An annuity pays €1 500 at the end of each year for 6 years. The interest rate is 3% per year. Find the present value PV of the annuity.
A = €1 500i = 3%n = 6
PV = €
2.Time value of moneyHard
You borrow €20 000 at 6% per year, compounded monthly, and repay €350 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
P = €20 000A = €3506% / 12
N = payments
3.NPV & IRRMedium
A machine costs €40 000 today, lasts 7 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 8%. Find its equivalent annual cost EAC.
C0 = €40 000M = €2 000n = 7r = 8%
EAC = €
4.NPV & IRRHard
Project A costs €30 000 and returns €17 000, €12 000, €7 000 at the end of years 1 to 3. Project B costs €35 000 and returns €11 000, €14 500, €18 000. Find the crossover rate r* at which both projects have the same NPV, in percent.
You buy a bond with 11 years to maturity (face value €1 000, annual coupon 8%) when its yield to maturity is 6%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 5%. Find your holding period return HPR in percent.
F = €1 000coupon = 8%n = 11y0 = 6%y1 = 5%
HPR = %
6.Bond pricing & yieldHard
A bond trades at €1 085 with a yield to maturity of 4% (annual compounding) and a Macaulay duration of 4.1 years. Use the modified duration to estimate the new price P if the yield rises by 50 basis points.
P0 = €1 085y = 4%D = 4.1Δy = +50 bp
P = €
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €4 000 000, preferred shares worth €500 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 8% and the pre-tax cost of debt 3%. The tax rate is 21%. Find the WACC.
A company's shares have a beta of 1.4 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 25%, the risk-free rate 4% and the market risk premium 7%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 17, 20, 14, 23, 15, 45, 52.
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 52.5.
μ = 60σ = 5x = 52.5
z =
11.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 7? The order does not matter.
n = 7k = 3
N =
12.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?
n = 5p = 0.1k = 2
P =
13.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.5 and 179.75 cm? Give P as a decimal.
μ = 178σ = 7x1 = 167.5x2 = 179.75
P =
14.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 11.8. What is the margin of error E₂ if the same study uses 400 observations instead?
95%n₁ = 100E₁ = 11.8n₂ = 400
E₂ =
16.Confidence intervalsHard
A pilot study suggests that about 0.7 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 0.01?
p ≈ 0.7E = 0.0199%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 120.4 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 120.4s = 5n = 64
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 75 of 150 patients on treatment A recover, and 133 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 75n₁ = 150x₂ = 133n₂ = 250
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
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. Predict ŷ for x = 7.
n = 6x = 7
ŷ =
3Solutions
1.Time value of money
Answer: PV = 8126 €
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 − 0.8375)/0.03 = 5.417
Present value
PV = 1 500·5.417 = €8 125.79
Answer
PV = €8 125.79
2.Time value of money
Answer: N = 68 payments
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.
Rearrange the annuity formula
i = 0.005
(1 + i)−N = 1 − 20 000·0.005/350 = 0.7143
Solve with logarithms
N = −ln(0.7143)/ln(1.005) = 67.46
Round up to whole payments
67 full payments leave a small balance, which the last, smaller payment clears.
N = 68
Answer
N = 68 payments
3.NPV & IRR
Answer: EAC = 9683 €
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 − 1.08−7)/0.08 = 5.206
Annualised purchase price
40 000/5.206 = 7 682.90
EAC
EAC = 7 682.90 + 2 000 = €9 682.90
Answer
EAC = €9 682.90
4.NPV & IRR
Answer: r* = 9.426 %
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 = −35 000 − (−30 000) = −5 000
ΔCF1 = 11 000 − 17 000 = −6 000
ΔCF2 = 14 500 − 12 000 = 2 500
ΔCF3 = 18 000 − 7 000 = 11 000
IRR of the difference
NPVΔ(9%) = 93.63
NPVΔ(10%) = −123.97
r* = 9.426%
Answer
r* = 9.426%
Check: at 9.426% both NPVs are €899.92.
5.Bond pricing & yield
Answer: HPR = 13.29 %
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.
Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.
Modified duration
Dmod = 4.1/1.04 = 3.942
Estimated price change
%ΔP ≈ −3.942·(0.005) = −1.971%
P ≈ 1085·(1 − 0.01971) = €1 063.61
Answer
P = €1 063.61
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 7.142 %
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).
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 = 1.4/(1 + 0.75·0.2) = 1.217
Relever at the new D/E
βL = 1.217·(1 + 0.75·1.5) = 2.587
Cost of equity (CAPM)
re = 4% + 2.587·7% = 22.11%
Answer
re = 22.11%
9.Descriptive statistics
Answer: s = 15.41
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 = (52.5 − 60)/5 = −1.5
Answer
z = −1.5
11.Probability
Answer: N = 35
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, 3) = (7·6·5)/(3·2·1) = 35
Answer
N = 35
12.Probability
Answer: P = 0.0729
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 = 2) = 10·0.12·0.93 = 10·0.01·0.729 = 0.0729
Answer
P = 0.0729
13.Normal distribution
Answer: P = 0.5319
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 = (167.5 − 178)/7 = −1.5
z₂ = (179.75 − 178)/7 = 0.25
Area between
P = Φ(0.25) − Φ(−1.5) = 0.5987 − 0.06681 = 0.5319
Answer
P = 0.5319
14.Normal distribution
Answer: P = 0.7389
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 difference
μ = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
15.Confidence intervals
Answer: E₂ = 5.9
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.
Ratio of margins
E₂/E₁ = (1.96/1.96)·√(100/400) = 0.5
New margin
E₂ = 11.8·0.5 = 5.9
Answer
E₂ = 5.9
16.Confidence intervals
Answer: n = 13936
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 = 2.576²·0.7·0.3/0.01² = 13935
Round up
n = 13936
Answer
n = 13936
17.Hypothesis tests & confidence intervals
Answer: t = 0.64
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 = 5/√64 = 5/8 = 0.625
Test statistic
t = (120.4 − 120)/0.625 = 0.64
Answer
t = 0.64
t0.025; df = 63 = 1.998
|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −0.6202
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 = 75/150 = 0.5
p̂₂ = 133/250 = 0.532
p̂ = 208/400 = 0.52
Standard error under H₀
SE = √(0.52·0.48·(1/150 + 1/250)) = 0.0516
Test statistic
z = (0.5 − 0.532)/0.0516 = −0.6202
Answer
z = −0.6202
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r² = 0.9785
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 = 27/6 = 4.5
ȳ = 402/6 = 67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 100
Syy = Σ(y − ȳ)² = 584
Coefficient of determination
r² = 100²/(17.5·584) = 0.9785
Answer
r² = 0.9785
About 97.85% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 55.5
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 = 333/6 = 55.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 203
Regression line
b = 203/70 = 2.9
a = 55.5 − 2.9·7 = 35.2
ŷ = 35.2 + 2.9·x
Prediction
ŷ = 35.2 + 2.9·7 = 55.5
Answer
ŷ = 55.5
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