You invest €20 000 at a nominal rate of 4.8% per year, compounded monthly. Find the value FV of the investment after 8 years.
PV = €20 0004.8% / 12n = 8
FV = €
2.Time value of moneyHard
A loan of €20 000 is repaid with equal monthly payments over 6 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.
P = €20 0003% / 12N = 72
A = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 3: €14 000, €13 000, €18 000. The cost of capital is 6%. Find the net present value NPV.
A project costs €50 000 today. Its cash flow is €5 000 at the end of year 1 and grows by 2% per year until the last cash flow at the end of year 10. The cost of capital is 10%. Find the net present value NPV.
C0 = €50 000CF1 = €5 000g = 2%n = 10r = 10%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 14 years. Its yield to maturity changes from 8% to 6.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 6%n = 14y1 = 8%y2 = 6.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 4 years. It sells for €769. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 2% (2×)n = 4P = €769
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €3 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 6% and the tax rate is 30%. Find the WACC.
A company has 1 000 000 shares trading at €23.00; the book value of its equity is €9 200 000. It just paid a dividend of €2.60 per share, and dividends grow by 4% per year. Its debt has a market value of €13 800 000 and a pre-tax cost of 4%. The tax rate is 30%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 51, 40, 15, 45, 24.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 46.
μ = 70σ = 20x = 46
z =
11.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 15? The order does not matter.
n = 15k = 5
N =
12.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 4 correct answers?
n = 11p = 0.2k = 4
P =
13.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 240.25 and 254 g? Give P as a decimal.
μ = 250σ = 5x1 = 240.25x2 = 254
P =
14.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 32σ = 6P = 97.5%
x = min
15.Confidence intervalsMedium
A sample of 21 items has mean 91 and standard deviation s = 4.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 21x̄ = 91s = 4.595%t(20) = 2.086
E =
16.Confidence intervalsHard
A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.05?
p ≈ 0.15E = 0.0599%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 197.1 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 197.1s = 8n = 36
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 50 of 250 patients on treatment A recover, and 24 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 50n₁ = 250x₂ = 24n₂ = 200
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.
n = 5x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 29340 €
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.
Monthly rate and number of payments
i = 0.03/12 = 0.0025
N = 12·6 = 72
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.8355
A = 20 000·0.0025/(1 − 0.8355) = €303.87
Answer
A = €303.87
3.NPV & IRR
Answer: NPV = −5109 €
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 = −45 000
CF1/(1 + 0.06)1 = 14 000/1.06 = 13 207.55
CF2/(1 + 0.06)2 = 13 000/1.124 = 11 569.95
CF3/(1 + 0.06)3 = 18 000/1.191 = 15 113.15
Add them up
NPV = €−5 109.35
Answer
NPV = €−5 109.35
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: NPV = −16873 €
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 the growing annuity
((1.02)/(1.1))10 = 0.47
PV = 5 000/(0.1 − 0.02)·(1 − 0.47) = 33 126.56
NPV
NPV = 33 126.56 − 50 000 = −€16 873.44
Answer
NPV = −€16 873.44
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: ΔP = 119.8 €
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.
Per period
C = €10.00, N = 8
Solve numerically
i = 4.503% per half-year
YTM
YTM = 2·4.503% = 9.005%
Answer
YTM = 9.005%
7.CAPM & WACC
Answer: WACC = 9.229 %
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 = €7 000 000, E/V = 0.5714, D/V = 0.4286
WACC
WACC = 0.5714·13% + 0.4286·6%·(1 − 0.3) = 9.229%
Answer
WACC = 9.229%
8.CAPM & WACC
Answer: WACC = 10.9 %
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).
Cost of equity
D1 = 2.6·1.04 = €2.70
re = 2.704/23 + 0.04 = 15.76%
Market values
E = 1 000 000·€23.00 = €23 000 000
V = €36 800 000, E/V = 0.625, D/V = 0.375
WACC
WACC = 0.625·15.76% + 0.375·4%·(1 − 0.3) = 10.9%
Answer
WACC = 10.9%
9.Descriptive statistics
Answer: s = 15.02
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 = (46 − 70)/20 = −1.2
Answer
z = −1.2
11.Probability
Answer: N = 3003
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(15, 5) = (15·14·13·12·11)/(5·4·3·2·1) = 3003
Answer
N = 3003
12.Probability
Answer: P = 0.1107
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.
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 = (240.25 − 250)/5 = −1.95
z₂ = (254 − 250)/5 = 0.8
Area between
P = Φ(0.8) − Φ(−1.95) = 0.7881 − 0.02559 = 0.7626
Answer
P = 0.7626
14.Normal distribution
Answer: x = 43.76 min
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) = 0.975 → z = 1.96
Back to x
x = 32 + 1.96·6 = 43.76 min
Answer
x = 43.76 min
15.Confidence intervals
Answer: E = 2.048
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 = 2.086
Margin of error
E = 2.086·4.5/√21 = 2.048
Answer
E = 2.048
16.Confidence intervals
Answer: n = 339
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 = 2.576²·0.15·0.85/0.05² = 338.4
Round up
n = 339
Answer
n = 339
17.Hypothesis tests & confidence intervals
Answer: t = −2.175
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 = 8/√36 = 8/6 = 1.333
Test statistic
t = (197.1 − 200)/1.333 = −2.175
Answer
t = −2.175
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: z = 2.275
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.
Sample proportions
p̂₁ = 50/250 = 0.2
p̂₂ = 24/200 = 0.12
p̂ = 74/450 = 0.1644
Standard error under H₀
SE = √(0.1644·0.8356·(1/250 + 1/200)) = 0.03517
Test statistic
z = (0.2 − 0.12)/0.03517 = 2.275
Answer
z = 2.275
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9694
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 = 66/6 = 11
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 98
Syy = Σ(y − ȳ)² = 146
Correlation coefficient
r = 98/√(70·146) = 0.9694
Answer
r = 0.9694
20.Linear regression & correlation
Answer: ŷ = 59.85
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 = 20/5 = 4
ȳ = 291/5 = 58.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33
Regression line
b = 33/10 = 3.3
a = 58.2 − 3.3·4 = 45
ŷ = 45 + 3.3·x
Prediction
ŷ = 45 + 3.3·4.5 = 59.85
Answer
ŷ = 59.85
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