You invest €8 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 6 years.
PV = €8 0004.8% / 4n = 6
FV = €
2.Time value of moneyHard
You save €100 at the end of every month for 19 years. The account pays 3.6% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €1003.6% / 12N = 228
FV = €
3.NPV & IRRMedium
A machine costs €80 000 today, lasts 5 years and costs €3 000 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.
C0 = €80 000M = €3 000n = 5r = 10%
EAC = €
4.NPV & IRRHard
A project costs €40 000 and returns €28 500, €19 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 28 500CF2 = 19 500
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 3 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 3%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4.5%. Find your holding period return HPR in percent.
F = €1 000coupon = 2%n = 3y0 = 3%y1 = 4.5%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 3 years and sells for €975. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 3P = €975
YTM = %
7.CAPM & WACCMedium
A share with a beta of 1.6 has a required return of 14.4%. The risk-free rate is 4%. What expected market return rm does CAPM imply?
β = 1.6re = 14.4%rf = 4%
rm = %
8.CAPM & WACCHard
A company has equity with a market value of €8 000 000 and a cost of equity of 10%. Its debt consists of 4 000 bonds (face value €1 000, annual coupon 6%, 11 years to maturity) that trade at €1 128 each. The tax rate is 20%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
Find the sample standard deviation s of the data: 12, 53, 39, 24, 24.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is above 428?
μ = 500σ = 60x = 428
P =
11.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 6? The order does not matter.
n = 6k = 3
N =
12.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?
n = 7p = 0.25k = 1
P =
13.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 70 and 118.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 70x2 = 118.75
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h70%
x = h
15.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 4.8. What is the margin of error E₂ of a 99% interval based on 576 observations?
90%n₁ = 64E₁ = 4.899%n₂ = 576
E₂ =
16.Confidence intervalsHard
The population standard deviation is 25. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?
σ = 25E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 51.2 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 51.2s = 4n = 64
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 95%. The standard deviation is 10. Find the minimum sample size n.
The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (3, 56) is made. Find its residual e = y − ŷ.
n = 6(3, 56)
e =
3Solutions
1.Time value of money
Answer: FV = 10652 €
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.
You paid in €22 800 yourself; the rest is interest.
3.NPV & IRR
Answer: EAC = 24104 €
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 − 1.1−5)/0.1 = 3.791
Annualised purchase price
80 000/3.791 = 21 103.80
EAC
EAC = 21 103.80 + 3 000 = €24 103.80
Answer
EAC = €24 103.80
4.NPV & IRR
Answer: IRR = 14.01 %
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(10%) = 2 024.79
NPV(15%) = −472.59
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.
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
€975 = €40·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³
Solve numerically
P(4%) = €1 000.00, P(5%) = €972.77
YTM = 4.917%
Answer
YTM = 4.917%
7.CAPM & WACC
Answer: rm = 10.5 %
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).
Market risk premium
rm − rf = (14.4% − 4%)/1.6 = 6.5%
Market return
rm = 4% + 6.5% = 10.5%
Answer
rm = 10.5%
8.CAPM & WACC
Answer: WACC = 7.692 %
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849
Answer
P = 0.8849
11.Probability
Answer: N = 20
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.
Combinations
C(6, 3) = (6·5·4)/(3·2·1) = 20
Answer
N = 20
12.Probability
Answer: P = 0.3115
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.
Binomial formula
P(X = 1) = 7·0.251·0.756 = 7·0.25·0.178 = 0.3115
Answer
P = 0.3115
13.Normal distribution
Answer: P = 0.8716
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-scores
z₁ = (70 − 100)/15 = −2
z₂ = (118.75 − 100)/15 = 1.25
Area between
P = Φ(1.25) − Φ(−2) = 0.8944 − 0.02275 = 0.8716
Answer
P = 0.8716
14.Normal distribution
Answer: x = 479 h
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) = 0.3
z from the table
z ≈ −0.52
Back to the original units
x = 500 − 0.52·40 = 479 h
Answer
x = 479 h
15.Confidence intervals
Answer: E₂ = 2.506
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.
Ratio of margins
E₂/E₁ = (2.576/1.645)·√(64/576) = 0.522
New margin
E₂ = 4.8·0.522 = 2.506
Answer
E₂ = 2.506
16.Confidence intervals
Answer: n = 752
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 = (1.645·25/1.5)² = 751.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 752
Answer
n = 752
17.Hypothesis tests & confidence intervals
Answer: t = 2.4
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/√64 = 4/8 = 0.5
Test statistic
t = (51.2 − 50)/0.5 = 2.4
Answer
t = 2.4
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 121
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 + 1.645)·10/3)² = 120.3
Round up
n = 121
Answer
n = 121
19.Linear regression & correlation
Answer: r = −0.9791
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 = 15/5 = 3
ȳ = 97/5 = 19.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −25
Syy = Σ(y − ȳ)² = 65.2
Correlation coefficient
r = −25/√(10·65.2) = −0.9791
Answer
r = −0.9791
20.Linear regression & correlation
Answer: e = −2.629
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 = 42/6 = 7
ȳ = 480/6 = 80
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 374
Regression line
b = 374/70 = 5.343
a = 80 − 5.343·7 = 42.6
ŷ = 42.6 + 5.343·x
Predicted value
ŷ = 42.6 + 5.343·3 = 58.63
Residual
e = 56 − 58.63 = −2.629
Answer
e = −2.629
The point lies below the line.
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