An annuity pays €2 000 at the end of each year for 19 years. The interest rate is 8% per year. Find the present value PV of the annuity.
A = €2 000i = 8%n = 19
PV = €
2.Time value of moneyHard
A loan of €250 000 is repaid with equal monthly payments over 30 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.
P = €250 0004.8% / 12N = 360
A = €
3.NPV & IRRMedium
A machine costs €80 000 today. It generates €16 000 at the end of each year for 7 years and can then be sold for €20 000. The cost of capital is 10%. Find the net present value NPV.
C0 = €80 000CF = €16 000S = €20 000n = 7r = 10%
NPV = €
4.NPV & IRRHard
A project costs €20 000 and returns €8 000, €8 500, €7 000, €8 000 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 5.5% per year. Find the price P of the bond.
F = €1 000coupon = 4%YTM = 5.5%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 5 years. It sells for €1 210. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 7% (2×)n = 5P = €1 210
YTM = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €6 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 3% and the tax rate is 30%. Find the WACC.
A company has 5 000 000 shares at €17.00 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 5% and the tax rate is 21%. The share’s beta is 1.2, the risk-free rate 1% and the market return 6.5%. Find the WACC.
β = 1.2rf = 1%rm = 6.5%rd = 5%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 55, 29, 15, 54, 33.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 76.25.
μ = 70σ = 5x = 76.25
z =
11.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 12? The order does not matter.
n = 12k = 2
N =
12.ProbabilityHard
12 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?
n = 12p = 0.7
P =
13.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 30 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30 min
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 10% of them lie (the 10th percentile).
μ = 178σ = 7P = 10%
x = cm
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 128 h, known standard deviation 19 h. Model 2: 30 phones, mean 143 h, known standard deviation 10 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.
Two fertilisers are tested. Plot group 1: 11 plots, mean yield 98 kg, standard deviation 8.5 kg. Plot group 2: 5 plots, mean 93 kg, standard deviation 1 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
A sample of 64 students has a mean score of 118.9 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 118.9s = 4n = 64
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 290%
n =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (19, 44) is made. Find its residual e = y − ŷ.
n = 5(19, 44)
e =
3Solutions
1.Time value of money
Answer: PV = 19207 €
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.
Annuity factor
(1 − (1 + i)−n)/i = (1 − 0.2317)/0.08 = 9.604
Present value
PV = 2 000·9.604 = €19 207.20
Answer
PV = €19 207.20
2.Time value of money
Answer: A = 1312 €
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 rate and number of payments
i = 0.048/12 = 0.004
N = 12·30 = 360
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.2376
A = 250 000·0.004/(1 − 0.2376) = €1 311.66
Answer
A = €1 311.66
3.NPV & IRR
Answer: NPV = 8158 €
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 the yearly cash flows
16 000·4.868 = 77 894.70
Present value of the salvage value
20 000/1.949 = 10 263.16
NPV
NPV = −80 000 + 77 894.70 + 10 263.16 = €8 157.86
Answer
NPV = €8 157.86
4.NPV & IRR
Answer: MIRR = 14.6 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
8 000·1.063 = 9 528.13
8 500·1.062 = 9 550.60
7 000·1.061 = 7 420.00
8 000·1.060 = 8 000.00
TV = 34 498.73
MIRR
MIRR = (34 498.73/20 000)1/4 − 1 = 14.6%
Answer
MIRR = 14.6%
5.Bond pricing & yield
Answer: P = 869.5 €
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 = €20.00, i = 0.0275, N = 24
Present value of the coupons
€20.00·(1 − 0.5215)/0.0275 = €348.02
Present value of the face value
€1 000·0.5215 = €521.48
Price
P = €869.49
Answer
P = €869.49
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 2.505 %
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, N = 10
Solve numerically
i = 1.253% per half-year
YTM
YTM = 2·1.253% = 2.505%
Answer
YTM = 2.505%
7.CAPM & WACC
Answer: WACC = 7.055 %
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 = €11 000 000, E/V = 0.4545, D/V = 0.5455
WACC
WACC = 0.4545·13% + 0.5455·3%·(1 − 0.3) = 7.055%
Answer
WACC = 7.055%
8.CAPM & WACC
Answer: WACC = 7.397 %
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 = (76.25 − 70)/5 = 1.25
Answer
z = 1.25
11.Probability
Answer: N = 66
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.
Combinations
C(12, 2) = (12·11)/(2·1) = 66
Answer
N = 66
12.Probability
Answer: P = 1.541 × 10⁻⁵
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) = 0.312 = 5.314 × 10⁻⁷
P(X = 1) = 12·0.7·0.311 = 1.488 × 10⁻⁵
Add them
P = 5.314 × 10⁻⁷ + 1.488 × 10⁻⁵ = 1.541 × 10⁻⁵
Answer
P = 1.541 × 10⁻⁵
13.Normal distribution
Answer: P = 0.7881
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 = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (30 − 32)/2.5 = −0.8
Upper tail
P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
14.Normal distribution
Answer: x = 169 cm
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 = 178 + −1.282·7 = 169.03 cm
Answer
x = 169.03 cm
15.Confidence intervals
Answer: lower limit = −22.88 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 = 128 − 143 = −15
Standard error
SE = √(19²/60 + 10²/30) = 3.058
Margin of error
E = 2.576·3.058 = 7.877
Interval
−15 ± 7.877 → [−22.88, −7.123]
Answer
lower limit = −22.88 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: lower limit = −3.334 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 = (10·8.5² + 4·1²)/14 = 51.89
sp = 7.204
Margin of error
E = 2.145·7.204·√(1/11 + 1/5) = 8.334
Interval
5 ± 8.334 → [−3.334, 13.33]
Answer
lower limit = −3.334 kg
17.Hypothesis tests & confidence intervals
Answer: t = −2.2
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 = 4/√64 = 4/8 = 0.5
Test statistic
t = (118.9 − 120)/0.5 = −2.2
Answer
t = −2.2
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 271
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 = (1.645·20/2)² = 270.6
Round up
n = 271
Answer
n = 271
19.Linear regression & correlation
Answer: r = 0.9879
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 = 42/6 = 7
ȳ = 374/6 = 62.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 278
Syy = Σ(y − ȳ)² = 1131
Correlation coefficient
r = 278/√(70·1131) = 0.9879
Answer
r = 0.9879
20.Linear regression & correlation
Answer: e = −4.1
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 = 110/5 = 22
ȳ = 287/5 = 57.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 124
Regression line
b = 124/40 = 3.1
a = 57.4 − 3.1·22 = −10.8
ŷ = −10.8 + 3.1·x
Predicted value
ŷ = −10.8 + 3.1·19 = 48.1
Residual
e = 44 − 48.1 = −4.1
Answer
e = −4.1
The point lies below the line.
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