A credit card charges a nominal interest rate of 22% per year, compounded monthly. Find the effective annual rate EAR in percent.
r = 22%m = 12
EAR = %
2.Time value of moneyHard
You save €50 at the end of every month for 17 years. The account pays 7.2% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €507.2% / 12N = 204
FV = €
3.NPV & IRRMedium
A machine costs €50 000 today. It generates €11 000 at the end of each year for 5 years and can then be sold for €5 000. The cost of capital is 10%. Find the net present value NPV.
C0 = €50 000CF = €11 000S = €5 000n = 5r = 10%
NPV = €
4.NPV & IRRHard
A project costs €25 000 and returns €13 000, €12 000, €10 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 12 years. Its yield to maturity changes from 9% to 8.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 6%n = 12y1 = 9%y2 = 8.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 6 years. The yield to maturity is 7%. Find the Macaulay duration D in years.
F = €1 000coupon = 3%n = 6YTM = 7%
D = years
7.CAPM & WACCMedium
A share with a beta of 2 has a required return of 13%. The risk-free rate is 4%. What expected market return rm does CAPM imply?
β = 2re = 13%rf = 4%
rm = %
8.CAPM & WACCHard
A company has 1 000 000 shares at €18.00 each and 40 000 bonds trading at €1 020 each. The pre-tax cost of debt is 4.5% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 3.5% and the market return 8%. Find the WACC.
β = 0.8rf = 3.5%rm = 8%rd = 4.5%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 26, 47, 18, 35, 49.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 620.
μ = 500σ = 60x = 620
z =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 2, 4, 6p(x): 0.15, 0.25, 0.3, 0.3
σ =
12.ProbabilityHard
9 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 9p = 0.1
P =
13.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 759.8 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 759.8 g
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 20% of them lie (the 20th percentile).
μ = 500σ = 40P = 20%
x = h
15.Confidence intervalsMedium
A sample of 16 items has mean 17.5 and standard deviation s = 17.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 16x̄ = 17.5s = 17.595%t(15) = 2.131
E =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 15 plots, mean yield 88 kg, standard deviation 4.5 kg. Plot group 2: 7 plots, mean 84.5 kg, standard deviation 3 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.
A sample of 100 students has a mean score of 202.5 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 202.5s = 12n = 100
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 18 of 120 patients on treatment A recover, and 50 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 18n₁ = 120x₂ = 50n₂ = 250
z =
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 = 107Σx² = 55Σy² = 2355Σxy = 298
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. For which value x does the line predict ŷ = 69?
n = 5ŷ = 69
x =
3Solutions
1.Time value of money
Answer: EAR = 24.36 %
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 = 0.22/12 = 0.01833
Compound over one year
EAR = (1 + 0.01833)12 − 1 = 1.244 − 1 = 24.36%
Answer
EAR = 24.36%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: FV = 19903 €
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.
You paid in €10 200 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = −5197 €
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 the yearly cash flows
11 000·3.791 = 41 698.65
Present value of the salvage value
5 000/1.611 = 3 104.61
NPV
NPV = −50 000 + 41 698.65 + 3 104.61 = −€5 196.74
Answer
NPV = −€5 196.74
4.NPV & IRR
Answer: MIRR = 15.1 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
13 000·1.082 = 15 163.20
12 000·1.081 = 12 960.00
10 000·1.080 = 10 000.00
TV = 38 123.20
MIRR
MIRR = (38 123.20/25 000)1/3 − 1 = 15.1%
Answer
MIRR = 15.1%
5.Bond pricing & yield
Answer: ΔP = 31.2 €
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: 30/1.07 = 28.04, t·PV = 28.04
t = 2: 30/1.145 = 26.20, t·PV = 52.41
t = 3: 30/1.225 = 24.49, t·PV = 73.47
t = 4: 30/1.311 = 22.89, t·PV = 91.55
t = 5: 30/1.403 = 21.39, t·PV = 106.95
t = 6: 1030/1.501 = 686.33, t·PV = 4 117.99
Duration
P = 809.34
D = 4 470.40/809.34 = 5.524
Answer
D = 5.524 years
7.CAPM & WACC
Answer: rm = 8.5 %
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 = (13% − 4%)/2 = 4.5%
Market return
rm = 4% + 4.5% = 8.5%
Answer
rm = 8.5%
8.CAPM & WACC
Answer: WACC = 4.671 %
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 = (620 − 500)/60 = 2
Answer
z = 2
11.Probability
Answer: σ = 2.086
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.
Mean
μ = 0·0.15 + 2·0.25 + 4·0.3 + 6·0.3 = 3.5
Mean of the squares
E(X²) = 0·0.15 + 4·0.25 + 16·0.3 + 36·0.3 = 16.6
Variance
Var(X) = 16.6 − 3.5² = 4.35
σ = √4.35 = 2.086
Answer
σ = 2.086
12.Probability
Answer: P = 0.7748
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.
Zero and one failure
P(X = 0) = 0.99 = 0.3874
P(X = 1) = 9·0.1·0.98 = 0.3874
Add them
P = 0.3874 + 0.3874 = 0.7748
Answer
P = 0.7748
13.Normal distribution
Answer: P = 0.1587
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (759.8 − 750)/9.849 = 1
Upper tail
P = 1 − Φ(1) = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
14.Normal distribution
Answer: x = 466.3 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) = 0.2 → z = −0.8416
Back to x
x = 500 + −0.8416·40 = 466.34 h
Answer
x = 466.34 h
15.Confidence intervals
Answer: E = 9.323
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 = 15, t = 2.131
Margin of error
E = 2.131·17.5/√16 = 9.323
Answer
E = 9.323
16.Confidence intervals
Answer: E = 3.922 kg
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.
Pooled standard deviation
sp² = (14·4.5² + 6·3²)/20 = 16.88
sp = 4.108
Margin of error
E = 2.086·4.108·√(1/15 + 1/7) = 3.922
Answer
E = 3.922 kg
17.Hypothesis tests & confidence intervals
Answer: t = 2.083
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 = (202.5 − 200)/1.2 = 2.083
Answer
t = 2.083
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 = −1.162
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 = 18/120 = 0.15
p̂₂ = 50/250 = 0.2
p̂ = 68/370 = 0.1838
Standard error under H₀
SE = √(0.1838·0.8162·(1/120 + 1/250)) = 0.04301
Test statistic
z = (0.15 − 0.2)/0.04301 = −1.162
Answer
z = −1.162
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = −0.9007
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 = 298 − 15·107/5 = −23
Sxx = 55 − 15²/5 = 10
Syy = 2355 − 107²/5 = 65.2
Correlation coefficient
r = −23/√(10·65.2) = −0.9007
Answer
r = −0.9007
20.Linear regression & correlation
Answer: x = 7.114
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 = 323/5 = 64.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 158
Regression line
b = 158/40 = 3.95
a = 64.6 − 3.95·6 = 40.9
ŷ = 40.9 + 3.95·x
Solve for x
x = (69 − 40.9)/3.95 = 7.114
Answer
x = 7.114
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