You invest €12 000 at a nominal rate of 3% per year, compounded quarterly. Find the value FV of the investment after 2 years.
PV = €12 0003% / 4n = 2
FV = €
2.Time value of moneyHard
You deposit €2 000 at the start of every year for 6 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 6? Find FV.
A = €2 000i = 5%n = 6
FV = €
3.NPV & IRRMedium
A project costs €60 000 today and returns €16 500, €22 500, €13 000, €17 500, €22 500 at the end of years 1 to 5. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
Project A costs €30 000 and returns €16 500, €11 500, €7 000 at the end of years 1 to 3. Project B costs €40 000 and returns €11 000, €16 000, €20 500. 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 a 7% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 6% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 6%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 10 years. It sells for €805. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 6% (2×)n = 10P = €805
YTM = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €1 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 6% and the tax rate is 21%. Find the WACC.
A company has 5 000 000 shares at €59.00 each and 40 000 bonds trading at €1 050 each. The pre-tax cost of debt is 7% and the tax rate is 20%. The share’s beta is 0.6, the risk-free rate 4% and the market return 11%. Find the WACC.
β = 0.6rf = 4%rm = 11%rd = 7%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 60, 17, 10, 26, 40, 34.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is below 82?
μ = 70σ = 12x = 82
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3p(x): 0.25, 0.1, 0.35, 0.3
σ =
12.ProbabilityHard
11 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 11p = 0.6
P =
13.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 76 and 115.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 76x2 = 115.75
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 10% of them lie (the 10th percentile).
μ = 250σ = 5P = 10%
x = g
15.Confidence intervalsMedium
A sample of 16 items has mean 106 and standard deviation s = 14.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 16x̄ = 106s = 14.595%t(15) = 2.131
lower limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 8 plots, mean yield 17.5 kg, standard deviation 10 kg. Plot group 2: 6 plots, mean 18.5 kg, standard deviation 2 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
The mean of a population is claimed to be 80; the population standard deviation is known to be 6. A random sample of 36 observations has a mean of 82.4. Test H₀: μ = 80 against H₁: μ > 80. Find the p-value p.
μ₀ = 80σ = 6n = 36x̄ = 82.4H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
In a trial, 52 of 80 patients on treatment A recover, and 78 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 52n₁ = 80x₂ = 78n₂ = 150
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. 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 = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 30.5.
n = 6x = 30.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 12739 €
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.
Ordinary annuity factor
((1 + 0.05)6 − 1)/0.05 = 6.802
Annuity due
Each deposit is invested one year longer.
FV = 2 000·6.802·1.05 = €14 284.02
Answer
FV = €14 284.02
3.NPV & IRR
Answer: DPB = 4.545 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: 16 500/1.12 = 14 732.14 → Σ = 14 732.14
year 2: 22 500/1.254 = 17 936.86 → Σ = 32 669.01
year 3: 13 000/1.405 = 9 253.14 → Σ = 41 922.15
year 4: 17 500/1.574 = 11 121.57 → Σ = 53 043.71
year 5: 22 500/1.762 = 12 767.10 → Σ = 65 810.82
Fraction of the last year
DPB = 4 + 6 956.29/12 767.10 = 4.545
Answer
DPB = 4.545 years
The discounted payback is always longer than the simple payback (3.457 years).
4.NPV & IRR
Answer: r* = 6.439 %
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 = −40 000 − (−30 000) = −10 000
ΔCF1 = 11 000 − 16 500 = −5 500
ΔCF2 = 16 000 − 11 500 = 4 500
ΔCF3 = 20 500 − 7 000 = 13 500
IRR of the difference
NPVΔ(6%) = 151.17
NPVΔ(7%) = −189.69
r* = 6.439%
Answer
r* = 6.439%
Check: at 6.439% both NPVs are €1 457.57.
5.Bond pricing & yield
Answer: P = 1085 €
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 = 0.03, N = 24
Present value of the coupons
€35.00·(1 − 0.4919)/0.03 = €592.74
Present value of the face value
€1 000·0.4919 = €491.93
Price
P = €1 084.68
Answer
P = €1 084.68
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 8.998 %
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 = €30.00, N = 20
Solve numerically
i = 4.499% per half-year
YTM
YTM = 2·4.499% = 8.998%
Answer
YTM = 8.998%
7.CAPM & WACC
Answer: WACC = 11.69 %
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 = €4 000 000, E/V = 0.75, D/V = 0.25
WACC
WACC = 0.75·14% + 0.25·6%·(1 − 0.21) = 11.69%
Answer
WACC = 11.69%
8.CAPM & WACC
Answer: WACC = 7.876 %
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 = 5 000 000·€59.00 = €295 000 000
D = 40 000·€1 050 = €42 000 000
Cost of equity (CAPM)
re = 4% + 0.6·7% = 8.2%
Weights
V = €337 000 000, E/V = 0.8754, D/V = 0.1246
WACC
WACC = 0.8754·8.2% + 0.1246·7%·(1 − 0.2) = 7.876%
Answer
WACC = 7.876%
9.Descriptive statistics
Answer: s = 17.85
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 = 0.8413
Answer
P = 0.8413
11.Probability
Answer: σ = 1.145
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 = 0·0.25 + 1·0.1 + 2·0.35 + 3·0.3 = 1.7
Mean of the squares
E(X²) = 0·0.25 + 1·0.1 + 4·0.35 + 9·0.3 = 4.2
Variance
Var(X) = 4.2 − 1.7² = 1.31
σ = √1.31 = 1.145
Answer
σ = 1.145
12.Probability
Answer: P = 0.000734
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) = 0.411 = 4.194 × 10⁻⁵
P(X = 1) = 11·0.6·0.410 = 0.0006921
Add them
P = 4.194 × 10⁻⁵ + 0.0006921 = 0.000734
Answer
P = 0.000734
13.Normal distribution
Answer: P = 0.7983
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 = (76 − 100)/15 = −1.6
z₂ = (115.75 − 100)/15 = 1.05
Area between
P = Φ(1.05) − Φ(−1.6) = 0.8531 − 0.0548 = 0.7983
Answer
P = 0.7983
14.Normal distribution
Answer: x = 243.6 g
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) = 0.1 → z = −1.282
Back to x
x = 250 + −1.282·5 = 243.59 g
Answer
x = 243.59 g
15.Confidence intervals
Answer: lower limit = 98.28
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 = 15, t = 2.131
Margin of error
E = 2.131·14.5/√16 = 7.725
Interval
106 ± 7.725 → [98.28, 113.7]
Answer
lower limit = 98.28
16.Confidence intervals
Answer: lower limit = −8.455 kg
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.
Pooled standard deviation
sp² = (7·10² + 5·2²)/12 = 60
sp = 7.746
Margin of error
E = 1.782·7.746·√(1/8 + 1/6) = 7.455
Interval
−1 ± 7.455 → [−8.455, 6.455]
Answer
lower limit = −8.455 kg
17.Hypothesis tests & confidence intervals
Answer: p = 0.008198
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.
Test statistic
SE = σ/√n = 6/√36 = 1
z = (82.4 − 80)/1 = 2.4
Tail area
Φ(2.4) = 0.9918
1 − Φ(2.4) = 0.0082
p-value
p = 0.008198
Answer
p = 0.008198
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: z = 1.894
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 = 52/80 = 0.65
p̂₂ = 78/150 = 0.52
p̂ = 130/230 = 0.5652
Standard error under H₀
SE = √(0.5652·0.4348·(1/80 + 1/150)) = 0.06863
Test statistic
z = (0.65 − 0.52)/0.06863 = 1.894
Answer
z = 1.894
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r² = 0.9312
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 = 118/6 = 19.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −39
Syy = Σ(y − ȳ)² = 93.33
Coefficient of determination
r² = −39²/(17.5·93.33) = 0.9312
Answer
r² = 0.9312
About 93.12% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 105.9
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 = 147/6 = 24.5
ȳ = 489/6 = 81.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 640.5
Regression line
b = 640.5/157.5 = 4.067
a = 81.5 − 4.067·24.5 = −18.13
ŷ = −18.13 + 4.067·x
Prediction
ŷ = −18.13 + 4.067·30.5 = 105.9
Answer
ŷ = 105.9
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