You invest €8 000 at a nominal rate of 2% per year, compounded quarterly. Find the value FV of the investment after 8 years.
PV = €8 0002% / 4n = 8
FV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 25 years at 5.4% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €200 0005.4% / 12N = 300
I = €
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 4: €13 000, €15 000, €11 000, €15 000. The cost of capital is 6%. Find the net present value NPV.
An investment of €50 000 returns €10 500 at the end of each year for 8 years. Find its internal rate of return IRR in percent.
C0 = €50 000CF = €10 500n = 8
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 4 years. The yield to maturity is 5.5% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 5.5%n = 4
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 9 years and sells for €930. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 9P = €930
YTM = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €5 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 7% and the tax rate is 25%. Find the WACC.
E = €8 000 000D = €5 000 000re = 8%rd = 7%T = 25%
WACC = %
8.CAPM & WACCHard
A company has 4 000 000 shares trading at €28.50; the book value of its equity is €34 200 000. It just paid a dividend of €1.60 per share, and dividends grow by 4% per year. Its debt has a market value of €91 200 000 and a pre-tax cost of 4.5%. 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: 15, 37, 26, 12, 54.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 140?
μ = 100σ = 20x = 140
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.2, 0.25, 0.3, 0.25
Var(X) =
12.ProbabilityHard
9 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 9p = 0.25
P =
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 759.8 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 759.8 g
P =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 90% of them lie (the 90th percentile).
μ = 100σ = 15P = 90%
x = points
15.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 11.2. What is the margin of error E₂ of a 90% interval from the same data?
99%n₁ = 25E₁ = 11.290%
E₂ =
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
20 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −7.31 and the standard deviation of the differences is 9.9. Find the paired test statistic t for H₀: μd = 0.
n = 20d̄ = −7.31sd = 9.9
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 90 of 150 patients on treatment A recover, and 96 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 90n₁ = 150x₂ = 96n₂ = 150
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 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 = 19?
n = 5ŷ = 19
x =
3Solutions
1.Time value of money
Answer: FV = 9384 €
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.
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.
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 = −30 000
CF1/(1 + 0.06)1 = 13 000/1.06 = 12 264.15
CF2/(1 + 0.06)2 = 15 000/1.124 = 13 349.95
CF3/(1 + 0.06)3 = 11 000/1.191 = 9 235.81
CF4/(1 + 0.06)4 = 15 000/1.262 = 11 881.40
Add them up
NPV = €16 731.31
Answer
NPV = €16 731.31
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 13.23 %
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)−8)/i = 50 000/10 500 = 4.762
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 13%: 4.799
i = 14%: 4.639
IRR = 13.23%
Answer
IRR = 13.23%
5.Bond pricing & yield
Answer: P = 1053 €
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 = €35.00, i = 0.0275, N = 8
Present value of the coupons
€35.00·(1 − 0.8049)/0.0275 = €248.30
Present value of the face value
€1 000·0.8049 = €804.91
Price
P = €1 053.21
Answer
P = €1 053.21
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 4.984 %
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
€930 = €40·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹
Solve numerically
P(4%) = €1 000.00, P(5%) = €928.92
YTM = 4.984%
Answer
YTM = 4.984%
7.CAPM & WACC
Answer: WACC = 6.942 %
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 = €13 000 000, E/V = 0.6154, D/V = 0.3846
WACC
WACC = 0.6154·8% + 0.3846·7%·(1 − 0.25) = 6.942%
Answer
WACC = 6.942%
8.CAPM & WACC
Answer: WACC = 6.906 %
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
11.Probability
Answer: Var(X) = 4.56
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 = 0·0.2 + 2·0.25 + 4·0.3 + 6·0.25 = 3.2
Mean of the squares
E(X²) = 0·0.2 + 4·0.25 + 16·0.3 + 36·0.25 = 14.8
Variance
Var(X) = 14.8 − 3.2² = 4.56
Answer
Var(X) = 4.56
12.Probability
Answer: P = 0.3003
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.
Zero and one failure
P(X = 0) = 0.759 = 0.07508
P(X = 1) = 9·0.25·0.758 = 0.2253
Add them
P = 0.07508 + 0.2253 = 0.3003
Answer
P = 0.3003
13.Normal distribution
Answer: P = 0.1587
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 = (759.8 − 750)/9.849 = 1
Upper tail
P = 1 − Φ(1) = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
14.Normal distribution
Answer: x = 119.2 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) = 0.9 → z = 1.282
Back to x
x = 100 + 1.282·15 = 119.22 points
Answer
x = 119.22 points
15.Confidence intervals
Answer: E₂ = 7.152
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.
Ratio of margins
E₂/E₁ = (1.645/2.576)·√(25/25) = 0.6386
New margin
E₂ = 11.2·0.6386 = 7.152
Answer
E₂ = 7.152
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 = −3.302
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 of the mean difference
SE = 9.9/√20 = 2.214
Test statistic
t = −7.31/2.214 = −3.302
Answer
t = −3.302
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: z = −0.7137
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.
Sample proportions
p̂₁ = 90/150 = 0.6
p̂₂ = 96/150 = 0.64
p̂ = 186/300 = 0.62
Standard error under H₀
SE = √(0.62·0.38·(1/150 + 1/150)) = 0.05605
Test statistic
z = (0.6 − 0.64)/0.05605 = −0.7137
Answer
z = −0.7137
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9971
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 = 2797 − 48·321/6 = 229
Sxx = 454 − 48²/6 = 70
Syy = 17927 − 321²/6 = 753.5
Correlation coefficient
r = 229/√(70·753.5) = 0.9971
Answer
r = 0.9971
20.Linear regression & correlation
Answer: x = 2.533
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.