An annuity pays €2 000 at the end of each year for 19 years. The interest rate is 7% per year. Find the present value PV of the annuity.
A = €2 000i = 7%n = 19
PV = €
2.Time value of moneyHard
You save €300 at the end of every month for 16 years. The account pays 4.2% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €3004.2% / 12N = 192
FV = €
3.NPV & IRRMedium
A machine costs €60 000 today, lasts 4 years and costs €1 500 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.
C0 = €60 000M = €1 500n = 4r = 12%
EAC = €
4.NPV & IRRHard
A project costs €50 000 and returns €33 000, €32 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 33 000CF2 = 32 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €734 when the yield to maturity is 5%. Find the coupon rate c in percent.
F = €1 000n = 12P = €734YTM = 5%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 5 years and sells for €960. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 5P = €960
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €2 000 000. Its WACC is 8.41%, its pre-tax cost of debt is 4.5% and the tax rate is 28%. Find the cost of equity re.
A company has 1 000 000 shares at €55.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 4.5% and the tax rate is 25%. The share’s beta is 0.9, the risk-free rate 4% and the market return 11%. Find the WACC.
β = 0.9rf = 4%rm = 11%rd = 4.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 35, 24, 17, 52, 13, 59, 21.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 5. Find the z-score z of a score of 65.
μ = 60σ = 5x = 65
z =
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
X has the distribution shown, and Y = 5X − 3. Find the standard deviation σY of Y.
Drag z so that the shaded area between −z and z under the standard normal curve is 0.4039. Find z to two decimals.
A = 0.4039
z =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points25%
x = points
15.Confidence intervalsMedium
A sample of 8 items has mean 112.5 and standard deviation s = 10. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 8x̄ = 112.5s = 1095%t(7) = 2.365
upper limit =
16.Confidence intervalsHard
A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.05?
p ≈ 0.6E = 0.0590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 200.8 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 200.8s = 4n = 81
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 295%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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. For which value x does the line predict ŷ = 24?
n = 6ŷ = 24
x =
3Solutions
1.Time value of money
Answer: PV = 20671 €
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.
Annuity factor
(1 − (1 + i)−n)/i = (1 − 0.2765)/0.07 = 10.34
Present value
PV = 2 000·10.34 = €20 671.19
Answer
PV = €20 671.19
2.Time value of money
Answer: FV = 81930 €
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 €57 600 yourself; the rest is interest.
3.NPV & IRR
Answer: EAC = 21254 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.12−4)/0.12 = 3.037
Annualised purchase price
60 000/3.037 = 19 754.07
EAC
EAC = 19 754.07 + 1 500 = €21 254.07
Answer
EAC = €21 254.07
4.NPV & IRR
Answer: IRR = 19.54 %
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.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(15%) = 2 892.25
NPV(20%) = −277.78
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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 + 0.05)−12 = €556.84
Coupon
C = (734 − 556.84)/8.863 = €19.99
Coupon rate
c = 19.99/1000 = 1.999%
Answer
c = 1.999%
6.Bond pricing & yield
Answer: YTM = 6.975 %
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
€960 = €60·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(6%) = €1 000.00, P(7%) = €959.00
YTM = 6.975%
Answer
YTM = 6.975%
7.CAPM & WACC
Answer: re = 11 %
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 = €6 000 000, E/V = 0.6667, D/V = 0.3333
Debt part of the WACC
0.3333·4.5%·(1 − 0.28) = 1.08%
Cost of equity
re = (8.41% − 1.08%)/0.6667 = 11%
Answer
re = 11%
8.CAPM & WACC
Answer: WACC = 7.453 %
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 = (65 − 60)/5 = 1
Answer
z = 1
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: σY = 58.09
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 = |5|·11.62 = 58.09
Answer
σY = 58.09
The mean does move: E(Y) = 5·25 − 3 = 122.
13.Normal distribution
Answer: z = 0.53
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 + 0.4039)/2 = 0.7019
Read the z-table backwards
Find 0.7019 inside the table and read off z.
z = 0.53
Answer
z = 0.53
14.Normal distribution
Answer: x = 110.1 points
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) = 0.75
z from the table
z ≈ 0.67
Back to the original units
x = 100 + 0.67·15 = 110.1 points
Answer
x = 110.1 points
15.Confidence intervals
Answer: upper limit = 120.9
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 = 7, t = 2.365
Margin of error
E = 2.365·10/√8 = 8.362
Interval
112.5 ± 8.362 → [104.1, 120.9]
Answer
upper limit = 120.9
16.Confidence intervals
Answer: n = 260
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.645²·0.6·0.4/0.05² = 259.8
Round up
n = 260
Answer
n = 260
17.Hypothesis tests & confidence intervals
Answer: t = 1.8
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 = 4/√81 = 4/9 = 0.4444
Test statistic
t = (200.8 − 200)/0.4444 = 1.8
Answer
t = 1.8
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 139
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 = (1.96·12/2)² = 138.3
Round up
n = 139
Answer
n = 139
19.Linear regression & correlation
Answer: r² = 0.9811
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 = 104/5 = 20.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 118
Syy = Σ(y − ȳ)² = 354.8
Coefficient of determination
r² = 118²/(40·354.8) = 0.9811
Answer
r² = 0.9811
About 98.11% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: x = 5.167
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 = 156/6 = 26
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −21
Regression line
b = −21/17.5 = −1.2
a = 26 − (−1.2)·3.5 = 30.2
ŷ = 30.2 − 1.2·x
Solve for x
x = (24 − 30.2)/(−1.2) = 5.167
Answer
x = 5.167
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