You need €20 000 in 7 years. An account pays 3.6% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €20 0003.6% / 4n = 7
PV = €
2.Time value of moneyHard
A loan of €20 000 is repaid with equal monthly payments over 3 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.
P = €20 0004.2% / 12N = 36
A = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 4: €18 000, €13 000, €13 000, €10 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €80 000 today. Its cash flow is €14 500 at the end of year 1 and grows by 4% per year until the last cash flow at the end of year 5. The cost of capital is 12%. Find the net present value NPV.
C0 = €80 000CF1 = €14 500g = 4%n = 5r = 12%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 14 years and pays annual coupons. It sells for €1 423 when the yield to maturity is 4%. Find the coupon rate c in percent.
F = €1 000n = 14P = €1 423YTM = 4%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 9 years. It sells for €1 144. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 7% (2×)n = 9P = €1 144
YTM = %
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €5 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 3% and the tax rate is 21%. Find the WACC.
A company has equity with a market value of €6 000 000 and a cost of equity of 14%. Its debt consists of 2 000 bonds (face value €1 000, annual coupon 7%, 8 years to maturity) that trade at €915 each. The tax rate is 25.8%. 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: 38, 59, 19, 28, 19, 29, 48.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 8. Find the z-score z of a score of 38.
μ = 50σ = 8x = 38
z =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3p(x): 0.15, 0.35, 0.35, 0.15
σ =
12.ProbabilityHard
A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 5 correct answers?
n = 8p = 0.7k = 5
P =
13.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 12 of them exceeds 302.6 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 12> 302.6 kg
P =
14.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
A sample of 15 items has mean 90.5 and standard deviation s = 6.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 15x̄ = 90.5s = 6.590%t(14) = 1.761
E =
16.Confidence intervalsHard
The population standard deviation is 21. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 0.5?
σ = 21E = 0.599%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 50 parts has mean length 132.5 mm, and the line's known standard deviation is 10 mm. Line 2: a sample of 40 parts has mean 137.5 mm, known standard deviation 5 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 299%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (16.5, 67) is made. Find its residual e = y − ŷ.
n = 6(16.5, 67)
e =
3Solutions
1.Time value of money
Answer: PV = 15562 €
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 and number of periods
i = 0.036000000000000004/4 = 0.009
N = 4·7 = 28
Present value
PV = FV/(1 + i)N = 20 000/1.285 = €15 562.43
Answer
PV = €15 562.43
2.Time value of money
Answer: A = 592.3 €
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.
Monthly rate and number of payments
i = 0.042/12 = 0.0035
N = 12·3 = 36
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.8818
A = 20 000·0.0035/(1 − 0.8818) = €592.26
Answer
A = €592.26
3.NPV & IRR
Answer: NPV = 482.2 €
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 = −45 000
CF1/(1 + 0.08)1 = 18 000/1.08 = 16 666.67
CF2/(1 + 0.08)2 = 13 000/1.166 = 11 145.40
CF3/(1 + 0.08)3 = 13 000/1.26 = 10 319.82
CF4/(1 + 0.08)4 = 10 000/1.36 = 7 350.30
Add them up
NPV = €482.19
Answer
NPV = €482.19
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: NPV = −23878 €
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.
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: c = 8.004 %
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 + 0.04)−14 = €577.48
Coupon
C = (1423 − 577.48)/10.56 = €80.04
Coupon rate
c = 80.04/1000 = 8.004%
Answer
c = 8.004%
6.Bond pricing & yield
Answer: YTM = 4.994 %
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 = €35.00, N = 18
Solve numerically
i = 2.497% per half-year
YTM
YTM = 2·2.497% = 4.994%
Answer
YTM = 4.994%
7.CAPM & WACC
Answer: WACC = 8.714 %
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 = €11 000 000, E/V = 0.5455, D/V = 0.4545
WACC
WACC = 0.5455·14% + 0.4545·3%·(1 − 0.21) = 8.714%
Answer
WACC = 8.714%
8.CAPM & WACC
Answer: WACC = 12.2 %
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 = (38 − 50)/8 = −1.5
Answer
z = −1.5
11.Probability
Answer: σ = 0.922
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 + 1·0.35 + 2·0.35 + 3·0.15 = 1.5
Mean of the squares
E(X²) = 0·0.15 + 1·0.35 + 4·0.35 + 9·0.15 = 3.1
Variance
Var(X) = 3.1 − 1.5² = 0.85
σ = √0.85 = 0.922
Answer
σ = 0.922
12.Probability
Answer: P = 0.2541
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 total
μT = 300 kg, σT = 0.5·√12 = 1.732 kg
Standardise
z = (302.6 − 300)/1.732 = 1.5
Probability
P = 1 − Φ(1.5) = 0.06681
Answer
P = 0.06681
14.Normal distribution
Answer: P = 0.8413
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 difference
μ = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
15.Confidence intervals
Answer: E = 2.955
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 = 14, t = 1.761
Margin of error
E = 1.761·6.5/√15 = 2.955
Answer
E = 2.955
16.Confidence intervals
Answer: n = 11706
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 = (2.576·21/0.5)² = 11706
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 11706
Answer
n = 11706
17.Hypothesis tests & confidence intervals
Answer: z = −3.086
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.
Standard error of the difference
SE = √(10²/50 + 5²/40) = √2.625 = 1.62
Test statistic
z = (132.5 − 137.5)/1.62 = −3.086
Answer
z = −3.086
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 166
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 = (2.576·10/2)² = 165.9
Round up
n = 166
Answer
n = 166
19.Linear regression & correlation
Answer: r = 0.9857
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 = 7593 − 110·333/5 = 267
Sxx = 2510 − 110²/5 = 90
Syy = 22993 − 333²/5 = 815.2
Correlation coefficient
r = 267/√(90·815.2) = 0.9857
Answer
r = 0.9857
20.Linear regression & correlation
Answer: e = −4.781
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 = 135/6 = 22.5
ȳ = 622/6 = 103.7
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 837
Regression line
b = 837/157.5 = 5.314
a = 103.7 − 5.314·22.5 = −15.9
ŷ = −15.9 + 5.314·x
Predicted value
ŷ = −15.9 + 5.314·16.5 = 71.78
Residual
e = 67 − 71.78 = −4.781
Answer
e = −4.781
The point lies below the line.
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