A credit card charges a nominal interest rate of 17.5% per year, compounded monthly. Find the effective annual rate EAR in percent.
r = 17.5%m = 12
EAR = %
2.Time value of moneyHard
A mortgage of €250 000 is repaid with equal monthly payments over 30 years at 2.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 3? Find the outstanding balance B.
P = €250 0002.4% / 12N = 360k = 36
B = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 4: €21 000, €26 000, €29 000, €20 000. The cost of capital is 5%. Find the net present value NPV.
A project costs €50 000 and returns €32 500, €23 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 32 500CF2 = 23 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 2% and matures in 12 years. Its yield to maturity changes from 5.5% to 7.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 2%n = 12y1 = 5.5%y2 = 7.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with 15 years to maturity (face value €1 000, annual coupon 5%) sells for €1 310. The issuer can call it in 3 years at a call price of €1 050. Find the yield to call YTC in percent.
F = €1 000coupon = 5%P = €1 310call price = €1 050call in 3 years
YTC = %
7.CAPM & WACCMedium
A share with a beta of 1.6 has a required return of 10.7%. The risk-free rate is 3.5%. What expected market return rm does CAPM imply?
β = 1.6re = 10.7%rf = 3.5%
rm = %
8.CAPM & WACCHard
A company's shares have a beta of 1.4 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 21%, the risk-free rate 3% and the market risk premium 5.5%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 46, 41, 51, 55, 20, 57.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 15. Find the z-score z of a score of 79.
μ = 70σ = 15x = 79
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.05, 0.35
Var(X) =
12.ProbabilityHard
A game costs €1 to play. You win €4 with probability 0.1 and €20 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€4: p = 0.1€20: p = 0.01
E(G) = €
13.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 27.2 and 41.3 min? Give P as a decimal.
μ = 32σ = 6x1 = 27.2x2 = 41.3
P =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points40%
x = points
15.Confidence intervalsMedium
A sample of 7 items has mean 52 and standard deviation s = 3. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 7x̄ = 52s = 395%t(6) = 2.447
E =
16.Confidence intervalsHard
The population standard deviation is 37. 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 1?
σ = 37E = 195%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 499 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 499s = 8n = 49
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.10 with a sample of 100; the standard deviation is known to be 5. In reality the mean is 50.6. Find the probability of a Type II error β in percent.
μ₀ = 50μ = 50.6σ = 5n = 100α = 0.10
β = %
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 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.
n = 5x = 3
ŷ =
3Solutions
1.Time value of money
Answer: EAR = 18.97 %
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 = 0.175/12 = 0.01458
Compound over one year
EAR = (1 + 0.01458)12 − 1 = 1.19 − 1 = 18.97%
Answer
EAR = 18.97%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: B = 232293 €
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 payment
i = 0.002, N = 360
A = 250 000·0.002/(1 − (1 + 0.002)−360) = €974.85
Present value of the remaining payments
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 360 − 36 = 324
B = 974.9·(1 − (1 + 0.002)−324)/0.002 = €232 293.17
Answer
B = €232 293.17
3.NPV & IRR
Answer: NPV = 10088 €
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 + 0.05)1 = 21 000/1.05 = 20 000.00
CF2/(1 + 0.05)2 = 26 000/1.103 = 23 582.77
CF3/(1 + 0.05)3 = 29 000/1.158 = 25 051.29
CF4/(1 + 0.05)4 = 20 000/1.216 = 16 454.05
Add them up
NPV = €10 088.11
Answer
NPV = €10 088.11
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 7.708 %
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.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(5%) = 1 814.06
NPV(10%) = −1 446.28
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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
1310 = 50·(1 − (1 + y)−3)/y + 1050/(1 + y)3
Solve numerically
y = 50%: €381.48
y = 50%: €381.48
YTC = 50%
Answer
YTC = 50%
The yield to maturity is 2.497%. 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 = (10.7% − 3.5%)/1.6 = 4.5%
Market return
rm = 3.5% + 4.5% = 8%
Answer
rm = 8%
8.CAPM & WACC
Answer: re = 15.78 %
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 + 0.79·0.4) = 1.064
Relever at the new D/E
βL = 1.064·(1 + 0.79·1.5) = 2.324
Cost of equity (CAPM)
re = 3% + 2.324·5.5% = 15.78%
Answer
re = 15.78%
9.Descriptive statistics
Answer: s = 13.58
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 = (79 − 70)/15 = 0.6
Answer
z = 0.6
11.Probability
Answer: Var(X) = 164
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.
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.
Net outcomes
G: 3 (p = 0.1), 19 (p = 0.01), −1 (p = 0.89)
Expected value
E(G) = 3·0.1 + 19·0.01 − 1·0.89 = −0.4 €
Answer
E(G) = −0.4 €
A negative value means you lose money on average.
13.Normal distribution
Answer: P = 0.7276
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 = (27.2 − 32)/6 = −0.8
z₂ = (41.3 − 32)/6 = 1.55
Area between
P = Φ(1.55) − Φ(−0.8) = 0.9394 − 0.2119 = 0.7276
Answer
P = 0.7276
14.Normal distribution
Answer: x = 67.5 points
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 65 + 0.25·10 = 67.5 points
Answer
x = 67.5 points
15.Confidence intervals
Answer: E = 2.775
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 = 6, t = 2.447
Margin of error
E = 2.447·3/√7 = 2.775
Answer
E = 2.775
16.Confidence intervals
Answer: n = 5260
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 = (1.96·37/1)² = 5259
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 5260
Answer
n = 5260
17.Hypothesis tests & confidence intervals
Answer: t = −0.875
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/√49 = 8/7 = 1.143
Test statistic
t = (499 − 500)/1.143 = −0.875
Answer
t = −0.875
t0.025; df = 48 = 2.011
|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: β = 53.27 %
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 = 5/√100 = 0.5
x̄c = 50 + 1.282·0.5 = 50.64
z under the true mean
z = (50.64 − 50.6)/0.5 = 0.082
Type II error
power = P(Z > 0.082) = 46.73%
β = 100% − 46.73% = 53.27%
Answer
β = 53.27%
19.Linear regression & correlation
Answer: r² = 0.9544
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 = 33/6 = 5.5
ȳ = 469/6 = 78.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 101.5
Syy = Σ(y − ȳ)² = 616.8
Coefficient of determination
r² = 101.5²/(17.5·616.8) = 0.9544
Answer
r² = 0.9544
About 95.44% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 12.6
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 = 120/5 = 24
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 152
Regression line
b = 152/40 = 3.8
a = 24 − 3.8·6 = 1.2
ŷ = 1.2 + 3.8·x
Prediction
ŷ = 1.2 + 3.8·3 = 12.6
Answer
ŷ = 12.6
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