You need €40 000 in 6 years. An account pays 3.6% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €40 0003.6% / 12n = 6
PV = €
2.Time value of moneyHard
You deposit €5 000 at the start of every year for 11 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 11? Find FV.
A = €5 000i = 5%n = 11
FV = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 4: €17 000, €17 000, €17 000, €11 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €100 000 today. Its cash flow is €17 000 at the end of year 1 and grows by 4% per year until the last cash flow at the end of year 7. The cost of capital is 12%. Find the net present value NPV.
C0 = €100 000CF1 = €17 000g = 4%n = 7r = 12%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 4% per year. Find the price P of the bond.
F = €1 000coupon = 4.5%YTM = 4%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 6 years. The yield to maturity is 5.5%. Find the Macaulay duration D in years.
F = €1 000coupon = 6%n = 6YTM = 5.5%
D = years
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €1 000 000. Its WACC is 8.3%, its pre-tax cost of debt is 4% and the tax rate is 20%. Find the cost of equity re.
A company has 2 500 000 shares at €27.00 each and 10 000 bonds trading at €980 each. The pre-tax cost of debt is 7% and the tax rate is 25.8%. The share’s beta is 1.2, the risk-free rate 2.5% and the market return 7%. Find the WACC.
β = 1.2rf = 2.5%rm = 7%rd = 7%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 58, 40, 46, 14, 20, 10, 15.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is above 200?
μ = 170σ = 15x = 200
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
σ =
12.ProbabilityHard
A game costs €10 to play. You win €40 with probability 0.25 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€40: p = 0.25€100: p = 0.05
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0179. Find z to two decimals.
A = 0.0179
z =
14.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 40 phones, mean 103.5 h, known standard deviation 15 h. Model 2: 80 phones, mean 109 h, known standard deviation 16 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 8. 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 2?
σ = 8E = 295%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 6 students, mean score 117.5, standard deviation 8.5. Group 2: 20 students, mean score 112.6, standard deviation 4. 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 = 6x̄₁ = 117.5s₁ = 8.5n₂ = 20x̄₂ = 112.6s₂ = 4
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 7.5. In reality the mean is 96.55. Find the probability of a Type II error β in percent.
μ₀ = 100μ = 96.55σ = 7.5n = 25α = 0.01
β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 22.
n = 5x = 22
ŷ =
3Solutions
1.Time value of money
Answer: PV = 32240 €
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 = 0.036000000000000004/12 = 0.003
N = 12·6 = 72
Present value
PV = FV/(1 + i)N = 40 000/1.241 = €32 239.84
Answer
PV = €32 239.84
2.Time value of money
Answer: FV = 74586 €
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.
Ordinary annuity factor
((1 + 0.05)11 − 1)/0.05 = 14.21
Annuity due
Each deposit is invested one year longer.
FV = 5 000·14.21·1.05 = €74 585.63
Answer
FV = €74 585.63
3.NPV & IRR
Answer: NPV = 6896 €
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 = 17 000/1.08 = 15 740.74
CF2/(1 + 0.08)2 = 17 000/1.166 = 14 574.76
CF3/(1 + 0.08)3 = 17 000/1.26 = 13 495.15
CF4/(1 + 0.08)4 = 11 000/1.36 = 8 085.33
Add them up
NPV = €6 895.98
Answer
NPV = €6 895.98
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: NPV = −13993 €
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: P = 1047 €
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 = €22.50, i = 0.02, N = 24
Present value of the coupons
€22.50·(1 − 0.6217)/0.02 = €425.56
Present value of the face value
€1 000·0.6217 = €621.72
Price
P = €1 047.28
Answer
P = €1 047.28
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: D = 5.223 years
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: 60/1.055 = 56.87, t·PV = 56.87
t = 2: 60/1.113 = 53.91, t·PV = 107.81
t = 3: 60/1.174 = 51.10, t·PV = 153.29
t = 4: 60/1.239 = 48.43, t·PV = 193.73
t = 5: 60/1.307 = 45.91, t·PV = 229.54
t = 6: 1060/1.379 = 768.76, t·PV = 4 612.56
Duration
P = 1 024.98
D = 5 353.81/1 024.98 = 5.223
Answer
D = 5.223 years
7.CAPM & WACC
Answer: re = 10 %
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 = €4 000 000, E/V = 0.75, D/V = 0.25
Debt part of the WACC
0.25·4%·(1 − 0.2) = 0.8%
Cost of equity
re = (8.3% − 0.8%)/0.75 = 10%
Answer
re = 10%
8.CAPM & WACC
Answer: WACC = 7.557 %
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
11.Probability
Answer: σ = 0.8367
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 = 0·0.35 + 1·0.3 + 2·0.35 = 1
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7
Variance
Var(X) = 1.7 − 1² = 0.7
σ = √0.7 = 0.8367
Answer
σ = 0.8367
12.Probability
Answer: E(G) = 5 €
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: 30 (p = 0.25), 90 (p = 0.05), −10 (p = 0.7)
Expected value
E(G) = 30·0.25 + 90·0.05 − 10·0.7 = 5 €
Answer
E(G) = 5 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = 2.1
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 − 0.0179 = 0.9821
Read the z-table backwards
Find 0.9821 inside the table and read off z.
z = 2.10
Answer
z = 2.10
14.Normal distribution
Answer: P = 0.7389
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 = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
15.Confidence intervals
Answer: upper limit = 2.152 h
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.
Point estimate
x̄₁ − x̄₂ = 103.5 − 109 = −5.5
Standard error
SE = √(15²/40 + 16²/80) = 2.971
Margin of error
E = 2.576·2.971 = 7.652
Interval
−5.5 ± 7.652 → [−13.15, 2.152]
Answer
upper limit = 2.152 h
The interval contains 0, so the data do not show a significant difference.
16.Confidence intervals
Answer: n = 62
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 = (1.96·8/2)² = 61.47
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 62
Answer
n = 62
17.Hypothesis tests & confidence intervals
Answer: t = 1.999
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 = (5·8.5² + 19·4²)/24 = 27.72
sp = 5.265
Standard error
SE = 5.265·√(1/6 + 1/20) = 2.451
Test statistic
t = (117.5 − 112.6)/2.451 = 1.999
Answer
t = 1.999
With df = 24, the two-sided 5% critical value is 2.064; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: β = 51.04 %
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 = 7.5/√25 = 1.5
x̄c = 100 − 2.326·1.5 = 96.51
z under the true mean
z = (96.51 − 96.55)/1.5 = −0.026
Type II error
power = P(Z < −0.026) = 48.96%
β = 100% − 48.96% = 51.04%
Answer
β = 51.04%
19.Linear regression & correlation
Answer: r = 0.989
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 = 2481 − 36·365/6 = 291
Sxx = 286 − 36²/6 = 70
Syy = 23441 − 365²/6 = 1237
Correlation coefficient
r = 291/√(70·1237) = 0.989
Answer
r = 0.989
20.Linear regression & correlation
Answer: ŷ = 49.15
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 = 95/5 = 19
ȳ = 212/5 = 42.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 90
Regression line
b = 90/40 = 2.25
a = 42.4 − 2.25·19 = −0.35
ŷ = −0.35 + 2.25·x
Prediction
ŷ = −0.35 + 2.25·22 = 49.15
Answer
ŷ = 49.15
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