A credit card charges a nominal interest rate of 21% per year, compounded quarterly. Find the effective annual rate EAR in percent.
r = 21%m = 4
EAR = %
2.Time value of moneyHard
You save €250 at the end of every month for 26 years. The account pays 7.2% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €2507.2% / 12N = 312
FV = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 4: €31 000, €24 000, €29 000, €16 000. The cost of capital is 9%. Find the net present value NPV.
Project A costs €40 000 and returns €22 000, €15 500, €9 500 at the end of years 1 to 3. Project B costs €45 000 and returns €12 000, €17 000, €25 500. 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 2% and matures in 6 years. Its yield to maturity changes from 5% to 5.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 2%n = 6y1 = 5%y2 = 5.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 4 years. The yield to maturity is 4%. Find the Macaulay duration D in years.
F = €1 000coupon = 7%n = 4YTM = 4%
D = years
7.CAPM & WACCMedium
A share with a beta of 0.8 has a required return of 7%. The risk-free rate is 3%. What expected market return rm does CAPM imply?
β = 0.8re = 7%rf = 3%
rm = %
8.CAPM & WACCHard
A company has 1 000 000 shares at €44.50 each and 40 000 bonds trading at €1 000 each. The pre-tax cost of debt is 5% and the tax rate is 25.8%. The share’s beta is 1.1, the risk-free rate 1.5% and the market return 6.5%. Find the WACC.
β = 1.1rf = 1.5%rm = 6.5%rd = 5%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 24, 52, 28, 16, 16, 33.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 158?
μ = 170σ = 12x = 158
P =
11.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 15? The order does not matter.
n = 15k = 2
N =
12.ProbabilityHard
X has the distribution shown, and Y = 3X + 1. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 3X + 1
σY =
13.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85 and 112.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 85x2 = 112.75
P =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 65σ = 10P = 80%
x = points
15.Confidence intervalsMedium
A sample of 15 items has mean 58.5 and standard deviation s = 16.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 15x̄ = 58.5s = 16.599%t(14) = 2.977
lower limit =
16.Confidence intervalsHard
The population standard deviation is 39. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?
σ = 39E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 40 parts has mean length 133.5 mm, and the line's known standard deviation is 4 mm. Line 2: a sample of 50 parts has mean 138 mm, known standard deviation 17 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 80%. The standard deviation is 17. Find the minimum sample size n.
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 5
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. A new observation (5, 71) is made. Find its residual e = y − ŷ.
n = 5(5, 71)
e =
3Solutions
1.Time value of money
Answer: EAR = 22.71 %
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.
Rate per period
r/m = 0.21/4 = 0.0525
Compound over one year
EAR = (1 + 0.0525)4 − 1 = 1.227 − 1 = 22.71%
Answer
EAR = 22.71%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: FV = 227709 €
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 €78 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 7369 €
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 = −75 000
CF1/(1 + 0.09)1 = 31 000/1.09 = 28 440.37
CF2/(1 + 0.09)2 = 24 000/1.188 = 20 200.32
CF3/(1 + 0.09)3 = 29 000/1.295 = 22 393.32
CF4/(1 + 0.09)4 = 16 000/1.412 = 11 334.80
Add them up
NPV = €7 368.81
Answer
NPV = €7 368.81
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: r* = 7.083 %
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 = 12 000 − 22 000 = −10 000
ΔCF2 = 17 000 − 15 500 = 1 500
ΔCF3 = 25 500 − 9 500 = 16 000
IRR of the difference
NPVΔ(7%) = 25.13
NPVΔ(8%) = −271.94
r* = 7.083%
Answer
r* = 7.083%
Check: at 7.083% both NPVs are €1 798.95.
5.Bond pricing & yield
Answer: ΔP = −22.57 €
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.
Present value of each cash flow
t = 1: 70/1.04 = 67.31, t·PV = 67.31
t = 2: 70/1.082 = 64.72, t·PV = 129.44
t = 3: 70/1.125 = 62.23, t·PV = 186.69
t = 4: 1070/1.17 = 914.64, t·PV = 3 658.56
Duration
P = 1 108.90
D = 4 042.00/1 108.90 = 3.645
Answer
D = 3.645 years
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 = (7% − 3%)/0.8 = 5%
Market return
rm = 3% + 5% = 8%
Answer
rm = 8%
8.CAPM & WACC
Answer: WACC = 5.443 %
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 = 1 000 000·€44.50 = €44 500 000
D = 40 000·€1 000 = €40 000 000
Cost of equity (CAPM)
re = 1.5% + 1.1·5% = 7%
Weights
V = €84 500 000, E/V = 0.5266, D/V = 0.4734
WACC
WACC = 0.5266·7% + 0.4734·5%·(1 − 0.258) = 5.443%
Answer
WACC = 5.443%
9.Descriptive statistics
Answer: s = 13.45
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
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
11.Probability
Answer: N = 105
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, 2) = (15·14)/(2·1) = 105
Answer
N = 105
12.Probability
Answer: σY = 2.414
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Spread of X
μ = 1.05, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |3|·0.8047 = 2.414
Answer
σY = 2.414
The mean does move: E(Y) = 3·1.05 + 1 = 4.15.
13.Normal distribution
Answer: P = 0.6437
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 = (85 − 100)/15 = −1
z₂ = (112.75 − 100)/15 = 0.85
Area between
P = Φ(0.85) − Φ(−1) = 0.8023 − 0.1587 = 0.6437
Answer
P = 0.6437
14.Normal distribution
Answer: x = 73.42 points
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) = 0.8 → z = 0.8416
Back to x
x = 65 + 0.8416·10 = 73.42 points
Answer
x = 73.42 points
15.Confidence intervals
Answer: lower limit = 45.82
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 = 14, t = 2.977
Margin of error
E = 2.977·16.5/√15 = 12.68
Interval
58.5 ± 12.68 → [45.82, 71.18]
Answer
lower limit = 45.82
16.Confidence intervals
Answer: n = 631
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 = (2.576·39/4)² = 630.8
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 631
Answer
n = 631
17.Hypothesis tests & confidence intervals
Answer: z = −1.81
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.
Standard error of the difference
SE = √(4²/40 + 17²/50) = √6.18 = 2.486
Test statistic
z = (133.5 − 138)/2.486 = −1.81
Answer
z = −1.81
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 72
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 + 0.842)·17/5)² = 71.5
Round up
n = 72
Answer
n = 72
19.Linear regression & correlation
Answer: r = 0.9801
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 = 120/5 = 24
ȳ = 402/5 = 80.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 360
Syy = Σ(y − ȳ)² = 1499
Correlation coefficient
r = 360/√(90·1499) = 0.9801
Answer
r = 0.9801
20.Linear regression & correlation
Answer: e = 3.1
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 = 366/5 = 73.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 212
Regression line
b = 212/40 = 5.3
a = 73.2 − 5.3·6 = 41.4
ŷ = 41.4 + 5.3·x
Predicted value
ŷ = 41.4 + 5.3·5 = 67.9
Residual
e = 71 − 67.9 = 3.1
Answer
e = 3.1
The point lies above the line.
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