Semester 2 2019 Midterm PDF

Title Semester 2 2019 Midterm
Course Principles of Econometrics
Institution University of Sydney
Pages 3
File Size 66.5 KB
File Type PDF
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Summary

Semester 2 2019 Midterm Exam...


Description

ECMT5001 Midterm

Practice

The University of Sydney Principles of Econometrics ECMT 5001 Mid-semester Test Practice

Name: SID:

Instructions: Total marks: Time permitted: Permitted materials:

40 90 minutes Calculators are permitted. Mobile phones are NOT permitted. Answer all questions in the booklet provided. Marks for each question are in parentheses (). Show all workings and provide explanations for your answers. State any assumptions that you need to make. Do not begin the test until instructed. Good luck!

ECMT5001 Midterm

Practice

1. The table below describes the joint distribution of phone ownership by brand for households in the town of Lottering. Rows refer to the number of Samsung phones owned and columns refer to the number of iPhones owned.

Samsung 0 1 2

0 0.1 0.1 0.2

iPhone 1 2 0.1 0.3 0.05 0.0 0.05 0.1

Answer the following questions related to the table. (a) What is the probability that a random household has 1 iPhone and 1 Samsung phone?

[1 mark]

(b) What is the probability that a random household has no Samsung phones and at least one iPhone? [2 marks] (c) What is the probability that a randomly selected household has 3 phones in total?

[2 marks]

(d) If a household has 1 Samsung phone, what is the probability that they have no iPhones?

[2 marks]

(e) If a household has 3 phones in total, what is the probability that they have 2 iPhones?

[2 marks]

(f) Calculate the mean and variance of the number of iPhones per household.

[4 marks]

(g) Let Y be a random variable describing the number of iPhones per household. Suppose two households are sampled at random. i. Find the expected value of the sample variance of Y , E(s2 ). ii. Find the variance of the sample variance, Var(s2 ). 2.

[2 marks] [5 marks]

(a) The random variable X is uniformly distributed with minimum value 0 and maximum value 5. Let x be the outcome of a random draw from X . Find the probability that 1 ≤ x ≤ 3. [2 marks]

(b) The random variable X is normally distributed with mean µ and variance σ2 . i. Let Y = X − 2µ. Find the distribution of Y . ii. The random variable W is defined as follows:

[3 marks]

W = a + bX , where a and b are constants, and X is the sample mean from a random sample of size n from the random variable X . Find the distribution of the random variable W . [3 marks] (c) Suppose that hourly wages in an industry are normally distributed with mean $25 and variance $225. i. Find the probability that a randomly selected person has hourly wages greater than $40. [3 marks] ii. 25 people are selected at random. Find the probability that average wages for these 25 people are less than $16. [3 marks] (d) Katie estimates a regression of the relationship between hourly wages and years of education for a random sample of 300 employees. She obtains the following results: b Y = 10 + 0.7X ,

where Y is the hourly wage in dollars, and X is the number of years of education. i. Carefully interpret Katie’s regression results. ii. Let Z = 2Y + 5. If Katie estimated the regression model

[3 marks]

Z = β0 + β1 X + u, what estimates would she obtain for β0 and β1 ? Explain. END OF THE TEST (Please see the last page for useful formulae and tables)

[3 marks]

ECMT5001 Midterm

Practice

Formulae and Tables Probability

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) P(A ∩ B) P(A|B) = P (B ) Distributions

Binomial : f (x; p, n) = px (1 − p)n−x Poisson : f (x; µ) =

µx e−µ x!

n! x!(n − x)!

Sample distributions x¯ − µ √ ∼ N(0, 1), σ/ n

x¯ − µ √ ∼ tn−1 , s/ n

D¯ − µ q 2 d 2 ∼ N(0, 1), s1 s2 n1 + n2

(n − 1)s2 ∼ χ2n−1 , σ2

p−π q ∼ N(0, 1) π(1−π) n 2 s1 /s22 σ21 /σ22

∼ Fn1 −1,n2 −1

Linear regression ∑n (xi − x)(yi − y) βˆ 1 = i=1 n , ∑ i=1 (xi − x)2

βˆ 0 = y − βˆ 1 x

Mini normal table P(Z < 1) = 0.841, P(Z < 2) = 0.977, P(Z < 3) = 0.9987, P(Z < 4) = 0.99997, P(Z < 1.645) = 0.95, P(Z < 1.96) = 0.975, P(Z < 2.326) = 0.99, P(Z < 2.576) = 0.995, where Z is the standard normal random variable, Z ∼ N(0, 1)....


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