You invest €20 000 at a nominal rate of 4% per year, compounded monthly. Find the value FV of the investment after 3 years.
PV = €20 0004% / 12n = 3
FV = €
2.Time value of moneyHard
You save €100 at the end of every month for 27 years. The account pays 3.6% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €1003.6% / 12N = 324
FV = €
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 4: €14 000, €11 000, €10 000, €13 000. The cost of capital is 6%. Find the net present value NPV.
Project A costs €40 000 and returns €21 500, €16 000, €9 500 at the end of years 1 to 3. Project B costs €45 000 and returns €13 000, €18 500, €23 000. Find the crossover rate r* at which both projects have the same NPV, in percent.
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 5.5% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 5.5%n = 7
P = €
6.Bond pricing & yieldHard
A bond with 16 years to maturity (face value €1 000, annual coupon 6%) sells for €1 233. The issuer can call it in 6 years at a call price of €1 060. Find the yield to call YTC in percent.
F = €1 000coupon = 6%P = €1 233call price = €1 060call in 6 years
YTC = %
7.CAPM & WACCMedium
A share with a beta of 0.6 has a required return of 6.4%. The risk-free rate is 4%. What expected market return rm does CAPM imply?
β = 0.6re = 6.4%rf = 4%
rm = %
8.CAPM & WACCHard
A company has 5 000 000 shares at €18.50 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 4% and the tax rate is 21%. The share’s beta is 1.8, the risk-free rate 2% and the market return 8%. Find the WACC.
β = 1.8rf = 2%rm = 8%rd = 4%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 53, 13, 16, 53, 46.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 38?
μ = 50σ = 10x = 38
P =
11.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 7? The order does not matter.
n = 7k = 2
N =
12.ProbabilityHard
9 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?
n = 9p = 0.7
P =
13.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 30.2 and 36.8 min? Give P as a decimal.
μ = 32σ = 6x1 = 30.2x2 = 36.8
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 10% of them lie (the 10th percentile).
μ = 250σ = 5P = 10%
x = g
15.Confidence intervalsMedium
The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −1 and standard deviation 7.6. Find the margin of error E of the 90% confidence interval for the mean difference.
n = 10d̄ = −1sd = 7.690%t(9) = 1.833
E = mmHg
16.Confidence intervalsHard
The population standard deviation is 17. 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 5?
σ = 17E = 595%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 120; the population standard deviation is known to be 15. A random sample of 25 observations has a mean of 125.7. Test H₀: μ = 120 against H₁: μ > 120. Find the p-value p.
μ₀ = 120σ = 15n = 25x̄ = 125.7H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 500 against H₁: μ < 500 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 10. In reality the mean is 497.2. Find the probability of a Type II error β in percent.
μ₀ = 500μ = 497.2σ = 10n = 25α = 0.10
β = %
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 = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 102?
n = 5ŷ = 102
x =
3Solutions
1.Time value of money
Answer: FV = 22545 €
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 €32 400 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 11691 €
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 = −30 000
CF1/(1 + 0.06)1 = 14 000/1.06 = 13 207.55
CF2/(1 + 0.06)2 = 11 000/1.124 = 9 789.96
CF3/(1 + 0.06)3 = 10 000/1.191 = 8 396.19
CF4/(1 + 0.06)4 = 13 000/1.262 = 10 297.22
Add them up
NPV = €11 690.92
Answer
NPV = €11 690.92
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: r* = 7.961 %
Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.
Difference in cash flows
ΔCF0 = −45 000 − (−40 000) = −5 000
ΔCF1 = 13 000 − 21 500 = −8 500
ΔCF2 = 18 500 − 16 000 = 2 500
ΔCF3 = 23 000 − 9 500 = 13 500
IRR of the difference
NPVΔ(7%) = 259.69
NPVΔ(8%) = −10.29
r* = 7.961%
Answer
r* = 7.961%
Check: at 7.961% both NPVs are €1 191.43.
5.Bond pricing & yield
Answer: P = 971.3 €
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 = €25.00, i = 0.0275, N = 14
Present value of the coupons
€25.00·(1 − 0.684)/0.0275 = €287.28
Present value of the face value
€1 000·0.684 = €684.00
Price
P = €971.27
Answer
P = €971.27
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTC = 2.68 %
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
1233 = 60·(1 − (1 + y)−6)/y + 1060/(1 + y)6
Solve numerically
y = 2%: €1 277.34
y = 3%: €1 212.76
YTC = 2.68%
Answer
YTC = 2.68%
The yield to maturity is 4%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.
7.CAPM & WACC
Answer: rm = 8 %
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 = (6.4% − 4%)/0.6 = 4%
Market return
rm = 4% + 4% = 8%
Answer
rm = 8%
8.CAPM & WACC
Answer: WACC = 11.02 %
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 values
E = 5 000 000·€18.50 = €92 500 000
D = 20 000·€1 050 = €21 000 000
Cost of equity (CAPM)
re = 2% + 1.8·6% = 12.8%
Weights
V = €113 500 000, E/V = 0.815, D/V = 0.185
WACC
WACC = 0.815·12.8% + 0.185·4%·(1 − 0.21) = 11.02%
Answer
WACC = 11.02%
9.Descriptive statistics
Answer: s = 18.28
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 = 21
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(7, 2) = (7·6)/(2·1) = 21
Answer
N = 21
12.Probability
Answer: P = 0.000433
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.
Zero and one failure
P(X = 0) = 0.39 = 1.968 × 10⁻⁵
P(X = 1) = 9·0.7·0.38 = 0.0004133
Add them
P = 1.968 × 10⁻⁵ + 0.0004133 = 0.000433
Answer
P = 0.000433
13.Normal distribution
Answer: P = 0.4061
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 = (30.2 − 32)/6 = −0.3
z₂ = (36.8 − 32)/6 = 0.8
Area between
P = Φ(0.8) − Φ(−0.3) = 0.7881 − 0.3821 = 0.4061
Answer
P = 0.4061
14.Normal distribution
Answer: x = 243.6 g
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) = 0.1 → z = −1.282
Back to x
x = 250 + −1.282·5 = 243.59 g
Answer
x = 243.59 g
15.Confidence intervals
Answer: E = 4.405 mmHg
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.
Critical value
df = 9, t = 1.833
Margin of error
E = 1.833·7.6/√10 = 4.405
Answer
E = 4.405 mmHg
16.Confidence intervals
Answer: n = 45
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 = (1.96·17/5)² = 44.41
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 45
Answer
n = 45
17.Hypothesis tests & confidence intervals
Answer: p = 0.02872
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 = 15/√25 = 3
z = (125.7 − 120)/3 = 1.9
Tail area
Φ(1.9) = 0.9713
1 − Φ(1.9) = 0.0287
p-value
p = 0.02872
Answer
p = 0.02872
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: β = 45.3 %
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 = 10/√25 = 2
x̄c = 500 − 1.282·2 = 497.4
z under the true mean
z = (497.4 − 497.2)/2 = 0.118
Type II error
power = P(Z < 0.118) = 54.7%
β = 100% − 54.7% = 45.3%
Answer
β = 45.3%
19.Linear regression & correlation
Answer: r = 0.9939
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 = 119/5 = 23.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 144
Syy = Σ(y − ȳ)² = 524.8
Correlation coefficient
r = 144/√(40·524.8) = 0.9939
Answer
r = 0.9939
20.Linear regression & correlation
Answer: x = 21
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 = 100/5 = 20
ȳ = 479/5 = 95.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 248
Regression line
b = 248/40 = 6.2
a = 95.8 − 6.2·20 = −28.2
ŷ = −28.2 + 6.2·x
Solve for x
x = (102 − (−28.2))/6.2 = 21
Answer
x = 21
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