You need €40 000 in 3 years. An account pays 6% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €40 0006% / 4n = 3
PV = €
2.Time value of moneyHard
You save €200 at the end of every month for 9 years. The account pays 3% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €2003% / 12N = 108
FV = €
3.NPV & IRRMedium
A project costs €100 000 today. It generates €9 000 at the end of the first year, and the cash flow then grows by 2% per year forever. The cost of capital is 12%. Find the net present value NPV.
C0 = €100 000CF1 = €9 000g = 2%r = 12%
NPV = €
4.NPV & IRRHard
An investment of €50 000 returns €16 500 at the end of each year for 4 years. Find its internal rate of return IRR in percent.
C0 = €50 000CF = €16 500n = 4
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 3 years and pays annual coupons. It sells for €906 when the yield to maturity is 5.5%. Find the coupon rate c in percent.
F = €1 000n = 3P = €906YTM = 5.5%
c = %
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 10 years. It sells for €1 293. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 7% (2×)n = 10P = €1 293
YTM = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €2 000 000. Its WACC is 9.28%, its pre-tax cost of debt is 3% and the tax rate is 20%. Find the cost of equity re.
A company has equity with a market value of €5 000 000 and a cost of equity of 12%. Its debt consists of 8 000 bonds (face value €1 000, annual coupon 5%, 9 years to maturity) that trade at €965 each. The tax rate is 25.8%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
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 754.9 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 754.9 g
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
15.Confidence intervalsMedium
The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −3.6 and standard deviation 2.9. Find the lower limit of the 99% confidence interval for the mean difference.
n = 16d̄ = −3.6sd = 2.999%t(15) = 2.947
lower limit = mmHg
16.Confidence intervalsHard
The population standard deviation is 19. 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 2?
σ = 19E = 299%
n =
17.Hypothesis tests & confidence intervalsMedium
7 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 0.97 and the standard deviation of the differences is 3.2. Find the paired test statistic t for H₀: μd = 0.
n = 7d̄ = 0.97sd = 3.2
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 75 of 250 patients on treatment A recover, and 59 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 75n₁ = 250x₂ = 59n₂ = 150
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 5
r =
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. Predict ŷ for x = 5.
n = 5x = 5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 33455 €
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 and number of periods
i = 0.06/4 = 0.015
N = 4·3 = 12
Present value
PV = FV/(1 + i)N = 40 000/1.196 = €33 455.50
Answer
PV = €33 455.50
2.Time value of money
Answer: FV = 24762 €
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.
You paid in €21 600 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = −10000 €
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 = 9 000/(0.12 − 0.02) = 90 000.00
NPV
NPV = 90 000.00 − 100 000 = −€10 000.00
Answer
NPV = −€10 000.00
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 12.11 %
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)−4)/i = 50 000/16 500 = 3.03
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 12%: 3.037
i = 13%: 2.974
IRR = 12.11%
Answer
IRR = 12.11%
5.Bond pricing & yield
Answer: c = 2.016 %
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 + 0.055)−3 = €851.61
Coupon
C = (906 − 851.61)/2.698 = €20.16
Coupon rate
c = 20.16/1000 = 2.016%
Answer
c = 2.016%
6.Bond pricing & yield
Answer: YTM = 3.502 %
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 = 20
Solve numerically
i = 1.751% per half-year
YTM
YTM = 2·1.751% = 3.502%
Answer
YTM = 3.502%
7.CAPM & WACC
Answer: re = 11 %
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 = €10 000 000, E/V = 0.8, D/V = 0.2
Debt part of the WACC
0.2·3%·(1 − 0.2) = 0.48%
Cost of equity
re = (9.28% − 0.48%)/0.8 = 11%
Answer
re = 11%
8.CAPM & WACC
Answer: WACC = 7.195 %
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
11.Probability
Answer: σ = 1.388
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 = 1.65, Var(X) = 1.627, σX = 1.276
Linear change
σY = |0.5|·1.276 = 0.6379
Answer
σY = 0.6379
The mean does move: E(Y) = 0.5·1.65 − 3 = −2.175.
13.Normal distribution
Answer: P = 0.3085
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 = (754.9 − 750)/9.849 = 0.5
Upper tail
P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085
Answer
P = 0.3085
14.Normal distribution
Answer: P = 0.8133
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 difference
μ = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm
Standardise
z = (0 − 0.02)/0.02236 = −0.89
Probability
P = 1 − Φ(−0.89) = 0.8133
Answer
P = 0.8133
15.Confidence intervals
Answer: lower limit = −5.737 mmHg
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.
Critical value
df = 15, t = 2.947
Margin of error
E = 2.947·2.9/√16 = 2.137
Interval
−3.6 ± 2.137 → [−5.737, −1.463]
Answer
lower limit = −5.737 mmHg
16.Confidence intervals
Answer: n = 599
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 = (2.576·19/2)² = 598.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 599
Answer
n = 599
17.Hypothesis tests & confidence intervals
Answer: t = 0.802
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 = 3.2/√7 = 1.209
Test statistic
t = 0.97/1.209 = 0.802
Answer
t = 0.802
With df = 6, the two-sided 5% critical value is 2.447; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: z = −1.915
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 = 75/250 = 0.3
p̂₂ = 59/150 = 0.3933
p̂ = 134/400 = 0.335
Standard error under H₀
SE = √(0.335·0.665·(1/250 + 1/150)) = 0.04875
Test statistic
z = (0.3 − 0.3933)/0.04875 = −1.915
Answer
z = −1.915
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9232
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 = 15/5 = 3
ȳ = 262/5 = 52.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 47
Syy = Σ(y − ȳ)² = 259.2
Correlation coefficient
r = 47/√(10·259.2) = 0.9232
Answer
r = 0.9232
20.Linear regression & correlation
Answer: ŷ = 16.95
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 = 101/5 = 20.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 130
Regression line
b = 130/40 = 3.25
a = 20.2 − 3.25·6 = 0.7
ŷ = 0.7 + 3.25·x
Prediction
ŷ = 0.7 + 3.25·5 = 16.95
Answer
ŷ = 16.95
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