You invest €3 000 at a nominal rate of 2% per year, compounded monthly. Find the value FV of the investment after 8 years.
PV = €3 0002% / 12n = 8
FV = €
2.Time value of moneyHard
You borrow €15 000 at 3.6% per year, compounded monthly, and repay €450 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
P = €15 000A = €4503.6% / 12
N = payments
3.NPV & IRRMedium
A machine costs €20 000 today, lasts 6 years and costs €1 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.
C0 = €20 000M = €1 000n = 6r = 12%
EAC = €
4.NPV & IRRHard
A project costs €25 000 and returns €16 000, €15 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 16 000CF2 = 15 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 9 years. Its yield to maturity changes from 8% to 7.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 5%n = 9y1 = 8%y2 = 7.5%
ΔP = €
6.Bond pricing & yieldHard
A bond trades at €920 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 7.8 years. Use the modified duration to estimate the new price P if the yield falls by 75 basis points.
P0 = €920y = 6%D = 7.8Δy = −75 bp
P = €
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €10 000 000, preferred shares worth €1 000 000 and debt worth €2 000 000. The cost of equity is 9%, the cost of preferred stock 7.5% and the pre-tax cost of debt 4.5%. The tax rate is 20%. Find the WACC.
A company's business has an unlevered (asset) beta of 0.9. It plans a debt-to-equity ratio of 1. The tax rate is 25.8%, the risk-free rate is 3.5% and the market risk premium is 5%. Find the cost of equity re at the new capital structure.
βU = 0.9D/E = 1T = 25.8%rf = 3.5%rm − rf = 5%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 11, 46, 45, 52, 42, 22, 22.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is below 66?
μ = 60σ = 15x = 66
P =
11.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 15? The order does not matter.
n = 15k = 2
N =
12.ProbabilityHard
A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 4 correct answers?
n = 10p = 0.2k = 4
P =
13.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 34 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 34 min
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm60%
x = cm
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 80 phones, mean 94 h, known standard deviation 12 h. Model 2: 50 phones, mean 104 h, known standard deviation 17 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 25. 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 3?
σ = 25E = 390%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 78 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 78s = 15n = 36
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 5. Find the smallest sample size n.
σ = 5E = 2.595%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 104Σx² = 55Σy² = 2210Σxy = 292
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
n = 6x = 2.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 3520 €
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.
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.
Rearrange the annuity formula
i = 0.003
(1 + i)−N = 1 − 15 000·0.003/450 = 0.9
Solve with logarithms
N = −ln(0.9)/ln(1.003) = 35.17
Round up to whole payments
35 full payments leave a small balance, which the last, smaller payment clears.
N = 36
Answer
N = 36 payments
3.NPV & IRR
Answer: EAC = 5865 €
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 − 1.12−6)/0.12 = 4.111
Annualised purchase price
20 000/4.111 = 4 864.51
EAC
EAC = 4 864.51 + 1 000 = €5 864.51
Answer
EAC = €5 864.51
4.NPV & IRR
Answer: IRR = 15.81 %
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(15%) = 255.20
NPV(20%) = −1 250.00
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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.
Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.
Modified duration
Dmod = 7.8/1.06 = 7.358
Estimated price change
%ΔP ≈ −7.358·(−0.0075) = 5.519%
P ≈ 920·(1 + 0.05519) = €970.77
Answer
P = €970.77
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 8.054 %
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).
Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.
Levered beta
βL = 0.9·(1 + 0.742·1) = 1.568
Cost of equity (CAPM)
re = 3.5% + 1.568·5% = 11.34%
Answer
re = 11.34%
9.Descriptive statistics
Answer: s = 15.65
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 0.6554
Answer
P = 0.6554
11.Probability
Answer: N = 105
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.
Combinations
C(15, 2) = (15·14)/(2·1) = 105
Answer
N = 105
12.Probability
Answer: P = 0.08808
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.
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (34 − 32)/2.5 = 0.8
Upper tail
P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119
Answer
P = 0.2119
14.Normal distribution
Answer: x = 176.3 cm
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) = 0.4
z from the table
z ≈ −0.25
Back to the original units
x = 178 − 0.25·7 = 176.3 cm
Answer
x = 176.3 cm
15.Confidence intervals
Answer: upper limit = −2.908 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 = 94 − 104 = −10
Standard error
SE = √(12²/80 + 17²/50) = 2.753
Margin of error
E = 2.576·2.753 = 7.092
Interval
−10 ± 7.092 → [−17.09, −2.908]
Answer
upper limit = −2.908 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 188
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 = (1.645·25/3)² = 187.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 188
Answer
n = 188
17.Hypothesis tests & confidence intervals
Answer: t = 1.2
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 = 15/√36 = 15/6 = 2.5
Test statistic
t = (78 − 75)/2.5 = 1.2
Answer
t = 1.2
t0.025; df = 35 = 2.03
|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 16
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).
Solve for n
n = (1.96·5/2.5)² = 15.37
Round up
n = 16
Answer
n = 16
19.Linear regression & correlation
Answer: r = −0.9245
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.
Centred sums
Sxy = 292 − 15·104/5 = −20
Sxx = 55 − 15²/5 = 10
Syy = 2210 − 104²/5 = 46.8
Correlation coefficient
r = −20/√(10·46.8) = −0.9245
Answer
r = −0.9245
20.Linear regression & correlation
Answer: ŷ = 19.07
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 = 21/6 = 3.5
ȳ = 100/6 = 16.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −42
Regression line
b = −42/17.5 = −2.4
a = 16.67 − (−2.4)·3.5 = 25.07
ŷ = 25.07 − 2.4·x
Prediction
ŷ = 25.07 + (−2.4)·2.5 = 19.07
Answer
ŷ = 19.07
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