A credit card charges a nominal interest rate of 25% per year, compounded semi-annually. Find the effective annual rate EAR in percent.
r = 25%m = 2
EAR = %
2.Time value of moneyHard
A mortgage of €150 000 is repaid with equal monthly payments over 25 years at 3% per year, compounded monthly. How much is still owed just after the payment at the end of year 17? Find the outstanding balance B.
P = €150 0003% / 12N = 300k = 204
B = €
3.NPV & IRRMedium
A project costs €500 000 today. It generates €65 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 11%. Find the net present value NPV.
C0 = €500 000CF1 = €65 000g = 1%r = 11%
NPV = €
4.NPV & IRRHard
An investment of €80 000 returns €19 000 at the end of each year for 10 years. Find its internal rate of return IRR in percent.
C0 = €80 000CF = €19 000n = 10
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 10 years. Its yield to maturity changes from 3.5% to 3%. By how much ΔP does the bond price change?
F = €1 000coupon = 3%n = 10y1 = 3.5%y2 = 3%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 9 years and sells for €935. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 9P = €935
YTM = %
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €4 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 3% and the tax rate is 25%. Find the WACC.
A company has 5 000 000 shares trading at €36.00; the book value of its equity is €108 000 000. It just paid a dividend of €2.90 per share, and dividends grow by 2% per year. Its debt has a market value of €108 000 000 and a pre-tax cost of 6%. The tax rate is 28%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 29, 52, 18, 15, 11, 15.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 73.2?
μ = 70σ = 8x = 73.2
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.35, 0.05, 0.3, 0.3
σ =
12.ProbabilityHard
A game costs €1 to play. You win €2 with probability 0.3 and €20 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€2: p = 0.3€20: p = 0.05
E(G) = €
13.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 440 and 532 h? Give P as a decimal.
μ = 500σ = 40x1 = 440x2 = 532
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 top 30% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points30%
x = points
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 80 phones, mean 140 h, known standard deviation 10 h. Model 2: 80 phones, mean 142 h, known standard deviation 6 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 13. 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 1.5?
σ = 13E = 1.599%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 16 students, mean score 60, standard deviation 9. Group 2: 5 students, mean score 71.7, standard deviation 6. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 16x̄₁ = 60s₁ = 9n₂ = 5x̄₂ = 71.7s₂ = 6
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 199.3. The population standard deviation is 12. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 199.3σ = 12n = 2595%
L =
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. A new observation (3, 8) is made. Find its residual e = y − ŷ.
n = 5(3, 8)
e =
3Solutions
1.Time value of money
Answer: EAR = 26.56 %
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 = 0.25/2 = 0.125
Compound over one year
EAR = (1 + 0.125)2 − 1 = 1.266 − 1 = 26.56%
Answer
EAR = 26.56%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: B = 60643 €
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 = 0.0025, N = 300
A = 150 000·0.0025/(1 − (1 + 0.0025)−300) = €711.32
Present value of the remaining payments
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 300 − 204 = 96
B = 711.3·(1 − (1 + 0.0025)−96)/0.0025 = €60 643.05
Answer
B = €60 643.05
3.NPV & IRR
Answer: NPV = 150000 €
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.
Value of the cash flows
PV = 65 000/(0.11 − 0.01) = 650 000.00
NPV
NPV = 650 000.00 − 500 000 = €150 000.00
Answer
NPV = €150 000.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 19.87 %
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)−10)/i = 80 000/19 000 = 4.211
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 19%: 4.339
i = 20%: 4.192
IRR = 19.87%
Answer
IRR = 19.87%
5.Bond pricing & yield
Answer: ΔP = 41.58 €
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
€935 = €60·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹
Solve numerically
P(6%) = €1 000.00, P(7%) = €934.85
YTM = 6.998%
Answer
YTM = 6.998%
7.CAPM & WACC
Answer: WACC = 9.929 %
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 = €14 000 000, E/V = 0.7143, D/V = 0.2857
WACC
WACC = 0.7143·13% + 0.2857·3%·(1 − 0.25) = 9.929%
Answer
WACC = 9.929%
8.CAPM & WACC
Answer: WACC = 8.005 %
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).
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 0.6554
Answer
P = 0.6554
11.Probability
Answer: σ = 12.44
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: 1 (p = 0.3), 19 (p = 0.05), −1 (p = 0.65)
Expected value
E(G) = 1·0.3 + 19·0.05 − 1·0.65 = 0.6 €
Answer
E(G) = 0.6 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.7213
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 = (440 − 500)/40 = −1.5
z₂ = (532 − 500)/40 = 0.8
Area between
P = Φ(0.8) − Φ(−1.5) = 0.7881 − 0.06681 = 0.7213
Answer
P = 0.7213
14.Normal distribution
Answer: x = 107.8 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) = 0.7
z from the table
z ≈ 0.52
Back to the original units
x = 100 + 0.52·15 = 107.8 points
Answer
x = 107.8 points
15.Confidence intervals
Answer: E = 3.359 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 = 140 − 142 = −2
Standard error
SE = √(10²/80 + 6²/80) = 1.304
Margin of error
E = 2.576·1.304 = 3.359
Answer
E = 3.359 h
The interval contains 0, so the data do not show a significant difference.
16.Confidence intervals
Answer: n = 499
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 = (2.576·13/1.5)² = 498.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 499
Answer
n = 499
17.Hypothesis tests & confidence intervals
Answer: t = −2.7
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.
Pooled variance
sp² = (15·9² + 4·6²)/19 = 71.53
sp = 8.457
Standard error
SE = 8.457·√(1/16 + 1/5) = 4.333
Test statistic
t = (60 − 71.7)/4.333 = −2.7
Answer
t = −2.7
With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: L = 194.6
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).
Margin of error
E = 1.96·12/√25 = 4.704
Interval limit
L = 199.3 − 4.704 = 194.6
Answer
L = 194.6
19.Linear regression & correlation
Answer: r² = 0.8274
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 = 360/6 = 60
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 53
Syy = Σ(y − ȳ)² = 194
Coefficient of determination
r² = 53²/(17.5·194) = 0.8274
Answer
r² = 0.8274
About 82.74% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = −1.5
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 = 30/5 = 6
ȳ = 103/5 = 20.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 148
Regression line
b = 148/40 = 3.7
a = 20.6 − 3.7·6 = −1.6
ŷ = −1.6 + 3.7·x
Predicted value
ŷ = −1.6 + 3.7·3 = 9.5
Residual
e = 8 − 9.5 = −1.5
Answer
e = −1.5
The point lies below the line.
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