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OCR (non mei) S2 Wednesday 15th June 2016

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Reply 40
Xsanda's solutions

1) Unbiased mean and variance of summarised data
E(V) = 268
Var(V) = 1912/13 = 147 [4]

2) Normal distribution, given P(M<1.00) = 5%
P(M>1.05) = 13.6% [6]

3) Approximated binomial
n large, np = 26 > 5, nq = 14 > 5
use normal distribution
P(X<=30) = 0.932 [7]

4) Poisson, P(X=4) = P(X=5)
λ = 5, P(X=5) = 0.175 [5]

5) School with 55% girls
i) P(G≥6) = 22% (above significance level), do not reject H₀ [7]
ii) Fixed prob of girl chosen, independent of previous Head Students [1]

6) Cars along a narrow road
i) Constant probability of cars throughout the minute, cars independent [2]
ii) λ = 6.5, P(4≤X<7) = 0.561[3]
iiia) N(30,30)
P(X>35) = 0.158 [6]
b) Not independent (narrow, slow road, so affected by car in front) [1]

7) Probability density function
i) x is a single sample value of X [1]
ii) a=1, b=1.5 [7]
iii) E(X) = 7/4 = 1.75 [3]

8) Animals dying at zoo
i) P(L<12.48) = 4.88% < 5%
reject H₀, substantial evidence zoos limit life expectancy [7]
ii) = 12.5
5.05% > 5%, do not reject H₀, insufficient evidence zoos limit life expectancy [5]
iii) yes, as the distribution of the original data is not given [1]

9) Poisson test, λ=11
i) reject if P(R≥20)
92% chance of Type II error [6]
(edited 7 years ago)
i reckon 67 for 90ums?
Original post by xsanda
Xsanda's solutions

1) Unbiased mean and variance of summarised data
E(V) = 268
Var(V) = 1912/13 = 147 [4]

2) Normal distribution, given P(M<1.00) = 5%
P(M>1.05) = 13.6% [6]

3) Approximated binomial
n large, np = 26 > 5, nq = 14 > 5
use normal distribution
P(X<=30) = 0.932 [7]

4) Poisson, P(X=4) = P(X=5)
λ = 5, P(X=5) = 0.175 [5]

5) School with 55% girls
i) P(G≥6) = 22% > 5%, do not reject H₀ [7]
ii) Fixed prob of girl chosen, independent of previous Head Students [1]

6) Cars along a narrow road
i) Constant probability of cars throughout the minute, cars independent [2]
ii) λ = 6.5, P(4≤X<7) = 0.561[3]
iiia) N(30,30)
P(X>35) = 0.157 [6]
b) Not independent (narrow, slow road, so affected by car in front) [1]

7) Probability density function
i) x is a single sample value of X [1]
ii) a=1, b=1.5 [7]
iii) E(X) = 7/4 = 1.75 [3]

8) Animals dying at zoo
i) P(L<12.48) = 4.88% < 5%
reject H₀, substantial evidence zoos limit life expectancy [7]
ii) = 12.5
5.05% > 5%, do not reject H₀, insufficient evidence zoos limit life expectancy [5]
iii) yes, as the distribution of the original data is not given [1]

9i) Poisson test, λ=11
reject if P(R≥20)
92% chance of Type II error [6]


i got 0.1578 for 6(iii)a which rounds to 0.158
Reply 43
Original post by Sid1234
i got 0.1578 for 6(iii)a which rounds to 0.158


You're right. I forgot to do continuity correction
For 5i it was one tailed at 10% so the comparison would be 0.22>0.1
Reply 45
Hey for question 8 ii) i completely forgot that you could actually work out the population variance and instead stated the value of the population variance where the conclusion would change from Ho rejected to Ho accepted - do you reckon I still might be able to get some marks from that?
Can someone send me the thread for S2 OCR MEI 2016 please as I haven't seen nothing come up yet


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Reply 47
Didnt q9 say any incorrect conclusion which would mean probability type i error added to the proability of a type ii error? If not, how many marks would i lose for this? Thanks
Reply 48
Original post by xsanda
Xsanda's solutions

1) Unbiased mean and variance of summarised data
E(V) = 268
Var(V) = 1912/13 = 147 [4]

2) Normal distribution, given P(M<1.00) = 5%
P(M>1.05) = 13.6% [6]

3) Approximated binomial
n large, np = 26 > 5, nq = 14 > 5
use normal distribution
P(X<=30) = 0.932 [7]

4) Poisson, P(X=4) = P(X=5)
λ = 5, P(X=5) = 0.175 [5]

5) School with 55% girls
i) P(G≥6) = 22% (above significance level), do not reject H₀ [7]
ii) Fixed prob of girl chosen, independent of previous Head Students [1]

6) Cars along a narrow road
i) Constant probability of cars throughout the minute, cars independent [2]
ii) λ = 6.5, P(4≤X<7) = 0.561[3]
iiia) N(30,30)
P(X>35) = 0.158 [6]
b) Not independent (narrow, slow road, so affected by car in front) [1]

7) Probability density function
i) x is a single sample value of X [1]
ii) a=1, b=1.5 [7]
iii) E(X) = 7/4 = 1.75 [3]

8) Animals dying at zoo
i) P(L<12.48) = 4.88% < 5%
reject H₀, substantial evidence zoos limit life expectancy [7]
ii) = 12.5
5.05% > 5%, do not reject H₀, insufficient evidence zoos limit life expectancy [5]
iii) yes, as the distribution of the original data is not given [1]

9) Poisson test, λ=11
i) reject if P(R≥20)
92% chance of Type II error [6]


I put all of these, except i put for 6(i) Cars come at an average constant rate instead of 'constant probability of cars througout the minute', is this wrong?
Also for 5) it says to use 'exact binomial' - does this mean that you have to work out P(G>=6) with the exact binomial expansion? (nCr e.t.c) or can you use tables.
Let's Compile the Mark Scheme and Questions in a central place. This will allow corrections to be made easily and so that there is only one central copy.
Request Access as Usual. I will credit people as I look through the thread.

https://docs.google.com/document/d/1hhJTTOT3lltnT6KiwEfYfy5hFjzxqkFsuvfnAOP69D0/edit?usp=sharing
Original post by SCalver
Didnt q9 say any incorrect conclusion which would mean probability type i error added to the proability of a type ii error? If not, how many marks would i lose for this? Thanks


For type 1 error, lambda must not have changed. As it explicitly said that the lambda is actually 14, its just type 2 errors that youd be considering
Reply 51
Original post by SCalver
Didnt q9 say any incorrect conclusion which would mean probability type i error added to the proability of a type ii error? If not, how many marks would i lose for this? Thanks


You can only have a type 1 error if the Null Hypothesis is actually correct, since it was incorrect you could only have a type 2 error. You'd probably still get some marks though if you worked out the probability of the type 2 :smile:
Original post by xsanda
Xsanda's solutions

1) Unbiased mean and variance of summarised data
E(V) = 268
Var(V) = 1912/13 = 147 [4]

2) Normal distribution, given P(M<1.00) = 5%
P(M>1.05) = 13.6% [6]

3) Approximated binomial
n large, np = 26 > 5, nq = 14 > 5
use normal distribution
P(X<=30) = 0.932 [7]

4) Poisson, P(X=4) = P(X=5)
λ = 5, P(X=5) = 0.175 [5]

5) School with 55% girls
i) P(G≥6) = 22% (above significance level), do not reject H₀ [7]
ii) Fixed prob of girl chosen, independent of previous Head Students [1]

6) Cars along a narrow road
i) Constant probability of cars throughout the minute, cars independent [2]
ii) λ = 6.5, P(4≤X<7) = 0.561[3]
iiia) N(30,30)
P(X>35) = 0.158 [6]
b) Not independent (narrow, slow road, so affected by car in front) [1]

7) Probability density function
i) x is a single sample value of X [1]
ii) a=1, b=1.5 [7]
iii) E(X) = 7/4 = 1.75 [3]

8) Animals dying at zoo
i) P(L<12.48) = 4.88% < 5%
reject H₀, substantial evidence zoos limit life expectancy [7]
ii) = 12.5
5.05% > 5%, do not reject H₀, insufficient evidence zoos limit life expectancy [5]
iii) yes, as the distribution of the original data is not given [1]

9) Poisson test, λ=11
i) reject if P(R≥20)
92% chance of Type II error [6]


Great solutions, thanks for being so quick. I said for why large lambda would maybe be an invalid Poisson distribution was because it said that lambda was particularly high at a particularly time of day and therefore the 'constant average rate' property wouldn't hold. Is this incorrect? Otherwise my solutions are the same as yours!
(edited 7 years ago)
Reply 53
Original post by klipsy
I put all of these, except i put for 6(i) Cars come at an average constant rate instead of 'constant probability of cars througout the minute', is this wrong?
Also for 5) it says to use 'exact binomial' - does this mean that you have to work out P(G>=6) with the exact binomial expansion? (nCr e.t.c) or can you use tables.


I used my calculator’s 1-BinomialCD(5,8,0.55) function, but using the tables is fine. Average constant rate would probably be allowed.
Reply 54
Original post by duncanjgraham
Great solutions, thanks for being so quick. I said for why large lambda would maybe be an invalid Poisson distribution because it said that lambda was particularly high at a particularly time of time and therefore the 'constant average rate' property wouldn't hold. Is this incorrect? Otherwise my solutions are the same as yours!


The constant average rate only applies through the minute, so poisson is still possible at peak times.
I messed up 8 i and ii completely
For the first one I didn't divide by n
The second one I divided by n
So my conclusions and all that are wrong
Do you think I'd get any marks at all :frown:
Original post by xsanda
The constant average rate only applies through the minute, so poisson is still possible at peak times.


Aha, thank you!
Original post by Yazmin123
I messed up 8 i and ii completely
For the first one I didn't divide by n
The second one I divided by n
So my conclusions and all that are wrong
Do you think I'd get any marks at all :frown:

I did exactly the same thing! I'm just so annoyed cause it's worth so many. We should get about 3/14 I think for our original h0,h1 and then correct evaluation to your outcome
Why must the answer for q2 be as a percentage?
Original post by JoshuaHope
I did exactly the same thing! I'm just so annoyed cause it's worth so many. We should get about 3/14 I think for our original h0,h1 and then correct evaluation to your outcome


I did the same too

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