5 and 10 in correct places

10
B1 for z = 2.3263 (or better)
1st A1 for awrt 19.7

2nd A1 for awrt 20.5
[Correct answer only scores M1B0A1A1]
(c)

1st B1 for rejection of the claim. Accept "unlikely" or "not correct"

2nd dB1 Dependent on scoring 1st B1 in this part for rejecting grower's claim
for an argument that supports this. Allow comment on their 98% CI from (b)



8371_9374 GCE Mathematics January 2009
138

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Question
Number
Scheme Marks
Q3 (a)






A B C D E F G H I J


BMI 1
6
3
8
4 5
7
2
9
10
M1



or
10 5 8 3 7 6
4
9
2
1


Finishing position
3 5 1 9 6 4 10
2
7
8


2
d
4 1 4 1 4 1
9 0 4 4





2
d = 32 (298)
M1



6x32
r

M1 A1ft

s = 1 - 10x99





133
A1
= 0.80606... (-0.80606) accept
awrt 0.806
(5)

165

(b)
B1 B1
H0 : = 0, H1 : > 0,








B1
Critical value is ( )0.5636


M1
(0.806 > 0.5636 therefore) in critical region/ reject H0
A1ft
The lower the BMI the higher the position in the race./ support for doctors belief
(5)
(c)





B1
The position is already ranked OR Position is not Normally distributed
(1)

[11]
(a) 1st M1 for attempt to rank BMI scores

2nd M1 for attempt at
2
d (must be using ranks)


No ranking
3rd M1 for use of the correct formula with their
2
d . If answer is not correct an can score 3rd
expression is required.
M1 only


1st A1ft for a correct expression. ft their
2
d but only if all 3 Ms are scored

2nd A1 awrt + 0.806 (but sign must be compatible with their
2
d )



(b)

2nd B1 for > 0 (or <0 but must be one tail and consistent with their ranking)
No 1
H
3rd B1 for critical value that is compatible with their
assume one-

1
H . If one-tail must be
tail for 3rd B1
+ 0.5636 if two-tail must be + 0.6485 [Condone wrong sign]
M1 for a correct statement relating their s
r with their cv.
e.g. "reject H0 ", "in critical region", "significant result"
May be implied by a correct comment

A1ft for correct comment in context. Must mention low/high BMI and
race/fitness or doctor's belief. Comment should be one-tailed.
Allow positive correlation between... but NOT ...positive relationship...
(c)

B1 for a correct and relevant comment either based on the fact that the data

was originally partially ordered or on the underlying normal assumption
"Quicker" or "easier" score B0
8371_9374 GCE Mathematics January 2009
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Question
Number
Scheme Marks



Q4
9

X N (55,32) therefore X N (55, )
B1 B1
8



57 - 55
P ( X > 57) = P ( Z >
) = P(Z > 1.8856...)
M1
9


8

= 1 - 0.9706
M1
= 0.0294 0.0294~0.0297
A1

[5]





1st B1 for X ~ normal and = 55, may be implied but must be X





2nd B1 for Var(X ) or st. dev of X e.g. X ~ N(55, ) or X ~ N 55,
for B1B1
8

( 8)2
9
3





9

Condone use of X if they clearly mean X so X ~ N (55, is OK for B1B1
8 )



1st M1 for an attempt to standardize with 57 and mean of 55 and their st. dev. 3


2nd M1 for 1 - tables value. Must be trying to find a probability < 0.5


A1 for answers in the range 0.0294~0.0297






8
2
ALT
X ~ N(8x55,8x3
i
)
1
1st B1 for X ~ normal and mean = 8x55
2nd B1 for variance =
2
8x3
1st M1 for attempt to standardise with 57 8
x , mean of 55 8
x and their st dev 3
8371_9374 GCE Mathematics January 2009
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Question
Number
Scheme Marks
Q5 (a)
0 40 1 33 2 14 3 8 4 5
M1 A1 (2)


x
+ x
+ x + x + x
=
= 1.05
100



-1.05
x
(b)
e
1.05
Using Expected frequency = 100 x P(X = x) = 100 x
gives
M1

x!

r = 36.743 awrt 36.743 or 36.744 A1
s = 19.290 19.29 or awrt 19.290 A1 (3)


(c)

H

0 : Poisson distribution is a suitable model
B1
H

1 : Poisson distribution is not a suitable model





Number of
Expected


Frequency

goals
frequency


0 40
34.994


1 33
36.743


2 14
19.290

3 8
6.752

8.972443
M1
> 4
5
2.221



= 4 - 1 - 1 = 2
B1ft
CR : 2
(0.05) > 5.991
B1
2

( - )2
O E

( -
)2
40 34.9937
( -
)2
13 8.972443
M1
=
+ ....+

E
34.9937
8.972443


[=0.7161...+0.3813...+1.4508...+1.80789..] A1
= 4.356. (ans in range 4.2 - 4.4)

Not in critical region
Number of goals scored can follow a Poisson distribution / managers claim is justified A1 ft (7)

[12]


(a) M1 for an attempt to find the mean- at least 2 terms on numerator seen

Correct answer only will score both marks


(b) M1 for use of correct formula (ft their mean). 1st A1 for r , 2nd A1 for s (19.29 OK)

(c) 1st B1 Must have both hypotheses and mention Poisson at least once
inclusion of their value for mean in hypotheses is B0 but condone in conclusion
1st M1 for an attempt to pool > 4
2nd B1ft for n -1-1 = 2 i.e realising that they must subtract 2 from their n
3rd B1 for 5.991 only
2nd M1 for an attempt at the test statistic, at least 2 correct expressions/values (to 3sf)
1st A1 for answers in the range 4.2~4.4
2nd A1 for correct comment in context based on their test statistic and their cv that
mentions goals or manager. Dependent on 2nd M1
Condone mention of Po(1.05) in conclusion
Score A0 for inconsistencies e.g. "significant" followed by "manager's
claim is justified"
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Question
Number
Scheme Marks



Q6 (a) ~ mean length of upper shore limpets, ~ mean length of lower shore limpets


U
L


H0 : u = L
B1

H1 : u < L

both






M1

2
2
0.42
0.67


s.e. =
+

A1

120
150


= 0.0668






5.05 - 4.97
dM1 A1
z =
= (
awrt +
)1.1975
1.20

0.0668


Critical region is z 1.6449, or probability = awrt (0.115 or 0.116) z = + 1.6449 B1






(1.1975 < 1.6449 ) therefore not in critical region / accept H0/not significant
M1

(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
A1

on the lower shore.
(8)




Assume the populations or variables are independent
B1
(b) Standard deviation of sample = standard deviation of population
B1

[Mention of Central Limit Theorem does NOT score the mark]
(2)


[10]



(a) 1st B1 If , used then it must be clear which refers to upper shore. Accept


1
2
sensible choice of letters such as u and l.

2
0.67
2
0.67 0.42
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
(b) limpets on the two shores.

1st B1 for one correct statement. Accept "samples are independent"
2nd B1 for both statements
8371_9374 GCE Mathematics January 2009
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Question
Number
Scheme Marks
Q7 (a)


600.9


Estimate of Mean =
= 120.18
M1A1

5


2
600.9
0.148
M1

Estimate of Variance = 1 { 72216.31 -
} or
= 0.037
4
A1ft A1

5
4
(5)



(b)


P(-0.05 < - < 0.05) = 0.90 or P(-0.05 < X - < 0.05) = 0.90 [ < is OK]
B1



0.05

=1.6449
0.2
M1 A1

n

2
2
1.6449 x 0.2
n =

dM1
2
0.05



n = 43.29...
A1


n = 44
A1

(6)
[11]




(a) 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




2
x (Accept 70 000 to 1sf) or (x- x)2 = 0.15(to 2 dp)


2nd A1ft for a correct expression with correct
2
x 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



(b)

B1 for a correct probability statement or "width of 90% CI = 0.05x 2 = 0.1"
1st B1 may
0.05
0.2
1st M1 for
= z value or 2x
x z = 0.1
be implied
0.2
n
by 1st A1
n
scored or
Condone 0.5 instead of 0.05 or missing 2 or 0.05 for 0.1 for M1
correct
1st A1 for a correct equation including 1.6449
equation.

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
8371_9374 GCE Mathematics January 2009
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Question
Number
Scheme Marks
Q8 (a)


E(4X -3Y )= 4E(X) - 3E(Y)
M1

= 4x30 - 3 x20


A1

= 60
(2)




(b) Var(4X -3Y) = 16 Var (X) + 9 Var (Y) 16 or 9; adding
M1; M1


= 16 x 9 + 9 x 4
A1
= 180
(3)
(c)

E(B) = 80
B1

Var (B) = 16
B1

E(B - A) = 20 E(B)-E(A)
M1


Var (B - A) = 196 ft on 180 and 16
A1ft




20
-


P (B - A >0) = P Z >
=

[P(Z > 1
- .428...)] stand. using their mean and var dM1

196

= 0.923 ... awrt 0.923 - 0.924
A1 (6)
[11]



(a) M1 for correct use of E(aX + bY) formula


(b) 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

(c) 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

8371_9374 GCE Mathematics January 2009
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