A credit card charges a nominal interest rate of 24% per year, compounded semi-annually. Find the effective annual rate EAR in percent.
r = 24%m = 2
EAR = %
2.Time value of moneyHard
A mortgage of €250 000 is repaid with equal monthly payments over 20 years at 5.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 13? Find the outstanding balance B.
P = €250 0005.4% / 12N = 240k = 156
B = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 3: €10 000, €11 000, €16 000. The cost of capital is 15%. Find the net present value NPV.
An investment of €20 000 returns €5 000 at the end of each year for 5 years. Find its internal rate of return IRR in percent.
C0 = €20 000CF = €5 000n = 5
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 13 years. Its yield to maturity changes from 9% to 8.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 4%n = 13y1 = 9%y2 = 8.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 7 years and sells for €1 090. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 7P = €1 090
YTM = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €4 000 000. Its WACC is 8.15%, its pre-tax cost of debt is 4.5% and the tax rate is 25.8%. Find the cost of equity re.
A company's business has an unlevered (asset) beta of 0.7. It plans a debt-to-equity ratio of 0.5. The tax rate is 25%, the risk-free rate is 1% and the market risk premium is 6%. Find the cost of equity re at the new capital structure.
βU = 0.7D/E = 0.5T = 25%rf = 1%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 11, 14, 21, 53, 20, 46, 12.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 58?
μ = 50σ = 8x = 58
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.1, 0.3, 0.3, 0.3
σ =
12.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?
n = 6p = 0.25k = 1
P =
13.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.8 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.8 mm
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h30%
x = h
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 100 phones, mean 40 h, known standard deviation 13 h. Model 2: 80 phones, mean 30.5 h, known standard deviation 10 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 38. 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 2?
σ = 38E = 290%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 124.7 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 124.7s = 20n = 81
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 95%. The standard deviation is 10. Find the minimum sample size n.
The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (16.5, 47) is made. Find its residual e = y − ŷ.
n = 6(16.5, 47)
e =
3Solutions
1.Time value of money
Answer: EAR = 25.44 %
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 = 0.24/2 = 0.12
Compound over one year
EAR = (1 + 0.12)2 − 1 = 1.254 − 1 = 25.44%
Answer
EAR = 25.44%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: B = 119086 €
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.
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 240 − 156 = 84
B = 1706·(1 − (1 + 0.0045)−84)/0.0045 = €119 086.34
Answer
B = €119 086.34
3.NPV & IRR
Answer: NPV = −17467 €
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 + 0.15)1 = 10 000/1.15 = 8 695.65
CF2/(1 + 0.15)2 = 11 000/1.322 = 8 317.58
CF3/(1 + 0.15)3 = 16 000/1.521 = 10 520.26
Add them up
NPV = €−17 466.51
Answer
NPV = €−17 466.51
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 7.931 %
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.
Required annuity factor
(1 − (1 + i)−5)/i = 20 000/5 000 = 4
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 7%: 4.1
i = 8%: 3.993
IRR = 7.931%
Answer
IRR = 7.931%
5.Bond pricing & yield
Answer: ΔP = 28.25 €
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
€1 090 = €60·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(4%) = €1 120.04, P(5%) = €1 057.86
YTM = 4.474%
Answer
YTM = 4.474%
7.CAPM & WACC
Answer: re = 12 %
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 = €9 000 000, E/V = 0.5556, D/V = 0.4444
Debt part of the WACC
0.4444·4.5%·(1 − 0.258) = 1.484%
Cost of equity
re = (8.15% − 1.484%)/0.5556 = 12%
Answer
re = 12%
8.CAPM & WACC
Answer: re = 6.775 %
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.
Levered beta
βL = 0.7·(1 + 0.75·0.5) = 0.9625
Cost of equity (CAPM)
re = 1% + 0.9625·6% = 6.775%
Answer
re = 6.775%
9.Descriptive statistics
Answer: s = 17.09
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) = 0.8413
P = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
11.Probability
Answer: σ = 0.9798
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.
Mean
μ = 1·0.1 + 2·0.3 + 3·0.3 + 4·0.3 = 2.8
Mean of the squares
E(X²) = 1·0.1 + 4·0.3 + 9·0.3 + 16·0.3 = 8.8
Variance
Var(X) = 8.8 − 2.8² = 0.96
σ = √0.96 = 0.9798
Answer
σ = 0.9798
12.Probability
Answer: P = 0.356
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 = 1) = 6·0.251·0.755 = 6·0.25·0.2373 = 0.356
Answer
P = 0.356
13.Normal distribution
Answer: P = 0.0548
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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (200.8 − 200)/0.5 = 1.6
Upper tail
P = 1 − Φ(1.6) = 1 − 0.9452 = 0.0548
Answer
P = 0.0548
14.Normal distribution
Answer: x = 479 h
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) = 0.3
z from the table
z ≈ −0.52
Back to the original units
x = 500 − 0.52·40 = 479 h
Answer
x = 479 h
15.Confidence intervals
Answer: upper limit = 12.32 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Point estimate
x̄₁ − x̄₂ = 40 − 30.5 = 9.5
Standard error
SE = √(13²/100 + 10²/80) = 1.715
Margin of error
E = 1.645·1.715 = 2.821
Interval
9.5 ± 2.821 → [6.679, 12.32]
Answer
upper limit = 12.32 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 977
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 = (1.645·38/2)² = 976.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 977
Answer
n = 977
17.Hypothesis tests & confidence intervals
Answer: t = 2.115
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 = 20/√81 = 20/9 = 2.222
Test statistic
t = (124.7 − 120)/2.222 = 2.115
Answer
t = 2.115
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 271
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 + 1.645)·10/2)² = 270.6
Round up
n = 271
Answer
n = 271
19.Linear regression & correlation
Answer: r = 0.9611
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.
Centred sums
Sxy = 1838 − 30·286/5 = 122
Sxx = 220 − 30²/5 = 40
Syy = 16762 − 286²/5 = 402.8
Correlation coefficient
r = 122/√(40·402.8) = 0.9611
Answer
r = 0.9611
20.Linear regression & correlation
Answer: e = 5.21
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 = 135/6 = 22.5
ȳ = 392/6 = 65.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 618
Regression line
b = 618/157.5 = 3.924
a = 65.33 − 3.924·22.5 = −22.95
ŷ = −22.95 + 3.924·x
Predicted value
ŷ = −22.95 + 3.924·16.5 = 41.79
Residual
e = 47 − 41.79 = 5.21
Answer
e = 5.21
The point lies above the line.
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