Exam 2015, questions and answers PDF

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UNIVERSITY OF LONDON

BSc EXAMINATION 2015

For Internal Students of Royal Holloway

MODEL ANSWERS EC1102: QUANTITATIVE METHODS IN ECONOMICS 1

Time Allowed: TWO hours Answer ALL questions EC calculators are permitted Distribution Tables and Formula sheet attached

c Royal Holloway, University of London 2015

Model Answer Copy

EC1102

SECTION A Answer ALL questions from this section.

QUESTION 1 Answer YES or NO to the following questions: (a) Is it true that

1 7

P5

i=2

i = 2?

(3 Marks)

Model Answer YES (b) You sell a project for $865, 000. Exactly one year ago you bought the project for $200, 000. Is it true that the annual rate of return on the project was 243%? (3 Marks)

Model Answer NO (c) Suppose f (x) = 4x2 + 89. Is it true that f ′ (x) = 8x?

(3 Marks)

Model Answer YES (d) Let f (K, L) = 10K 1/3 L2/3 . Is f (K, L) homogeneous of degree 1?

(3 Marks)

Model Answer YES

Marks so far: 15

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EC1102

(e) Does limx→0

ex −1 x

exist?

(3 Marks)

Model Answer YES (f) Is it true that

R

x5 dx = 14 x4 + C, where C is an arbitrary constant?

(3 Marks)

Model Answer NO (g) Suppose f (x, y) = αx3 + βx2 + γy 2 + ηy + θx2 y where α, β, γ , η and θ are constants. ∂ f (x, y) = αx2 + β + θy? (3 Marks) Is it true that ∂x

Model Answer NO (h) Suppose f (K) = 12K 0.5 . Is it true that ElK f (K) = 0.5?

(4 Marks)

Model Answer YES (Total 25 marks)

QUESTION 2 Answer YES or NO to the following questions: (a) The sum of the estimated residuals in any given regression analysis is zero. (4 Marks)

Marks so far: 29

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Model Answer YES (b) “If a population is uniformly distributed, the means of the samples of size 52 drawn randomly from this population are also uniformly distributed.” Is this statement true? (4 Marks)

Model Answer NO. It is (approximately) normally distributed as the CLT applies. (c) “For an exponential distribution, the mean and the variance are always equal.” Is this statement true? (4 Marks)

Model Answer NO (d) An exponential distribution, which is continuous, is closely related to the Poisson distribution, which is discrete. Is this statement true? (4 Marks)

Model Answer YES (e) Suppose the alternative hypothesis in a hypothesis test is: the population mean is smaller than 70. If the sample size is 80 and alpha = .01. Is it true that the critical value of z is -2.33? (4 Marks)

Model Answer YES (f) Output across firms is normally distributed with a mean of 78 and a variance of 25. Is it true that the z-value for an output of 90 is 0.48? (5 Marks)

Marks so far: 50

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EC1102

Model Answer NO, it is 90−5 78 = 2.4 (Total 25 marks)

SECTION B Answer ALL questions from this section.

QUESTION 3 A firm produces two different kinds A and B of a commodity. The daily cost of producing x units of A and y units of B is C(x, y) = 2x2 − 4xy + 4y 2 − 40x − 20y + 514. Suppose that the firm sells all its output at a price per unit of $24 for A and $12 for B . (a) Find the firm’s daily profit function

(2 Marks)

Model Answer π(x, y) = 24x + 12y − (2x2 − 4xy + 4y 2 − 40x − 20y + 514)

(b) Find the daily production levels x and y that maximize profit

(4 Marks)

Model Answer FOC: ∂ π(x, y) = 64 − 4x + 4y = 0 ∂x ∂ π(x, y) = 32 + 4x − 8y = 0. ∂y Marks so far: 56

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EC1102

Solving this system yields the unique solution x∗ = 40 y ∗ = 24. SOC: ∂2 π(x, y) = −4 ∂x2 ∂2 π(x, y) = −8 ∂y 2 ∂2 π(x, y) = 4 ∂x∂y The determinant of the Hessian matrix is (−4) · (−8) − 42 = 16 > 0. Hence, (x∗ , y ∗ ) = (40, 24) maximizes profit. (c) The firm is required to produce exactly 54 units per day of the two kinds combined. What now is the optimal production plan? (4 Marks)

Model Answer Under the constraint x + y = 54, the firm’s constrained profit function reads π e (x) = −10x2 + 680x − 10450 FOC:

d π e (x) = −20x + 680 = 0. dx Hence, the unique stationary point in the constrained profit function is x∗ = 34 y ∗ = 20.

SOC:

d2 π e (x) = −20 < 0. dx2 Hence, (x∗ , y ∗ ) = (34, 20) is the optimal production plan. Note: It is perfectly fine, albeit unnecessarily cumbersome in my opinion, to use the Lagrange multiplier method to answer this question. (Total 10 marks)

Marks so far: 60

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EC1102

QUESTION 4 A firm produces and sells two commodities. By selling x tons of the first commodity the firm gets a price per ton given by p = 96 − 4x. By selling y tons of the second commodity the price per ton given by q = 84 − 2y. The total cost of producing and selling x tons of the first commodity and y tons of the second is given by C(x, y) = 2x2 + 2xy + y 2 . (a) Show that the firm’s profit function is P (x, y) = −6x2 − 3y 2 − 2xy + 96x + 84y . (2 Marks)

Model Answer P (x, y) = px + qy − (2x2 + 2xy + y 2 ) = (96 − 4x)x + (84 − 2y)y − (2x2 + 2xy + y 2 ) = −6x2 − 3y 2 − 2xy + 96x + 84y

(b) Compute the first-order partial derivatives of P , and find its only stationary point. (4 Marks)

Model Answer FOC ∂ P (x, y) = −12x − 2y + 96 = 0 ∂x ∂ P (x, y) = −2x − 6y + 84 = 0. ∂x Solving this system yields the unique stationary point (x∗ , y ∗ ) = (6, 12).

Marks so far: 72

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(c) Suppose the firm’s production activity causes so much pollution that the authorities limit its output to 11 tons in total. Solve the firm’s constrained maximization problem in this case. Hint: Assume the constraint reads x + y = 11 and use the Lagrange multiplier method. Do not check second order conditions. (6 Marks)

Model Answer The Lagrangian is L(x, y) = −6x2 − 3y 2 − 2xy + 96x + 84y − λ(x + y − 11). FOC and the constraint reads ∂ L(x, y) = −12x − 2y + 96 − λ = 0 ∂x ∂ L(x, y) = −6y − 2x + 84 − λ = 0 ∂x x + y = 11. Solving this system of equations yields (x∗ , y ∗ , λ∗ ) = (4, 7, 34). (d) Does the production restriction impact the firm’s profit?

(3 Marks)

Model Answer Yes. λ∗ = 34 > 0, production restriction is binding and profit is reduced. (Total 15 marks)

QUESTION 5 On average, high school students are truant (i.e. missing school without reason) 5 days per year. (a) If truancy is uniformly distributed across all students between 0 and 10 days, what is the probability that a student is truant between 2 and 4 days? (4 Marks)

Model Answer P (2 < X < 4) =

Marks so far: 82

4−2 10−0

= 0.2

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(b) Empirically, would you expect truancy to be uniformly distributed? Discuss and make an alternative suggestion if applicable. (3 Marks)

Model Answer One would expect a small number of kids to be skipping school very often and the remainder to skip infrequently or potentially a few number of days a year, leading to either a normal distribution or a right skewed one. It is highly unlikely that it is uniformly distributed making skipping often and seldomly equally likely. (Total 7 marks)

QUESTION 6 Students at an excellent College in Surrey get better jobs than national graduates on average and thus earn higher salaries. The starting salary at the Surrey College is on average $30,000 with a standard deviation of $10,000. A histogram of the starting salaries reveals that the data are not normally distributed. (a) Apply Chebyshev’s theorem to find out within what interval of starting salaries at least 80% of the graduates’ starting salaries would fall. (4 Marks)

Model Answer Chebyshev’s theorem states that at least 1 − 1/k 2 % of the values are within the interval µ ± kσ. We are interested in the interval of starting salaries such that the proportion is 80%. We know µ and σ, so we search for k.√We know that 1 − 1/k 2 = 0.8, so solving for k yields: k 2 = 1/0.2 such that k = 5 = 2.24. The interval is thus: µ ± kσ = [7, 600; 52, 400]. Note: please give incremental points for stating the theorem, for computing k, and please only deduct minor points for incorrect rounding or incorrect computation of the square root of 5. (b) If you knew that starting salaries were normally distributed, would your estimate of the interval into which 80% of starting salaries fall change? Would it remain equal, become narrower or wider? Discuss briefly. (4 Marks)

Model Answer Marks so far: 90

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EC1102

The confidence interval would become narrower as we now have additional knowledge about the distribution of the random variable ‘starting salary’. You would substitute z for k where P r(0 < X < z) = 0.4 so that z ≈ 1.29. Hence, the confidence interval would be much narrower: [17, 100; 42, 900] (Total 8 marks)

QUESTION 7 A report in 2002 indicated that the prevalence of cigarette smoking among American adults was 21.1%. Data on smoking in a sample of n=3,536 participants indicated that 13.6% of the respondents were currently smoking. (a) Is there evidence of a statistically significant lower prevalence of smoking than indicated in the 2002 report on prevalence among all Americans? Make sure that you i) list the hypotheses, ii) compute the tests statistic and critical value, and iii) state your conclusion after the computation. Please choose a significance level of α = 0.05 (6 Marks)

Model Answer H0 : π ≥ 0.211 Ha : π < 0.211

Critical value: α = 0.05, one-tailed test: zc = −1.645 Test statistic: z = √ p−π = √ 0.211−0.136 = −10.93 π (1−π )/n

0.211·0.789/3536

Conclusion: we reject the null hypothesis as -10.93 falls into the rejection region (b) Which significance level would you need to choose in a) to be able to reject the Null? (4 Marks)

Model Answer p-value: Our distribution table stops at 3.99, so the P(z¡10.93)¡0.00003 (i.e. you can basically almost never reject this hypothesis) (Total 10 marks)

Marks so far: 100

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END OF QUESTIONS. PLEASE TURN OVER FOR THE DISTRIBUTION TABLES AND FORMULAS.

Total marks: 100

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

STUDENT’S t DISTRIBUTION

EC1102

END OF STATISTICAL TABLES AND FORMULA SHEET.

END

Total marks: 100

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