A credit card charges a nominal interest rate of 22% per year, compounded daily. Find the effective annual rate EAR in percent.
r = 22%m = 365
EAR = %
2.Time value of moneyHard
You save €100 at the end of every month for 19 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €1005.4% / 12N = 228
FV = €
3.NPV & IRRMedium
A project costs €40 000 today and returns €13 000 at the end of each year for 4 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 €30 000 and returns €16 000, €11 500, €7 000 at the end of years 1 to 3. Project B costs €35 000 and returns €9 500, €13 500, €18 000. 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 8% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 7%n = 5
P = €
6.Bond pricing & yieldHard
A bond with 20 years to maturity (face value €1 000, annual coupon 7%) sells for €1 115. The issuer can call it in 5 years at a call price of €1 070. Find the yield to call YTC in percent.
F = €1 000coupon = 7%P = €1 115call price = €1 070call in 5 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €12 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 3% and the tax rate is 30%. Find the WACC.
A company has 4 000 000 shares at €56.50 each and 40 000 bonds trading at €1 020 each. The pre-tax cost of debt is 4% and the tax rate is 20%. The share’s beta is 0.6, the risk-free rate 2% and the market return 8%. Find the WACC.
β = 0.6rf = 2%rm = 8%rd = 4%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 55, 51, 54, 19, 51.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 48?
μ = 60σ = 10x = 48
P =
11.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
12.ProbabilityHard
A game costs €2 to play. You win €6 with probability 0.25 and €40 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€6: p = 0.25€40: p = 0.05
E(G) = €
13.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 735.2 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 735.2 g
P =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points10%
x = points
15.Confidence intervalsMedium
A sample of 13 items has mean 115 and standard deviation s = 9. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 13x̄ = 115s = 995%t(12) = 2.179
E =
16.Confidence intervalsHard
The population standard deviation is 27. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 5?
σ = 27E = 590%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 30 parts has mean length 82.5 mm, and the line's known standard deviation is 11 mm. Line 2: a sample of 30 parts has mean 92.6 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 390%
n =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. 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 = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 21?
n = 6ŷ = 21
x =
3Solutions
1.Time value of money
Answer: EAR = 24.6 %
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 = 0.22/365 = 0.0006027
Compound over one year
EAR = (1 + 0.0006027)365 − 1 = 1.246 − 1 = 24.6%
Answer
EAR = 24.6%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: FV = 39633 €
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 €22 800 yourself; the rest is interest.
3.NPV & IRR
Answer: DPB = 3.51 years
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: 13 000/1.06 = 12 264.15 → Σ = 12 264.15
year 2: 13 000/1.124 = 11 569.95 → Σ = 23 834.10
year 3: 13 000/1.191 = 10 915.05 → Σ = 34 749.16
year 4: 13 000/1.262 = 10 297.22 → Σ = 45 046.37
Fraction of the last year
DPB = 3 + 5 250.84/10 297.22 = 3.51
Answer
DPB = 3.51 years
The discounted payback is always longer than the simple payback (3.077 years).
4.NPV & IRR
Answer: r* = 5.515 %
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 = 9 500 − 16 000 = −6 500
ΔCF2 = 13 500 − 11 500 = 2 000
ΔCF3 = 18 000 − 7 000 = 11 000
IRR of the difference
NPVΔ(5%) = 125.80
NPVΔ(6%) = −116.27
r* = 5.515%
Answer
r* = 5.515%
Check: at 5.515% both NPVs are €1 451.98.
5.Bond pricing & yield
Answer: P = 1042 €
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 = €40.00, i = 0.035, N = 10
Present value of the coupons
€40.00·(1 − 0.7089)/0.035 = €332.66
Present value of the face value
€1 000·0.7089 = €708.92
Price
P = €1 041.58
Answer
P = €1 041.58
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 5.556 %
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
1115 = 70·(1 − (1 + y)−5)/y + 1070/(1 + y)5
Solve numerically
y = 5%: €1 141.44
y = 6%: €1 094.43
YTC = 5.556%
Answer
YTC = 5.556%
The yield to maturity is 5.998%. 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: WACC = 7.367 %
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 = €18 000 000, E/V = 0.6667, D/V = 0.3333
WACC
WACC = 0.6667·10% + 0.3333·3%·(1 − 0.3) = 7.367%
Answer
WACC = 7.367%
8.CAPM & WACC
Answer: WACC = 5.233 %
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 = 4 000 000·€56.50 = €226 000 000
D = 40 000·€1 020 = €40 800 000
Cost of equity (CAPM)
re = 2% + 0.6·6% = 5.6%
Weights
V = €266 800 000, E/V = 0.8471, D/V = 0.1529
WACC
WACC = 0.8471·5.6% + 0.1529·4%·(1 − 0.2) = 5.233%
Answer
WACC = 5.233%
9.Descriptive statistics
Answer: s = 15.2
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.
Mean
x̄ = 230/5 = 46
Squared deviations
Σ(x − x̄)² = 9² + 5² + 8² + (−27)² + 5² = 924
Standard deviation
s = √(924/4) = 15.2
Answer
s = 15.2
10.Descriptive statistics
Answer: P = 0.8849
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849
Answer
P = 0.8849
11.Probability
Answer: P = 0.7674
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)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
12.Probability
Answer: E(G) = 1.5 €
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: 4 (p = 0.25), 38 (p = 0.05), −2 (p = 0.7)
Expected value
E(G) = 4·0.25 + 38·0.05 − 2·0.7 = 1.5 €
Answer
E(G) = 1.5 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.9332
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (735.2 − 750)/9.849 = −1.5
Upper tail
P = 1 − Φ(−1.5) = 1 − 0.06681 = 0.9332
Answer
P = 0.9332
14.Normal distribution
Answer: x = 80.8 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) = 0.1
z from the table
z ≈ −1.28
Back to the original units
x = 100 − 1.28·15 = 80.8 points
Answer
x = 80.8 points
15.Confidence intervals
Answer: E = 5.439
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 = 12, t = 2.179
Margin of error
E = 2.179·9/√13 = 5.439
Answer
E = 5.439
16.Confidence intervals
Answer: n = 79
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 = (1.645·27/5)² = 78.91
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 79
Answer
n = 79
17.Hypothesis tests & confidence intervals
Answer: z = −3.107
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 = √(11²/30 + 14²/30) = √10.57 = 3.251
Test statistic
z = (82.5 − 92.6)/3.251 = −3.107
Answer
z = −3.107
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 31
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).
Solve for n
n = (1.645·10/3)² = 30.07
Round up
n = 31
Answer
n = 31
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 = 153/6 = 25.5
ȳ = 426/6 = 71
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 495
Syy = Σ(y − ȳ)² = 1630
Coefficient of determination
r² = 495²/(157.5·1630) = 0.9544
Answer
r² = 0.9544
About 95.44% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: x = 2.563
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 = 21/6 = 3.5
ȳ = 108/6 = 18
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −56
Regression line
b = −56/17.5 = −3.2
a = 18 − (−3.2)·3.5 = 29.2
ŷ = 29.2 − 3.2·x
Solve for x
x = (21 − 29.2)/(−3.2) = 2.563
Answer
x = 2.563
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