You invest €3 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 7 years.
PV = €3 0004.8% / 4n = 7
FV = €
2.Time value of moneyHard
A mortgage of €200 000 is repaid with equal monthly payments over 30 years at 4.2% per year, compounded monthly. How much is still owed just after the payment at the end of year 15? Find the outstanding balance B.
P = €200 0004.2% / 12N = 360k = 180
B = €
3.NPV & IRRMedium
A project costs €200 000 today. It generates €11 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 8%. Find the net present value NPV.
C0 = €200 000CF1 = €11 000g = 3%r = 8%
NPV = €
4.NPV & IRRHard
A project costs €80 000 today. Its cash flow is €13 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 7. The cost of capital is 8%. Find the net present value NPV.
C0 = €80 000CF1 = €13 500g = 3%n = 7r = 8%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 2.5% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 2.5%n = 12
P = €
6.Bond pricing & yieldHard
A bond trades at €1 085 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 10.4 years. Use the modified duration to estimate the new price P if the yield falls by 50 basis points.
P0 = €1 085y = 6%D = 10.4Δy = −50 bp
P = €
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €4 000 000, preferred shares worth €2 000 000 and debt worth €6 000 000. The cost of equity is 11%, the cost of preferred stock 8% and the pre-tax cost of debt 4.5%. The tax rate is 30%. Find the WACC.
A company has 4 000 000 shares at €51.50 each and 5 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 30%. The share’s beta is 1.5, the risk-free rate 3.5% and the market return 8%. Find the WACC.
β = 1.5rf = 3.5%rm = 8%rd = 3%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 49, 38, 26, 56, 39, 45.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 46.
μ = 50σ = 5x = 46
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
12.ProbabilityHard
X has the distribution shown, and Y = 3X + 4. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.2, 0.1, 0.35, 0.35Y = 3X + 4
σY =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1539. Find z to two decimals.
A = 0.1539
z =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).
μ = 65σ = 10P = 95%
x = points
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 30 phones, mean 79 h, known standard deviation 3 h. Model 2: 80 phones, mean 77 h, known standard deviation 4 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 31. 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 1?
σ = 31E = 199%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 501.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 501.4s = 20n = 100
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 75 of 250 patients on treatment A recover, and 19 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 75n₁ = 250x₂ = 19n₂ = 80
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 68?
n = 5ŷ = 68
x =
3Solutions
1.Time value of money
Answer: FV = 4190 €
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.
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 payment
i = 0.0035, N = 360
A = 200 000·0.0035/(1 − (1 + 0.0035)−360) = €978.03
Present value of the remaining payments
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 360 − 180 = 180
B = 978·(1 − (1 + 0.0035)−180)/0.0035 = €130 448.00
Answer
B = €130 448.00
3.NPV & IRR
Answer: NPV = 20000 €
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.
Value of the cash flows
PV = 11 000/(0.08 − 0.03) = 220 000.00
NPV
NPV = 220 000.00 − 200 000 = €20 000.00
Answer
NPV = €20 000.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: NPV = −3757 €
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 = 1464 €
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, i = 0.0125, N = 24
Present value of the coupons
€35.00·(1 − 0.7422)/0.0125 = €721.85
Present value of the face value
€1 000·0.7422 = €742.20
Price
P = €1 464.05
Answer
P = €1 464.05
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: P = 1138 €
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.
Modified duration
Dmod = 10.4/1.06 = 9.811
Estimated price change
%ΔP ≈ −9.811·(−0.005) = 4.906%
P ≈ 1085·(1 + 0.04906) = €1 138.23
Answer
P = €1 138.23
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 6.575 %
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 = (46 − 50)/5 = −0.8
Answer
z = −0.8
11.Probability
Answer: Var(X) = 0.7
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
Answer
Var(X) = 0.7
12.Probability
Answer: σY = 3.324
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 = 2.85, Var(X) = 1.228, σX = 1.108
Linear change
σY = |3|·1.108 = 3.324
Answer
σY = 3.324
The mean does move: E(Y) = 3·2.85 + 4 = 12.55.
13.Normal distribution
Answer: z = 1.02
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 − 0.1539 = 0.8461
Read the z-table backwards
Find 0.8461 inside the table and read off z.
z = 1.02
Answer
z = 1.02
14.Normal distribution
Answer: x = 81.45 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 = 65 + 1.645·10 = 81.45 points
Answer
x = 81.45 points
15.Confidence intervals
Answer: lower limit = 0.1785 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 = 79 − 77 = 2
Standard error
SE = √(3²/30 + 4²/80) = 0.7071
Margin of error
E = 2.576·0.7071 = 1.822
Interval
2 ± 1.822 → [0.1785, 3.822]
Answer
lower limit = 0.1785 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 6377
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 = (2.576·31/1)² = 6377
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 6377
Answer
n = 6377
17.Hypothesis tests & confidence intervals
Answer: t = 0.7
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 = 20/√100 = 20/10 = 2
Test statistic
t = (501.4 − 500)/2 = 0.7
Answer
t = 0.7
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = 1.078
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 = 75/250 = 0.3
p̂₂ = 19/80 = 0.2375
p̂ = 94/330 = 0.2848
Standard error under H₀
SE = √(0.2848·0.7152·(1/250 + 1/80)) = 0.05798
Test statistic
z = (0.3 − 0.2375)/0.05798 = 1.078
Answer
z = 1.078
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9624
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 = 59/5 = 11.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 58
Syy = Σ(y − ȳ)² = 90.8
Correlation coefficient
r = 58/√(40·90.8) = 0.9624
Answer
r = 0.9624
20.Linear regression & correlation
Answer: x = 5.458
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 = 20/5 = 4
ȳ = 305/5 = 61
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 48
Regression line
b = 48/10 = 4.8
a = 61 − 4.8·4 = 41.8
ŷ = 41.8 + 4.8·x
Solve for x
x = (68 − 41.8)/4.8 = 5.458
Answer
x = 5.458
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