A credit card charges a nominal interest rate of 24% per year, compounded monthly. Find the effective annual rate EAR in percent.
r = 24%m = 12
EAR = %
2.Time value of moneyHard
You want €10 000 in 13 years. You save the same amount at the end of every month in an account paying 6% per year, compounded monthly. Find the monthly deposit A.
FV = €10 0006% / 12N = 156
A = €
3.NPV & IRRMedium
A machine costs €60 000 today, lasts 10 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.
C0 = €60 000M = €2 000n = 10r = 10%
EAC = €
4.NPV & IRRHard
Project A costs €10 000 and returns €5 500, €4 000, €2 500 at the end of years 1 to 3. Project B costs €12 000 and returns €3 000, €5 500, €6 000. Find the crossover rate r* at which both projects have the same NPV, in percent.
You buy a bond with 8 years to maturity (face value €1 000, annual coupon 6%) when its yield to maturity is 3%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 3.5%. Find your holding period return HPR in percent.
F = €1 000coupon = 6%n = 8y0 = 3%y1 = 3.5%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 9 years and sells for €1 080. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 9P = €1 080
YTM = %
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €3 000 000. Its WACC is 7.9%, its pre-tax cost of debt is 5% and the tax rate is 25.8%. Find the cost of equity re.
A company has 2 000 000 shares at €38.50 each and 5 000 bonds trading at €980 each. The pre-tax cost of debt is 4.5% and the tax rate is 25%. The share’s beta is 0.9, the risk-free rate 3% and the market return 10%. Find the WACC.
β = 0.9rf = 3%rm = 10%rd = 4.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 37, 14, 58, 27, 13, 28, 58.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 100. Find the z-score z of a score of 600.
μ = 500σ = 100x = 600
z =
11.ProbabilityMedium
A bag holds 5 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 8
P =
12.ProbabilityHard
X has the distribution shown, and Y = 10X + 10. Find the standard deviation σY of Y.
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 462 and 548 h? Give P as a decimal.
μ = 500σ = 40x1 = 462x2 = 548
P =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points80%
x = points
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 30 phones, mean 59 h, known standard deviation 19 h. Model 2: 30 phones, mean 47.5 h, known standard deviation 8 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 31. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 0.5?
σ = 31E = 0.590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 497.4 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 497.4s = 12n = 100
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 135 of 300 patients on treatment A recover, and 153 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 135n₁ = 300x₂ = 153n₂ = 300
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.
n = 6x = 9
ŷ =
3Solutions
1.Time value of money
Answer: EAR = 26.82 %
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Rate per period
r/m = 0.24/12 = 0.02
Compound over one year
EAR = (1 + 0.02)12 − 1 = 1.268 − 1 = 26.82%
Answer
EAR = 26.82%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: A = 42.47 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.1−10)/0.1 = 6.145
Annualised purchase price
60 000/6.145 = 9 764.72
EAC
EAC = 9 764.72 + 2 000 = €11 764.72
Answer
EAC = €11 764.72
4.NPV & IRR
Answer: r* = 5.034 %
Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.
Difference in cash flows
ΔCF0 = −12 000 − (−10 000) = −2 000
ΔCF1 = 3 000 − 5 500 = −2 500
ΔCF2 = 5 500 − 4 000 = 1 500
ΔCF3 = 6 000 − 2 500 = 3 500
IRR of the difference
NPVΔ(5%) = 3.02
NPVΔ(6%) = −84.83
r* = 5.034%
Answer
r* = 5.034%
Check: at 5.034% both NPVs are €1 019.71.
5.Bond pricing & yield
Answer: HPR = 0.1877 %
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Equation for the yield
€1 080 = €40·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹
Solve numerically
P(2%) = €1 163.24, P(3%) = €1 077.86
YTM = 2.974%
Answer
YTM = 2.974%
7.CAPM & WACC
Answer: re = 9.995 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Weights
V = €9 000 000, E/V = 0.6667, D/V = 0.3333
Debt part of the WACC
0.3333·5%·(1 − 0.258) = 1.237%
Cost of equity
re = (7.9% − 1.237%)/0.6667 = 9.995%
Answer
re = 9.995%
8.CAPM & WACC
Answer: WACC = 8.946 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = (600 − 500)/100 = 1
Answer
z = 1
11.Probability
Answer: P = 0.1282
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
First and second draw
P = 5/13 · 4/12 = 20/156 = 0.1282
Answer
P = 0.1282
12.Probability
Answer: σY = 120.3
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Spread of X
μ = 24.5, Var(X) = 144.8, σX = 12.03
Linear change
σY = |10|·12.03 = 120.3
Answer
σY = 120.3
The mean does move: E(Y) = 10·24.5 + 10 = 255.
13.Normal distribution
Answer: P = 0.7139
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (462 − 500)/40 = −0.95
z₂ = (548 − 500)/40 = 1.2
Area between
P = Φ(1.2) − Φ(−0.95) = 0.8849 − 0.1711 = 0.7139
Answer
P = 0.7139
14.Normal distribution
Answer: x = 56.6 points
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.2
z from the table
z ≈ −0.84
Back to the original units
x = 65 − 0.84·10 = 56.6 points
Answer
x = 56.6 points
15.Confidence intervals
Answer: upper limit = 17.69 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Point estimate
x̄₁ − x̄₂ = 59 − 47.5 = 11.5
Standard error
SE = √(19²/30 + 8²/30) = 3.764
Margin of error
E = 1.645·3.764 = 6.192
Interval
11.5 ± 6.192 → [5.308, 17.69]
Answer
upper limit = 17.69 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 10402
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Solve for n
n = (1.645·31/0.5)² = 10402
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 10402
Answer
n = 10402
17.Hypothesis tests & confidence intervals
Answer: t = −2.167
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Standard error
SE = 12/√100 = 12/10 = 1.2
Test statistic
t = (497.4 − 500)/1.2 = −2.167
Answer
t = −2.167
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −1.471
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Sample proportions
p̂₁ = 135/300 = 0.45
p̂₂ = 153/300 = 0.51
p̂ = 288/600 = 0.48
Standard error under H₀
SE = √(0.48·0.52·(1/300 + 1/300)) = 0.04079
Test statistic
z = (0.45 − 0.51)/0.04079 = −1.471
Answer
z = −1.471
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.966
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 35/5 = 7
ȳ = 371/5 = 74.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 150
Syy = Σ(y − ȳ)² = 602.8
Correlation coefficient
r = 150/√(40·602.8) = 0.966
Answer
r = 0.966
20.Linear regression & correlation
Answer: ŷ = 32.27
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 42/6 = 7
ȳ = 153/6 = 25.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 237
Regression line
b = 237/70 = 3.386
a = 25.5 − 3.386·7 = 1.8
ŷ = 1.8 + 3.386·x
Prediction
ŷ = 1.8 + 3.386·9 = 32.27
Answer
ŷ = 32.27
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