You invest €8 000 at a nominal rate of 3.6% per year, compounded quarterly. Find the value FV of the investment after 4 years.
PV = €8 0003.6% / 4n = 4
FV = €
2.Time value of moneyHard
You want €75 000 in 3 years. You save the same amount at the end of every month in an account paying 4.2% per year, compounded monthly. Find the monthly deposit A.
FV = €75 0004.2% / 12N = 36
A = €
3.NPV & IRRMedium
A project costs €30 000 today and returns €8 000 at the end of each year for 5 years. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
Project A costs €40 000 and returns €22 500, €16 000, €9 000 at the end of years 1 to 3. Project B costs €45 000 and returns €11 500, €17 000, €26 000. 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 an annual coupon of 8% and matures in 4 years. Its yield to maturity changes from 2.5% to 1%. By how much ΔP does the bond price change?
F = €1 000coupon = 8%n = 4y1 = 2.5%y2 = 1%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 3% matures in 3 years and sells for €960. Find its yield to maturity YTM in percent.
F = €1 000coupon = 3%n = 3P = €960
YTM = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €2 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 7% and the tax rate is 21%. Find the WACC.
A company's shares have a beta of 1.1 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 1. The tax rate is 25.8%, the risk-free rate 3% and the market risk premium 4%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 50, 10, 20, 58, 31, 20, 46.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 34?
μ = 50σ = 8x = 34
P =
11.ProbabilityMedium
A fair die is rolled 7 times. What is the probability P of getting at least one six?
n = 7
P =
12.ProbabilityHard
11 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 11p = 0.2
P =
13.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 98.6 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 98.6 s
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 90% of them lie (the 90th percentile).
μ = 500σ = 40P = 90%
x = h
15.Confidence intervalsMedium
A sample of 21 items has mean 23 and standard deviation s = 13. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 21x̄ = 23s = 1399%t(20) = 2.845
E =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 90 kg, standard deviation 1 kg. Plot group 2: 7 plots, mean 93 kg, standard deviation 1.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
A sample of 36 students has a mean score of 124.8 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 124.8s = 10n = 36
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 95%. The standard deviation is 7. Find the minimum sample size n.
α = 0.01power = 95%δ = 3σ = 7zα = 2.326zβ = 1.645
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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. A new observation (20, 119) is made. Find its residual e = y − ŷ.
n = 5(20, 119)
e =
3Solutions
1.Time value of money
Answer: FV = 9233 €
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.
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 8 000/1.06 = 7 547.17 → Σ = 7 547.17
year 2: 8 000/1.124 = 7 119.97 → Σ = 14 667.14
year 3: 8 000/1.191 = 6 716.95 → Σ = 21 384.10
year 4: 8 000/1.262 = 6 336.75 → Σ = 27 720.84
year 5: 8 000/1.338 = 5 978.07 → Σ = 33 698.91
Fraction of the last year
DPB = 4 + 2 279.16/5 978.07 = 4.381
Answer
DPB = 4.381 years
The discounted payback is always longer than the simple payback (3.75 years).
4.NPV & IRR
Answer: r* = 5.347 %
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 = 11 500 − 22 500 = −11 000
ΔCF2 = 17 000 − 16 000 = 1 000
ΔCF3 = 26 000 − 9 000 = 17 000
IRR of the difference
NPVΔ(5%) = 116.08
NPVΔ(6%) = −213.83
r* = 5.347%
Answer
r* = 5.347%
Check: at 5.347% both NPVs are €3 473.01.
5.Bond pricing & yield
Answer: ΔP = 66.23 €
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
€960 = €30·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³
Solve numerically
P(4%) = €972.25, P(5%) = €945.54
YTM = 4.454%
Answer
YTM = 4.454%
7.CAPM & WACC
Answer: WACC = 10.61 %
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 = €5 000 000, E/V = 0.6, D/V = 0.4
WACC
WACC = 0.6·14% + 0.4·7%·(1 − 0.21) = 10.61%
Answer
WACC = 10.61%
8.CAPM & WACC
Answer: re = 9.674 %
Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.
Unlever the beta
βU = 1.1/(1 + 0.742·0.2) = 0.9579
Relever at the new D/E
βL = 0.9579·(1 + 0.742·1) = 1.669
Cost of equity (CAPM)
re = 3% + 1.669·4% = 9.674%
Answer
re = 9.674%
9.Descriptive statistics
Answer: s = 18.04
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
11.Probability
Answer: P = 0.7209
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)7 = 0.2791
P = 1 − 0.2791 = 0.7209
Answer
P = 0.7209
12.Probability
Answer: P = 0.3221
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) = 0.811 = 0.0859
P(X = 1) = 11·0.2·0.810 = 0.2362
Add them
P = 0.0859 + 0.2362 = 0.3221
Answer
P = 0.3221
13.Normal distribution
Answer: P = 0.8413
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 sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (98.6 − 105)/6.403 = −1
Upper tail
P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
14.Normal distribution
Answer: x = 551.3 h
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) = 0.9 → z = 1.282
Back to x
x = 500 + 1.282·40 = 551.26 h
Answer
x = 551.26 h
15.Confidence intervals
Answer: E = 8.071
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 = 20, t = 2.845
Margin of error
E = 2.845·13/√21 = 8.071
Answer
E = 8.071
16.Confidence intervals
Answer: E = 1.214 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 = (6·1² + 6·1.5²)/12 = 1.625
sp = 1.275
Margin of error
E = 1.782·1.275·√(1/7 + 1/7) = 1.214
Answer
E = 1.214 kg
17.Hypothesis tests & confidence intervals
Answer: t = 2.88
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 = 10/√36 = 10/6 = 1.667
Test statistic
t = (124.8 − 120)/1.667 = 2.88
Answer
t = 2.88
t0.025; df = 35 = 2.03
|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 86
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 = ((2.326 + 1.645)·7/3)² = 85.85
Round up
n = 86
Answer
n = 86
19.Linear regression & correlation
Answer: r² = 0.7931
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 = 105/5 = 21
ȳ = 232/5 = 46.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 122
Syy = Σ(y − ȳ)² = 469.2
Coefficient of determination
r² = 122²/(40·469.2) = 0.7931
Answer
r² = 0.7931
About 79.31% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = 3.45
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 = 95/5 = 19
ȳ = 552/5 = 110.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 206
Regression line
b = 206/40 = 5.15
a = 110.4 − 5.15·19 = 12.55
ŷ = 12.55 + 5.15·x
Predicted value
ŷ = 12.55 + 5.15·20 = 115.5
Residual
e = 119 − 115.5 = 3.45
Answer
e = 3.45
The point lies above the line.
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