Sample exam answers

Sample exam answers

Q1) a) 38 + 12 = 208

12 = 170

= 14.16

-0.5t = 2.65

t = -5.30

b) - = 0

= 0

= 1

Using quadratic formula or factorization method,

x = 1 or 2

c) + -

=

= 0

Q2) and

a) At equilibrium = = or = = Q

=

= 28

Q = 8

P = 50 12 = 38

b) when tax is imposed, we assume that the supplier will be affected directly (and the consumer indirectly).

New supply function:

Solving this with the demand function, new equilibrium values are Q = 3.7 and P = 44.45.

In order to study how tax is distributed among consumer and supplier, we make the following assumptions:

1. Consumer pays the equilibrium price = $44.452. Supplier receives equilibrium price tax = $29.4.

If you compare these with part (a) answers,

the consumer now pays $6.45 more and the supplier pays $8.60 less. This is how the tax is distributed between the consumer and the supplier.

Q3) TC =

MC =

First order condition:

= 2Q 18 = 0

Therefore Q = 9 is a turning point.

= 2 > 0

Therefore Q = 9 minimizes the MC function

Q4) a) 3

b) Overall significance testing is expected for this part.

H0: = 0 against Ha: at least one of the parameters 0.

F test value = 3.557 with p-value = 0.040.

Since p-value (0.04) < 0.05, we reject H0 at the 5% level of significance. Therefore the model seems reliable overall.

c) Conflicting t test and F test results indicate the presence of multi-collinearity. In order to overcome multi-collinearity problem,

1. Find out the independent variables which are highly correlated with each other.Retain the variable which is highly correlated with the dependent variable and discard the other.

2. Model selection techniques such as Forward Selection also helps to avoid this problem.

d) First-order autocorrelation is defined by = ; where is an independent error term.

against

Decision rule: Reject if DW test statistic <

Accept if DW test statistic >

Inconclusive if < DW <

Conclusion: Since DW test statistic is < , we reject and accept . Thus positive autocorrelation exists.

Violated assumption: errors of the model are independent.

Methods to overcome this problem: First difference approach, auto regression and adding more independent regressors.e) Regress

1) y on and

2) y on and

3) y on and

Tabulate t and p values of , and and the corresponding adj. values. If none of these predictors are significant then is the only best predictor. Otherwise retain the best predictor and proceed as above.

Q5)

a)

b)

d)

c)

Q6) Plant I: TC = 10 + 21Q

TR =

= TR TC =

First-order condition:

= 0

-6Q + 60 = 0

Q = 10 is a turning point

Second derivative:

= -6 < 0

Therefore Q = 10 maximises profit.

Max = = $290

Repeat the same for Plant II. You can either use MR = MC approach or workout the TC via integration.

TC =

= TR TC =

Solving first-order conditions will give turning points Q = 3 or 33

(Second derivative) = -2Q + 36

When Q = 3, = 30 > 0, therefore minimum

When Q= 33, = -30, therefore maximum.

However when Q = 33, P = 81 3(33) = -18 < 0. Therefore Plant II is not feasible. Only Plant I is reliable.Q7)

a) First-derivatives:

1] =

2] =

Second-derivatives:

3] =

4] =

5] = = +

1] 4p = 100 q

p = 25 0.25q

2] = TR TC

= 25q 0.25 -

= - 0.25-

= + ---

First-order conditions:

= 23 0.5- 0.5 = 0

(1)

= 25-- = 0

(2)

= = 1, = = 42 and = = 4

Therefore = 42 and = 4 are turning points.

In order to decide the nature of the turning points, consider second derivatives:

1] = -0.5 < 0

2] = -1 < 0

3] = -0.5 Therefore

EMBED Equation.DSMT4 >

All conditions for a maximum hold. Therefore = 42 and = 4 maximise profit.

Q8) a) - + = 12

b) Only BA is possible.

BA =

b) Write down the system of equations into matrix form

EMBED Equation.DSMT4 = ,where A = is known as Left Hand Side (LHS) coefficients matrix and is known as the Right Hand Side (RHS) values.Consider

Note: L.H.S coefficients matrix and R.H.S values are enough to answer Gaussian or Gaussian-Jordan elimination process. An identity matrix is included here for the purpose of finding the inverse of L.H.S coefficients matrix, A. Step 1: Transform LHS coefficients matrix as an upper triangular matrix (that is, along the diagonal and upwards you can have elements, rest should be zeros).

New Eq3 = Eq3 + Eq1

New Eq3 = Eq3 Eq2

Note: If Gaussian elimination process was asked above upper triangular version is enough to answer that question.Step 2: Gaussian-Jordan process is an extension of Gaussian process. This process requires you to transform the above upper triangular matrix into an identity matrix (that is, diagonal elements are ones, rest zeros).New Eq3 = Eq3/2

New Eq3 = Eq3 Eq2

NewEq1 = Eq1 + Eq2

New Eq1 = Eq1 2*Eq3

New Eq2 = Eq2 Eq3

(*)LHS coefficients matrix is now transformed into an identity matrix. Therefore we stop the elimination process and re-write the system of equations as before:

EMBED Equation.DSMT4 =

a) Matrix multiplication leads to x = -27, y = -13 and z = 14

Note: You do not have to follow the same matrix transformations/operations explained above. Any suitable transformations are acceptable.

b) Inverse matrix method:

EMBED Equation.DSMT4 =

Let A = , = and b =

Above is known as A = b format.

Multiplying by both sides leads to

Therefore (reason: )Thus = (reason: I plays the role of 1 in scalar multiplication).

Hence inverse matrix method amounts to finding the unknowns by multiplying by b. Consider the above step indicated by (*). When LHS coefficients matrix becomes an identity matrix (at this stage) the far right side matrix is the required inverse matrix, . That is,

= .Thus the solution can be obtained by multiplying and .

EMBED Equation.DSMT4 = .Q9)

a) = +cb) = = + + cc) At the equilibrium,

=

Divide LHS by RHS (or vice-versa).

= 1

= 1

= 40

=

0.4q = 3.69

q = 9.22

Therefore p = 19.94

CS =

= - 183.898= -79.8 + 800-183.898 = 536.34

Q10)

Arithmetic series will have common difference d between the numbers.

That is,

2 + d = x

and

x + d = 15.

Thus 2+d+d = 15

2d = 13

d = 6.5

Therefore x = 2+6.5 = 8.5.

Geometric series will have common multiple r between numbers:

That is,

2r = xand

xr = 15

Therefore = 15

= 7.5

r = 2.74

Thus x = 5.48

b) Insurance is involved Future value formula should be used.Regular payments are involved annuity.

Payment period = compounding period ordinary annuity.

Payment is made at the start of the payment period ordinary annuity due.

Given all these, the correct formula to use is future value ordinary annuity due formula:

i = = 0.0054 and n = 84

= 350(1.0054) = $37278_1372526804.unknown

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