You invest €5 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 3 years.
PV = €5 0004.8% / 4n = 3
FV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 20 years at 5.4% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €200 0005.4% / 12N = 240
I = €
3.NPV & IRRMedium
A machine costs €80 000 today. It generates €11 500 at the end of each year for 7 years and can then be sold for €8 000. The cost of capital is 6%. Find the net present value NPV.
C0 = €80 000CF = €11 500S = €8 000n = 7r = 6%
NPV = €
4.NPV & IRRHard
A project costs €10 000 and returns €7 000, €7 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 7 000CF2 = 7 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 13 years. Its yield to maturity changes from 5.5% to 3.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 4%n = 13y1 = 5.5%y2 = 3.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 4.5% and matures in 3 years. The yield to maturity is 6%. Find the Macaulay duration D in years.
F = €1 000coupon = 4.5%n = 3YTM = 6%
D = years
7.CAPM & WACCMedium
A share with a beta of 0.9 has a required return of 5.95%. The risk-free rate is 1%. What expected market return rm does CAPM imply?
β = 0.9re = 5.95%rf = 1%
rm = %
8.CAPM & WACCHard
A company has equity with a market value of €15 000 000 and a cost of equity of 14%. Its debt consists of 8 000 bonds (face value €1 000, annual coupon 7%, 12 years to maturity) that trade at €925 each. The tax rate is 21%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
Find the sample standard deviation s of the data: 27, 19, 38, 44, 30.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 106?
μ = 100σ = 15x = 106
P =
11.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 9? The order does not matter.
n = 9k = 5
N =
12.ProbabilityHard
A game costs €2 to play. You win €10 with probability 0.15 and €40 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€10: p = 0.15€40: p = 0.01
E(G) = €
13.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29 and 39.5 min? Give P as a decimal.
μ = 32σ = 6x1 = 29x2 = 39.5
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 75% of them lie (the 75th percentile).
μ = 500σ = 40P = 75%
x = h
15.Confidence intervalsMedium
A sample of 11 items has mean 36.5 and standard deviation s = 15.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 11x̄ = 36.5s = 15.595%t(10) = 2.228
lower limit =
16.Confidence intervalsHard
The diameters of 30 machined shafts have a sample standard deviation of 4.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.
n = 30s = 4.6 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 40. A random sample of 100 observations has a mean of 255.44. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.
μ₀ = 250σ = 40n = 100x̄ = 255.44H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
In a trial, 105 of 150 patients on treatment A recover, and 160 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 105n₁ = 150x₂ = 160n₂ = 200
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 362Σx² = 91Σy² = 22154Σxy = 1337
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
n = 5x = 2.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 5769 €
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.
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.
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(25%) = 80.00
NPV(30%) = −473.37
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 25 + 5·80.00/(80.00 − (−473.37)) = 25.72%
exact: NPV(25.69%) = 0 → IRR = 25.69%
Answer
IRR = 25.69%
5.Bond pricing & yield
Answer: ΔP = 188.3 €
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.
Present value of each cash flow
t = 1: 45/1.06 = 42.45, t·PV = 42.45
t = 2: 45/1.124 = 40.05, t·PV = 80.10
t = 3: 1045/1.191 = 877.40, t·PV = 2 632.21
Duration
P = 959.90
D = 2 754.76/959.90 = 2.87
Answer
D = 2.87 years
7.CAPM & WACC
Answer: rm = 6.5 %
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).
Market risk premium
rm − rf = (5.95% − 1%)/0.9 = 5.5%
Market return
rm = 1% + 5.5% = 6.5%
Answer
rm = 6.5%
8.CAPM & WACC
Answer: WACC = 11.46 %
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446
Answer
P = 0.3446
11.Probability
Answer: N = 126
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(9, 5) = (9·8·7·6·5)/(5·4·3·2·1) = 126
Answer
N = 126
12.Probability
Answer: E(G) = −0.1 €
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.
Net outcomes
G: 8 (p = 0.15), 38 (p = 0.01), −2 (p = 0.84)
Expected value
E(G) = 8·0.15 + 38·0.01 − 2·0.84 = −0.1 €
Answer
E(G) = −0.1 €
A negative value means you lose money on average.
13.Normal distribution
Answer: P = 0.5858
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 = (29 − 32)/6 = −0.5
z₂ = (39.5 − 32)/6 = 1.25
Area between
P = Φ(1.25) − Φ(−0.5) = 0.8944 − 0.3085 = 0.5858
Answer
P = 0.5858
14.Normal distribution
Answer: x = 527 h
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 for the percentile
Φ(z) = 0.75 → z = 0.6745
Back to x
x = 500 + 0.6745·40 = 526.98 h
Answer
x = 526.98 h
15.Confidence intervals
Answer: lower limit = 26.09
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.
Critical value
df = n − 1 = 10, t = 2.228
Margin of error
E = 2.228·15.5/√11 = 10.41
Interval
36.5 ± 10.41 → [26.09, 46.91]
Answer
lower limit = 26.09
16.Confidence intervals
Answer: σ = 6.184 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 29·4.6² = 613.6
Limits for σ²
613.6/45.722 = 13.42
613.6/16.047 = 38.24
Limit for σ
σ = √38.24 = 6.184 µm
Answer
σ = 6.184 µm
95% interval for σ: [3.663, 6.184] µm.
17.Hypothesis tests & confidence intervals
Answer: p = 0.08692
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.
Test statistic
SE = σ/√n = 40/√100 = 4
z = (255.44 − 250)/4 = 1.36
Tail area
Φ(1.36) = 0.9131
1 − Φ(1.36) = 0.0869
p-value
p = 0.08692
Answer
p = 0.08692
p > 0.05, so H₀ is not rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: z = −2.159
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 = 105/150 = 0.7
p̂₂ = 160/200 = 0.8
p̂ = 265/350 = 0.7571
Standard error under H₀
SE = √(0.7571·0.2429·(1/150 + 1/200)) = 0.04632
Test statistic
z = (0.7 − 0.8)/0.04632 = −2.159
Answer
z = −2.159
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9453
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 = 1337 − 21·362/6 = 70
Sxx = 91 − 21²/6 = 17.5
Syy = 22154 − 362²/6 = 313.3
Correlation coefficient
r = 70/√(17.5·313.3) = 0.9453
Answer
r = 0.9453
20.Linear regression & correlation
Answer: ŷ = 37.05
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 = 20/5 = 4
ȳ = 210/5 = 42
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33
Regression line
b = 33/10 = 3.3
a = 42 − 3.3·4 = 28.8
ŷ = 28.8 + 3.3·x
Prediction
ŷ = 28.8 + 3.3·2.5 = 37.05
Answer
ŷ = 37.05
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