You need €40 000 in 3 years. An account pays 2% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €40 0002% / 4n = 3
PV = €
2.Time value of moneyHard
A loan of €150 000 is repaid with equal monthly payments over 25 years. The interest rate is 6% per year, compounded monthly. Find the monthly payment A.
P = €150 0006% / 12N = 300
A = €
3.NPV & IRRMedium
A machine costs €50 000 today. It generates €15 500 at the end of each year for 4 years and can then be sold for €7 500. The cost of capital is 12%. Find the net present value NPV.
C0 = €50 000CF = €15 500S = €7 500n = 4r = 12%
NPV = €
4.NPV & IRRHard
Project A costs €10 000 and returns €5 500, €4 000, €2 500 at the end of years 1 to 3. Project B costs €12 000 and returns €4 000, €4 500, €6 000. Find the crossover rate r* at which both projects have the same NPV, in percent.
You buy a bond with 8 years to maturity (face value €1 000, annual coupon 5%) when its yield to maturity is 6%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 7%. Find your holding period return HPR in percent.
F = €1 000coupon = 5%n = 8y0 = 6%y1 = 7%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 10 years and sells for €925. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 10P = €925
YTM = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €4 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 6% and the tax rate is 25.8%. Find the WACC.
A company has 4 000 000 shares at €31.50 each and 5 000 bonds trading at €960 each. The pre-tax cost of debt is 4% and the tax rate is 21%. The share’s beta is 1.1, the risk-free rate 3.5% and the market return 7.5%. Find the WACC.
β = 1.1rf = 3.5%rm = 7.5%rd = 4%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 25, 17, 48, 10, 14, 25, 13.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is above 162.5?
μ = 170σ = 5x = 162.5
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.2, 0.15, 0.1, 0.25
Var(X) =
12.ProbabilityHard
X has the distribution shown, and Y = 2X + 10. Find the standard deviation σY of Y.
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0207. Find z to two decimals.
A = 0.0207
z =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 5% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g5%
x = g
15.Confidence intervalsMedium
A sample of 16 items has mean 60.5 and standard deviation s = 17.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 16x̄ = 60.5s = 17.599%t(15) = 2.947
E =
16.Confidence intervalsHard
The population standard deviation is 19. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?
σ = 19E = 395%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 5 students, mean score 78.5, standard deviation 8. Group 2: 9 students, mean score 84.5, standard deviation 8.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 5x̄₁ = 78.5s₁ = 8n₂ = 9x̄₂ = 84.5s₂ = 8.5
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 100 measurements has mean 499.4. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 499.4σ = 8n = 10099%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
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. A new observation (5.5, 16) is made. Find its residual e = y − ŷ.
n = 6(5.5, 16)
e =
3Solutions
1.Time value of money
Answer: PV = 37676 €
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 = 0.02/4 = 0.005
N = 4·3 = 12
Present value
PV = FV/(1 + i)N = 40 000/1.062 = €37 676.21
Answer
PV = €37 676.21
2.Time value of money
Answer: A = 966.5 €
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 = 0.06/12 = 0.005
N = 12·25 = 300
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 = 0.224
A = 150 000·0.005/(1 − 0.224) = €966.45
Answer
A = €966.45
3.NPV & IRR
Answer: NPV = 1845 €
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
15 500·3.037 = 47 078.91
Present value of the salvage value
7 500/1.574 = 4 766.39
NPV
NPV = −50 000 + 47 078.91 + 4 766.39 = €1 845.30
Answer
NPV = €1 845.30
4.NPV & IRR
Answer: r* = 5.601 %
Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.
Difference in cash flows
ΔCF0 = −12 000 − (−10 000) = −2 000
ΔCF1 = 4 000 − 5 500 = −1 500
ΔCF2 = 4 500 − 4 000 = 500
ΔCF3 = 6 000 − 2 500 = 3 500
IRR of the difference
NPVΔ(5%) = 48.37
NPVΔ(6%) = −31.43
r* = 5.601%
Answer
r* = 5.601%
Check: at 5.601% both NPVs are €918.09.
5.Bond pricing & yield
Answer: HPR = 0.4598 %
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.
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.
Equation for the yield
€925 = €50·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰
Solve numerically
P(6%) = €926.40, P(7%) = €859.53
YTM = 6.02%
Answer
YTM = 6.02%
7.CAPM & WACC
Answer: WACC = 8.09 %
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 = €9 000 000, E/V = 0.5556, D/V = 0.4444
WACC
WACC = 0.5556·11% + 0.4444·6%·(1 − 0.258) = 8.09%
Answer
WACC = 8.09%
8.CAPM & WACC
Answer: WACC = 7.726 %
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332
Answer
P = 0.9332
11.Probability
Answer: Var(X) = 2.46
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.
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.
Spread of X
μ = 25, Var(X) = 135, σX = 11.62
Linear change
σY = |2|·11.62 = 23.24
Answer
σY = 23.24
The mean does move: E(Y) = 2·25 + 10 = 60.
13.Normal distribution
Answer: z = 2.04
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 of z
Φ(z) = 1 − 0.0207 = 0.9793
Read the z-table backwards
Find 0.9793 inside the table and read off z.
z = 2.04
Answer
z = 2.04
14.Normal distribution
Answer: x = 241.8 g
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) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 250 − 1.64·5 = 241.8 g
Answer
x = 241.8 g
15.Confidence intervals
Answer: E = 12.89
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 = 2.947
Margin of error
E = 2.947·17.5/√16 = 12.89
Answer
E = 12.89
16.Confidence intervals
Answer: n = 155
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 = (1.96·19/3)² = 154.1
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 155
Answer
n = 155
17.Hypothesis tests & confidence intervals
Answer: t = −1.29
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.
Pooled variance
sp² = (4·8² + 8·8.5²)/12 = 69.5
sp = 8.337
Standard error
SE = 8.337·√(1/5 + 1/9) = 4.65
Test statistic
t = (78.5 − 84.5)/4.65 = −1.29
Answer
t = −1.29
With df = 12, the two-sided 5% critical value is 2.179; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: U = 501.5
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).
Margin of error
E = 2.576·8/√100 = 2.061
Interval limit
U = 499.4 + 2.061 = 501.5
Answer
U = 501.5
19.Linear regression & correlation
Answer: r = 0.9904
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 = 132/6 = 22
ȳ = 692/6 = 115.3
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 440
Syy = Σ(y − ȳ)² = 2819
Correlation coefficient
r = 440/√(70·2819) = 0.9904
Answer
r = 0.9904
20.Linear regression & correlation
Answer: e = 3.933
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 = 106/6 = 17.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −49
Regression line
b = −49/17.5 = −2.8
a = 17.67 − (−2.8)·3.5 = 27.47
ŷ = 27.47 − 2.8·x
Predicted value
ŷ = 27.47 + (−2.8)·5.5 = 12.07
Residual
e = 16 − 12.07 = 3.933
Answer
e = 3.933
The point lies above the line.
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