A credit card charges a nominal interest rate of 20% per year, compounded monthly. Find the effective annual rate EAR in percent.
r = 20%m = 12
EAR = %
2.Time value of moneyHard
You borrow €8 000 at 3.6% per year, compounded monthly, and repay €350 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 = €8 000A = €3503.6% / 12
N = payments
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 3: €4 000, €9 000, €4 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €50 000 and returns €30 000, €31 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 30 000CF2 = 31 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 14 years and pays annual coupons. It sells for €907 when the yield to maturity is 6%. Find the coupon rate c in percent.
F = €1 000n = 14P = €907YTM = 6%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 10 years. It sells for €1 327. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 8% (2×)n = 10P = €1 327
YTM = %
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €2 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 20%. Find the WACC.
A company has 2 000 000 shares at €52.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 5% and the tax rate is 25%. The share’s beta is 1.1, the risk-free rate 2% and the market return 6%. Find the WACC.
β = 1.1rf = 2%rm = 6%rd = 5%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 36, 14, 56, 23, 26, 25, 44.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 77.2.
μ = 70σ = 12x = 77.2
z =
11.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 10? The order does not matter.
n = 10k = 5
N =
12.ProbabilityHard
A game costs €3 to play. You win €12 with probability 0.15 and €30 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€12: p = 0.15€30: p = 0.05
E(G) = €
13.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 4 of them exceeds 47.2 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 4> 47.2 kg
P =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 95% of them lie (the 95th percentile).
μ = 100σ = 15P = 95%
x = points
15.Confidence intervalsMedium
The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.1 and standard deviation 1.5. Find the upper limit of the 90% confidence interval for the mean difference.
n = 12d̄ = 8.1sd = 1.590%t(11) = 1.796
upper limit = mmHg
16.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.04?
p = 0.5E = 0.0499%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 120; the population standard deviation is known to be 10. A random sample of 25 observations has a mean of 124.28. Test H₀: μ = 120 against H₁: μ ≠ 120. Find the p-value p.
μ₀ = 120σ = 10n = 25x̄ = 124.28H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 9. In reality the mean is 55.85. Find the probability of a Type II error β in percent.
μ₀ = 50μ = 55.85σ = 9n = 36α = 0.01
β = %
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 15) is made. Find its residual e = y − ŷ.
n = 5(3.5, 15)
e =
3Solutions
1.Time value of money
Answer: EAR = 21.94 %
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 = 0.2/12 = 0.01667
Compound over one year
EAR = (1 + 0.01667)12 − 1 = 1.219 − 1 = 21.94%
Answer
EAR = 21.94%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: N = 24 payments
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 − 8 000·0.003/350 = 0.9314
Solve with logarithms
N = −ln(0.9314)/ln(1.003) = 23.71
Round up to whole payments
23 full payments leave a small balance, which the last, smaller payment clears.
N = 24
Answer
N = 24 payments
3.NPV & IRR
Answer: NPV = −5920 €
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.
Present value of each cash flow
CF0 = −20 000
CF1/(1 + 0.1)1 = 4 000/1.1 = 3 636.36
CF2/(1 + 0.1)2 = 9 000/1.21 = 7 438.02
CF3/(1 + 0.1)3 = 4 000/1.331 = 3 005.26
Add them up
NPV = €−5 920.36
Answer
NPV = €−5 920.36
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 14.85 %
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(10%) = 3 305.79
NPV(15%) = −94.52
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.
Present value of the face value
1000·(1 + 0.06)−14 = €442.30
Coupon
C = (907 − 442.30)/9.295 = €49.99
Coupon rate
c = 49.99/1000 = 4.999%
Answer
c = 4.999%
6.Bond pricing & yield
Answer: YTM = 4 %
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.
Per period
C = €40.00, N = 20
Solve numerically
i = 2% per half-year
YTM
YTM = 2·2% = 4%
Answer
YTM = 4%
7.CAPM & WACC
Answer: WACC = 8.933 %
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).
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 = (77.2 − 70)/12 = 0.6
Answer
z = 0.6
11.Probability
Answer: N = 252
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(10, 5) = (10·9·8·7·6)/(5·4·3·2·1) = 252
Answer
N = 252
12.Probability
Answer: E(G) = 0.3 €
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.
Net outcomes
G: 9 (p = 0.15), 27 (p = 0.05), −3 (p = 0.8)
Expected value
E(G) = 9·0.15 + 27·0.05 − 3·0.8 = 0.3 €
Answer
E(G) = 0.3 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.6915
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 total
μT = 48 kg, σT = 0.8·√4 = 1.6 kg
Standardise
z = (47.2 − 48)/1.6 = −0.5
Probability
P = 1 − Φ(−0.5) = 0.6915
Answer
P = 0.6915
14.Normal distribution
Answer: x = 124.7 points
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) = 0.95 → z = 1.645
Back to x
x = 100 + 1.645·15 = 124.67 points
Answer
x = 124.67 points
15.Confidence intervals
Answer: upper limit = 8.878 mmHg
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.
Critical value
df = 11, t = 1.796
Margin of error
E = 1.796·1.5/√12 = 0.7777
Interval
8.1 ± 0.7777 → [7.322, 8.878]
Answer
upper limit = 8.878 mmHg
16.Confidence intervals
Answer: n = 1037
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 2.576²·0.5·0.5/0.04² = 1037
Round up
n = 1037
Answer
n = 1037
17.Hypothesis tests & confidence intervals
Answer: p = 0.03235
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 = 10/√25 = 2
z = (124.28 − 120)/2 = 2.14
Tail area
Φ(2.14) = 0.9838
1 − Φ(2.14) = 0.0162
p-value
p = 2·0.0162 = 0.03235
Answer
p = 0.03235
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: β = 5.774 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 9/√36 = 1.5
x̄c = 50 + 2.326·1.5 = 53.49
z under the true mean
z = (53.49 − 55.85)/1.5 = −1.574
Type II error
power = P(Z > −1.574) = 94.23%
β = 100% − 94.23% = 5.774%
Answer
β = 5.774%
19.Linear regression & correlation
Answer: r² = 0.9828
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 = 30/5 = 6
ȳ = 49/5 = 9.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80
Syy = Σ(y − ȳ)² = 162.8
Coefficient of determination
r² = 80²/(40·162.8) = 0.9828
Answer
r² = 0.9828
About 98.28% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = −4.2
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 = 15/5 = 3
ȳ = 100/5 = 20
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −16
Regression line
b = −16/10 = −1.6
a = 20 − (−1.6)·3 = 24.8
ŷ = 24.8 − 1.6·x
Predicted value
ŷ = 24.8 + (−1.6)·3.5 = 19.2
Residual
e = 15 − 19.2 = −4.2
Answer
e = −4.2
The point lies below the line.
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