ECMT5001: Mid-semester Examination (2022s2)
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ECMT5001: Mid-semester Examination (2022s2)
1. [Total: 9 marks]
(a) [3 marks]
P(10N > 5000) = P(N > 500)
= P(Z > )
= P(Z > 2)
s 1 _ 0:9772
= 0:0228 s 0:02:
(b) [1 mark] P = 10N _ C .
(c) [2 marks]
E(P) = E(10N _ C)
= 10E(N) _ E(C)
= 10(400) _ (1000)
= 3000:
(d) [3 marks] First compute the covariance of N and C:
Cov(N;C) = Cor(N;C)sd(N)sd(C)
= (0:8)(50)(300)
= 12000:
Next compute the variance of P :
Var(P) = Var(10N _ C)
= Var(10N) + Var(C) _ 2Cov(10N;C) = 100Var(N) + Var(C) _ 2(10)Cov(N;C) = 100(50)2 + 3002 _ 2(10)(12000) = 250000 + 90000 _ 240000
= 100000;
so that sd(P) = ^100000 s 316:23:
2. [Total: 16 marks]
(a) [3 marks] Y has a binomial distribution with parameters n = 5 and p = 0:2, i.e.,
Y S Bin(n = 5;p = 0:2).
(b) [3 marks] We obtain the following binomial probabilities:
P(Y = 0) = (0:8)5 s 0:3277;
P(Y = 1) = / 、1(5) (0:2)(0:8)4 = 5(0:2)(0:8)4 s 0:4096:
It follows that
P(Y > 1) = 1 _ P(Y = 0) _ P(Y = 1)
s 1 _ 0:3277 _ 0:4096
s 0:2627 s 0:26: (1)
[Alternatively, student may obtain (1) by the following computation:
P(Y > 1) |
= s s |
P(Y = 2) + P(Y = 3) + P(Y = 4) + P(Y = 5) 0:2048 + 0:0512 + 0:00064 + 0:0003 0:2627 s 0:26:] |
(c) [4 marks] The required conditional probability is given by
P(Y > 2 and Y > 1)
Note that P(Y > 2 and Y > 1) = P(Y > 2). This is because {Y > 2} = {Y = 3; 4 or 5} and {Y > 1} = {Y = 2; 3; 4 or 5}, and their intersection is {Y > 2} = {Y = 3; 4 or 5}.
Also note that
P(Y = 2) = / 、2(5) (0:2)2 (0:8)3 = 10(0:2)2 (0:8)3 s 0:2048:
We therefore obtain
P(Y > 2 and Y > 1) = P(Y > 2)
= 1 _ P(Y = 0) _ P(Y = 1) _ P(Y = 2) = P(Y > 1) _ P(Y = 2)
s 0:2627 _ 0:2048
s 0:0579: (2)
Consequently, we have
P(Y > 2IY > 1) =
s
s
P(Y > 2 and Y > 1)
P(Y > 1)
0:0579
0:2627
0:2205 = 0:22:
[Alternatively, student may obtain (2) by the following computation:
P(Y > 2 and Y > 1) = P(Y > 2)
= P(Y = 3) + P(Y = 4) + P(Y = 5) s 0:0512 + 0:0064 + 0:0003 s 0:0579:]
(d) [6 marks] The null and alternative hypotheses are:
H0 : p = 0:2 vs Ha : p > 0:2:
Under H0, the z statistic z = approximately follows the N(0; 1) distribution
for large n (by the central limit theorem).
Rejection rule: reject H0 if z > z0.05 = 1:645.
Given the sample, we have p^ = = 0:27, and the realised z statistic is given by
z* = = = 1:75 > 1:645 = z0.05 :
We therefore reject H0 . We have strong enough evidence to conclude that the infection rate is signiÖcantly higher than 0.2 at the 5% level.
[Alternatively, student may use the p-value approach.
Rejection rule: reject H0 if P(z > z* ) < 0:05.
From the sample, the p-value is P(z > 1:75) s 0:0401 < 0:05.
Same conclusion.]
3. [Total: 15 marks]
(a) [6 marks]
i. [2 marks] E(X) = (_2)(0:02) + (_1)(0:08) + 0(0:8) + 1(0:08) + 2(0:02) = 0.
ii. [2 marks] E(X2) = (_2)2 (0:02) + (_1)2 (0:08) + 02 (0:8) + 12 (0:08) + 22 (0:02) = 0:32.
iii. [2 marks] Var(X) = E(X2) _ [E(X)]2 = 0:32. sd(X) =^Var(X) = ^0:32 s 0:5657 s 0:57.
(b) [9 marks]
i. [3 marks] The key is to note that S IXI = X . The covariance of S and IXI is given by
Cov(S;IXI) = E(S IXI) _ E(S)E(IXI)
= E(X) _ E(S)E(IXI):
But then
E(S) = (_1)(0:1) + 0(0:8) + 1(0:1) = 0:
It follows that Cov(S;IXI) = 0 and hence Cor(S;IXI) = = 0.
ii. [4 marks] The key is to note that SX = IXI. First, we evaluate
E(IXI) = 2(0:02 + 0:02) + 1(0:08 + 0:08) + 0(0:8) = 0:24:
The covariance of S and X is thus given by
Cov(S;X) = E(SX) _ E(S)E(X)
= E(IXI) _ E(S)E(X)
= 0:24 _ 0
= 0:24:
To compute the correlation, we need to Önd sd(S). The variance of S is given by
Var(S) = E(S2) _ [E(S)]2
= E(S2) _ 0 (from (b)(i))
= (_1)2 (0:1) + 02 (0:8) + 12 (0:1)
= 0:2,
so that sd(S) = ^0:2. The required correlation is computed as follows
Cor(S;X) =
=
s
s
Cov(S;X)
sd(S)sd(X)
0:24
^0:2^0:32
0:24
(0:4472)(0:5657)
0:9487 s 0:95:
iii. [2 marks] S and X are dependent because Cor(S;X) 0.
2022-09-12