You need €15 000 in 10 years. An account pays 5% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €15 0005% / 12n = 10
PV = €
2.Time value of moneyHard
You save €50 at the end of every month for 25 years. The account pays 2.4% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €502.4% / 12N = 300
FV = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 3: €28 000, €18 000, €30 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €25 000 and returns €11 500, €7 500, €8 500 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 4.5% and matures in 8 years. Its yield to maturity changes from 5.5% to 7%. By how much ΔP does the bond price change?
F = €1 000coupon = 4.5%n = 8y1 = 5.5%y2 = 7%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 3% matures in 10 years and sells for €880. Find its yield to maturity YTM in percent.
F = €1 000coupon = 3%n = 10P = €880
YTM = %
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €3 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 4% and the tax rate is 25.8%. Find the WACC.
A company has 4 000 000 shares at €33.00 each and 20 000 bonds trading at €1 000 each. The pre-tax cost of debt is 3.5% and the tax rate is 20%. The share’s beta is 1.3, the risk-free rate 1% and the market return 5%. Find the WACC.
β = 1.3rf = 1%rm = 5%rd = 3.5%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 60, 59, 31, 26, 26, 17.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 75?
μ = 60σ = 10x = 75
P =
11.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
12.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 5 correct answers?
n = 7p = 0.6k = 5
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9441. Find z to two decimals.
A = 0.9441
z =
14.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
A sample of 10 items has mean 85.5 and standard deviation s = 18. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 10x̄ = 85.5s = 1899%t(9) = 3.25
E =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 6 plots, mean yield 71.5 kg, standard deviation 7.5 kg. Plot group 2: 6 plots, mean 80.5 kg, standard deviation 7.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The mean of a population is claimed to be 120; the population standard deviation is known to be 20. A random sample of 25 observations has a mean of 125.2. Test H₀: μ = 120 against H₁: μ ≠ 120. Find the p-value p.
μ₀ = 120σ = 20n = 25x̄ = 125.2H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 90%. The standard deviation is 14. Find the minimum sample size n.
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 132Σx² = 364Σy² = 3556Σxy = 1136
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 44) is made. Find its residual e = y − ŷ.
n = 6(3.5, 44)
e =
3Solutions
1.Time value of money
Answer: PV = 9107 €
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 = 0.05/12 = 0.004167
N = 12·10 = 120
Present value
PV = FV/(1 + i)N = 15 000/1.647 = €9 107.42
Answer
PV = €9 107.42
2.Time value of money
Answer: FV = 20526 €
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 €15 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 4004 €
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 = −60 000
CF1/(1 + 0.09)1 = 28 000/1.09 = 25 688.07
CF2/(1 + 0.09)2 = 18 000/1.188 = 15 150.24
CF3/(1 + 0.09)3 = 30 000/1.295 = 23 165.50
Add them up
NPV = €4 003.82
Answer
NPV = €4 003.82
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: MIRR = 6.282 %
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
11 500·1.082 = 13 413.60
7 500·1.081 = 8 100.00
8 500·1.080 = 8 500.00
TV = 30 013.60
MIRR
MIRR = (30 013.60/25 000)1/3 − 1 = 6.282%
Answer
MIRR = 6.282%
5.Bond pricing & yield
Answer: ΔP = −85.94 €
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
€880 = €30·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰
Solve numerically
P(4%) = €918.89, P(5%) = €845.57
YTM = 4.518%
Answer
YTM = 4.518%
7.CAPM & WACC
Answer: WACC = 7.608 %
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 = €13 000 000, E/V = 0.7692, D/V = 0.2308
WACC
WACC = 0.7692·9% + 0.2308·4%·(1 − 0.258) = 7.608%
Answer
WACC = 7.608%
8.CAPM & WACC
Answer: WACC = 5.753 %
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) = 0.9332
P = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
11.Probability
Answer: P = 0.6651
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.
Complement
P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
12.Probability
Answer: P = 0.2613
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.
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 − 0.9441 = 0.0559
Read the z-table backwards
Find 0.0559 inside the table and read off z.
z = −1.59
Answer
z = −1.59
14.Normal distribution
Answer: P = 0.7881
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 = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
15.Confidence intervals
Answer: E = 18.5
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 = 9, t = 3.25
Margin of error
E = 3.25·18/√10 = 18.5
Answer
E = 18.5
16.Confidence intervals
Answer: upper limit = 4.722 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 = (5·7.5² + 5·7.5²)/10 = 56.25
sp = 7.5
Margin of error
E = 3.169·7.5·√(1/6 + 1/6) = 13.72
Interval
−9 ± 13.72 → [−22.72, 4.722]
Answer
upper limit = 4.722 kg
17.Hypothesis tests & confidence intervals
Answer: p = 0.1936
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.
Test statistic
SE = σ/√n = 20/√25 = 4
z = (125.2 − 120)/4 = 1.3
Tail area
Φ(1.3) = 0.9032
1 − Φ(1.3) = 0.0968
p-value
p = 2·0.0968 = 0.1936
Answer
p = 0.1936
p > 0.05, so H₀ is not rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: n = 420
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 1.282)·14/2)² = 419.8
Round up
n = 420
Answer
n = 420
19.Linear regression & correlation
Answer: r = 0.9923
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 = 1136 − 42·132/6 = 212
Sxx = 364 − 42²/6 = 70
Syy = 3556 − 132²/6 = 652
Correlation coefficient
r = 212/√(70·652) = 0.9923
Answer
r = 0.9923
20.Linear regression & correlation
Answer: e = 4.062
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 = 33/6 = 5.5
ȳ = 301/6 = 50.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 89.5
Regression line
b = 89.5/17.5 = 5.114
a = 50.17 − 5.114·5.5 = 22.04
ŷ = 22.04 + 5.114·x
Predicted value
ŷ = 22.04 + 5.114·3.5 = 39.94
Residual
e = 44 − 39.94 = 4.062
Answer
e = 4.062
The point lies above the line.
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