An annuity pays €1 000 at the end of each year for 20 years. The interest rate is 8% per year. Find the present value PV of the annuity.
A = €1 000i = 8%n = 20
PV = €
2.Time value of moneyHard
A loan of €150 000 is repaid with equal monthly payments over 20 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.
P = €150 0003% / 12N = 240
A = €
3.NPV & IRRMedium
A machine costs €80 000 today. It generates €19 000 at the end of each year for 5 years and can then be sold for €8 000. The cost of capital is 6%. Find the net present value NPV.
C0 = €80 000CF = €19 000S = €8 000n = 5r = 6%
NPV = €
4.NPV & IRRHard
A project costs €100 000 today. Its cash flow is €13 000 at the end of year 1 and grows by 5% per year until the last cash flow at the end of year 7. The cost of capital is 10%. Find the net present value NPV.
C0 = €100 000CF1 = €13 000g = 5%n = 7r = 10%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 15 years and pays annual coupons. It sells for €909 when the yield to maturity is 7%. Find the coupon rate c in percent.
F = €1 000n = 15P = €909YTM = 7%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 8 years. It sells for €929. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 2% (2×)n = 8P = €929
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 1.5. Its cost of equity is 13%, its pre-tax cost of debt is 7% and the tax rate is 25%. Find the WACC.
D/E = 1.5re = 13%rd = 7%T = 25%
WACC = %
8.CAPM & WACCHard
A company has 2 000 000 shares trading at €24.00; the book value of its equity is €14 400 000. It just paid a dividend of €1.00 per share, and dividends grow by 4% per year. Its debt has a market value of €24 000 000 and a pre-tax cost of 3%. The tax rate is 28%. Find the WACC, using the dividend growth model for the cost of equity.
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 70 and 123.25 points? Give P as a decimal.
μ = 100σ = 15x1 = 70x2 = 123.25
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 top 60% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h60%
x = h
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 1.2. What is the margin of error E₂ of a 90% interval from the same data?
95%n₁ = 50E₁ = 1.290%
E₂ =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 5 plots, mean yield 24 kg, standard deviation 3.5 kg. Plot group 2: 9 plots, mean 34 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
12 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 6.1 and the standard deviation of the differences is 6.6. Find the paired test statistic t for H₀: μd = 0.
n = 12d̄ = 6.1sd = 6.6
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 32 of 80 patients on treatment A recover, and 72 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 32n₁ = 80x₂ = 72n₂ = 200
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
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. Predict ŷ for x = 9.
n = 5x = 9
ŷ =
3Solutions
1.Time value of money
Answer: PV = 9818 €
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 − 0.2145)/0.08 = 9.818
Present value
PV = 1 000·9.818 = €9 818.15
Answer
PV = €9 818.15
2.Time value of money
Answer: A = 831.9 €
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 = 0.03/12 = 0.0025
N = 12·20 = 240
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 = 0.5492
A = 150 000·0.0025/(1 − 0.5492) = €831.90
Answer
A = €831.90
3.NPV & IRR
Answer: NPV = 6013 €
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
19 000·4.212 = 80 034.91
Present value of the salvage value
8 000/1.338 = 5 978.07
NPV
NPV = −80 000 + 80 034.91 + 5 978.07 = €6 012.98
Answer
NPV = €6 012.98
4.NPV & IRR
Answer: NPV = −27737 €
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 growing annuity
((1.05)/(1.1))7 = 0.7221
PV = 13 000/(0.1 − 0.05)·(1 − 0.7221) = 72 263.10
NPV
NPV = 72 263.10 − 100 000 = −€27 736.90
Answer
NPV = −€27 736.90
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: c = 6.001 %
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.
Present value of the face value
1000·(1 + 0.07)−15 = €362.45
Coupon
C = (909 − 362.45)/9.108 = €60.01
Coupon rate
c = 60.01/1000 = 6.001%
Answer
c = 6.001%
6.Bond pricing & yield
Answer: YTM = 3.005 %
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 = €10.00, N = 16
Solve numerically
i = 1.503% per half-year
YTM
YTM = 2·1.503% = 3.005%
Answer
YTM = 3.005%
7.CAPM & WACC
Answer: WACC = 8.35 %
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 from D/E
E/V = 1/(1 + 1.5) = 0.4
D/V = 1.5/(1 + 1.5) = 0.6
WACC
WACC = 0.4·13% + 0.6·7%·(1 − 0.25) = 8.35%
Answer
WACC = 8.35%
8.CAPM & WACC
Answer: WACC = 6.276 %
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 = (428 − 500)/60 = −1.2
Answer
z = −1.2
11.Probability
Answer: Var(X) = 1.947
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.
Spread of X
μ = 2.35, Var(X) = 1.428, σX = 1.195
Linear change
σY = |10|·1.195 = 11.95
Answer
σY = 11.95
The mean does move: E(Y) = 10·2.35 + 5 = 28.5.
13.Normal distribution
Answer: P = 0.9167
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 = (70 − 100)/15 = −2
z₂ = (123.25 − 100)/15 = 1.55
Area between
P = Φ(1.55) − Φ(−2) = 0.9394 − 0.02275 = 0.9167
Answer
P = 0.9167
14.Normal distribution
Answer: x = 490 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) = 0.4
z from the table
z ≈ −0.25
Back to the original units
x = 500 − 0.25·40 = 490 h
Answer
x = 490 h
15.Confidence intervals
Answer: E₂ = 1.007
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 = (1.645/1.96)·√(50/50) = 0.8393
New margin
E₂ = 1.2·0.8393 = 1.007
Answer
E₂ = 1.007
16.Confidence intervals
Answer: lower limit = −14.53 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 = (4·3.5² + 8·5²)/12 = 20.75
sp = 4.555
Margin of error
E = 1.782·4.555·√(1/5 + 1/9) = 4.528
Interval
−10 ± 4.528 → [−14.53, −5.472]
Answer
lower limit = −14.53 kg
17.Hypothesis tests & confidence intervals
Answer: t = 3.202
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 = 6.6/√12 = 1.905
Test statistic
t = 6.1/1.905 = 3.202
Answer
t = 3.202
With df = 11, the two-sided 5% critical value is 2.201; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: z = 0.6258
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 = 32/80 = 0.4
p̂₂ = 72/200 = 0.36
p̂ = 104/280 = 0.3714
Standard error under H₀
SE = √(0.3714·0.6286·(1/80 + 1/200)) = 0.06392
Test statistic
z = (0.4 − 0.36)/0.06392 = 0.6258
Answer
z = 0.6258
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9333
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 = 5014 − 90·271/5 = 136
Sxx = 1660 − 90²/5 = 40
Syy = 15219 − 271²/5 = 530.8
Correlation coefficient
r = 136/√(40·530.8) = 0.9333
Answer
r = 0.9333
20.Linear regression & correlation
Answer: ŷ = 32.35
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 = 110/5 = 22
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 138
Regression line
b = 138/40 = 3.45
a = 22 − 3.45·6 = 1.3
ŷ = 1.3 + 3.45·x
Prediction
ŷ = 1.3 + 3.45·9 = 32.35
Answer
ŷ = 32.35
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