Past Exam Questions

Practice with real ECMT5001 and ECMT1020-style exam material. Work each question before revealing the answer.

ECMT5001 In-Semester Exam (2022 S1)

Question 1

[Module: M01 — Random Variables, Normal Distribution, Variance Rules]

Bob is the proud owner of the restaurant "Hungry Bob." The only product Hungry Bob sells is Bob's burger, which is priced at $10 each. The number of Bob's burgers sold on a day, denoted $N$, follows a normal distribution with mean 400 and standard deviation 50.

(a) What is the probability that the daily revenue exceeds $5,000$?

It is known that the total daily cost, denoted $C$, follows a normal distribution with mean $1,000$ and standard deviation $300$. The correlation between $C$ and $N$ is $0.8$. Let $P$ denote total daily profit.

(b) Express $P$ in terms of $C$ and $N$.

(c) Compute $E(P)$.

(d) Compute $Var(P)$.

Show Answer

(a) Daily revenue is $10N$. Revenue exceeds $5,000$ when:

\[10N > 5000 \Rightarrow N > 500\]

Standardise:

\[z = \frac{500 - 400}{50} = 2\]

Therefore:

\[P(N > 500) = P(Z > 2) \approx 0.0228\]

Answer: about 2.28%.

(b) Profit equals revenue minus cost:

\[P = 10N - C\]

(c) Use linearity of expectation:

\[E(P) = 10E(N) - E(C) = 10(400) - 1000 = 3000\]

Answer: expected daily profit is $3,000.

(d) Use the variance rule:

\[Var(aX+bY)=a^2Var(X)+b^2Var(Y)+2abCov(X,Y)\]

Here $P=10N-C$, so:

\[Var(P)=100Var(N)+Var(C)-20Cov(N,C)\]

First compute:

\[Var(N)=50^2=2500,\quad Var(C)=300^2=90000\]

\[Cov(N,C)=Corr(N,C)\sigma_N\sigma_C=0.8(50)(300)=12000\]

Therefore:

\[Var(P)=100(2500)+90000-20(12000)=100000\]

Answer: $Var(P)=100,000$. The standard deviation of profit is $\sqrt{100000}\approx316.23$.

Question 2

[Module: M01 — Binomial Distribution and Hypothesis Testing]

The government reported that the infection rate of COVID-19 is $0.2$. Let $Y$ denote the number of people infected with COVID-19 in a random sample of 5 individuals.

(a) What is the distribution of $Y$? Name the distribution and specify its parameter(s).

(b) Compute $P(Y>1)$.

(c) Compute $P(Y>2 \mid Y>1)$.

(d) Simon wanted to test whether the true infection rate is higher than $0.2$. He collected a random sample of 100 individuals. It was found that 27 individuals were infected with COVID-19. Carry out a hypothesis test for Simon at the 5 percent significance level. Show all steps.

Show Answer

(a) Each person is infected or not infected, with probability $p=0.2$, across $n=5$ people:

\[Y \sim Binomial(5,0.2)\]

(b) Use the complement:

\[P(Y>1)=1-P(Y=0)-P(Y=1)\]

\[P(Y=0)=0.8^5=0.32768\]

\[P(Y=1)=\binom{5}{1}(0.2)(0.8)^4=0.40960\]

\[P(Y>1)=1-0.32768-0.40960=0.26272\]

Answer: $0.2627$.

(c) Conditional probability:

\[P(Y>2 \mid Y>1)=\frac{P(Y>2)}{P(Y>1)}\]

Compute:

\[P(Y=2)=\binom{5}{2}(0.2)^2(0.8)^3=0.20480\]

\[P(Y>2)=1-0.32768-0.40960-0.20480=0.05792\]

Therefore:

\[P(Y>2 \mid Y>1)=\frac{0.05792}{0.26272}=0.2205\]

Answer: about 0.22.

(d) Let $p$ be the true infection rate.

\[H_0:p=0.2,\quad H_1:p>0.2\]

The sample proportion is:

\[\hat{p}=\frac{27}{100}=0.27\]

Under $H_0$, use the one-sample proportion test:

\[z=\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}\]

\[z=\frac{0.27-0.20}{\sqrt{0.2(0.8)/100}}=\frac{0.07}{0.04}=1.75\]

For a one-sided 5% test, the critical value is approximately $1.645$.

Since $1.75>1.645$, reject $H_0$.

Conclusion: there is evidence at the 5% significance level that the true infection rate is higher than 0.2.

Question 3

[Module: M01 — Discrete Distributions, Moments, Correlation, Independence]

Carol is a trader for an investment bank on Wall Street. She is studying the tick movement of a blue chip stock. Let $X$ denote the price change, measured in number of ticks. The probability density function of $X$ is:

$x$	$f(x)$
-2	0.02
-1	0.08
0	0.80
1	0.08
2	0.02

(a) Compute:

$E(X)$
$E(X^2)$
$sd(X)$

Let $S$ denote the sign of $X$:

\[ S= \begin{cases} -1 & \text{if } X \text{ is negative}\\ 0 & \text{if } X=0\\ 1 & \text{if } X \text{ is positive} \end{cases} \]

Let $|X|$ denote the absolute value of $X$.

(b) Compute:

$Corr(S,|X|)$
$Corr(S,X)$
Are $S$ and $|X|$ independent? Why or why not?

Show Answer

(a)(i) By symmetry:

\[E(X)=(-2)(0.02)+(-1)(0.08)+0(0.80)+1(0.08)+2(0.02)=0\]

(a)(ii)

\[E(X^2)=4(0.02)+1(0.08)+0+1(0.08)+4(0.02)=0.32\]

(a)(iii) Since $E(X)=0$:

\[Var(X)=E(X^2)-[E(X)]^2=0.32\]

\[sd(X)=\sqrt{0.32}=0.5657\]

(b)(i) $S$ indicates direction while $|X|$ indicates size. The distribution is symmetric, so positive and negative signs are equally likely at each non-zero magnitude. Therefore:

\[Cov(S,|X|)=0\]

and:

\[Corr(S,|X|)=0\]

(b)(ii) Because $SX=|X|$:

\[Cov(S,X)=E(SX)-E(S)E(X)=E(|X|)\]

\[E(|X|)=2(0.02)+1(0.08)+1(0.08)+2(0.02)=0.24\]

Also:

\[Var(S)=E(S^2)=0.02+0.08+0.08+0.02=0.20\]

\[sd(S)=\sqrt{0.20}=0.4472,\quad sd(X)=\sqrt{0.32}=0.5657\]

Therefore:

\[Corr(S,X)=\frac{0.24}{0.4472 \times 0.5657}=0.9487\]

(b)(iii) $S$ and $|X|$ are independent in this symmetric distribution. For example:

\[P(S=1)=0.10,\quad P(|X|=2)=0.04\]

\[P(S=1,|X|=2)=P(X=2)=0.02\]

But:

\[P(S=1)P(|X|=2)=0.10(0.04)=0.004\]

This does not match $0.02$, so by this check they are not independent. The zero correlation in part (b)(i) does not imply independence.

(Exam trap: zero correlation is not the same as independence.)

Key Exam Formulas (Quick Reference)

Topic	Formula
Expectation	$E(aX+bY)=aE(X)+bE(Y)$
Variance	$Var(aX+bY)=a^2Var(X)+b^2Var(Y)+2abCov(X,Y)$
Covariance	$Cov(X,Y)=E(XY)-E(X)E(Y)$
Correlation	$Corr(X,Y)=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}$
Binomial probability	$P(Y=y)=\binom{n}{y}p^y(1-p)^{n-y}$
Proportion test	$z=\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}$
Simple OLS slope	$\hat{\beta}_1=\frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2}$
Simple OLS intercept	$\hat{\beta}_0=\bar{y}-\hat{\beta}_1\bar{x}$
$R^2$	$R^2=1-\frac{SSR}{SST}$
Adjusted $R^2$	$\bar{R}^2=1-\frac{SSR/(n-k-1)}{SST/(n-1)}$
Omitted variable bias	$E(\tilde{\beta}_1)=\beta_1+\beta_2\tilde{\delta}_1$
Robust inference	Use heteroskedasticity-robust SEs for cross-sectional data
Serial correlation	For AR(1), $u_t=\rho u_{t-1}+e_t$
IV estimator	$\hat{\beta}^{IV}_1=\frac{Cov(z,y)}{Cov(z,x)}$

Topic	Formula
Expectation	\(E(aX+bY)=aE(X)+bE(Y)\)
Variance	\(Var(aX+bY)=a^2Var(X)+b^2Var(Y)+2abCov(X,Y)\)
Covariance	\(Cov(X,Y)=E(XY)-E(X)E(Y)\)
Correlation	\(Corr(X,Y)=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}\)
Binomial probability	\(P(Y=y)=\binom{n}{y}p^y(1-p)^{n-y}\)
Proportion test	\(z=\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}\)
Simple OLS slope	\(\hat{\beta}_1=\frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2}\)
Simple OLS intercept	\(\hat{\beta}_0=\bar{y}-\hat{\beta}_1\bar{x}\)
\(R^2\)	\(R^2=1-\frac{SSR}{SST}\)
Adjusted \(R^2\)	\(\bar{R}^2=1-\frac{SSR/(n-k-1)}{SST/(n-1)}\)
Omitted variable bias	\(E(\tilde{\beta}_1)=\beta_1+\beta_2\tilde{\delta}_1\)
Robust inference	Use heteroskedasticity-robust SEs for cross-sectional data
Serial correlation	For AR(1), \(u_t=\rho u_{t-1}+e_t\)
IV estimator	\(\hat{\beta}^{IV}_1=\frac{Cov(z,y)}{Cov(z,x)}\)