You need €40 000 in 3 years. An account pays 4% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €40 0004% / 4n = 3
PV = €
2.Time value of moneyHard
You want €75 000 in 8 years. You save the same amount at the end of every month in an account paying 2.4% per year, compounded monthly. Find the monthly deposit A.
FV = €75 0002.4% / 12N = 96
A = €
3.NPV & IRRMedium
A machine costs €30 000 today. It generates €8 500 at the end of each year for 4 years and can then be sold for €4 500. The cost of capital is 10%. Find the net present value NPV.
C0 = €30 000CF = €8 500S = €4 500n = 4r = 10%
NPV = €
4.NPV & IRRHard
A project costs €20 000 and returns €10 000, €12 000, €8 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 10 000CF2 = 12 000CF3 = 8 500
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 5 years to maturity (face value €1 000, annual coupon 3%) when its yield to maturity is 7%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 7.5%. Find your holding period return HPR in percent.
F = €1 000coupon = 3%n = 5y0 = 7%y1 = 7.5%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 7 years and sells for €1 060. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 7P = €1 060
YTM = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €3 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 4.5% and the tax rate is 20%. Find the WACC.
A company has equity with a market value of €6 000 000 and a cost of equity of 11%. Its debt consists of 8 000 bonds (face value €1 000, annual coupon 6%, 4 years to maturity) that trade at €1 035 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.
Find the sample standard deviation s of the data: 47, 18, 54, 38, 49.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 150.
μ = 170σ = 10x = 150
z =
11.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
12.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 4 correct answers?
n = 5p = 0.3k = 4
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7357. Find z to two decimals.
A = 0.7357
z =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points40%
x = points
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 119 h, known standard deviation 12 h. Model 2: 60 phones, mean 119.5 h, known standard deviation 17 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 31. 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 1.5?
σ = 31E = 1.599%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 4. A random sample of 64 observations has a mean of 49.24. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.
μ₀ = 50σ = 4n = 64x̄ = 49.24H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
In a trial, 113 of 250 patients on treatment A recover, and 42 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 113n₁ = 250x₂ = 42n₂ = 120
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
20.Linear regression & correlationHard
The table gives 6 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 = 3.
n = 6x = 3
ŷ =
3Solutions
1.Time value of money
Answer: PV = 35498 €
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 = 0.04/4 = 0.01
N = 4·3 = 12
Present value
PV = FV/(1 + i)N = 40 000/1.127 = €35 497.97
Answer
PV = €35 497.97
2.Time value of money
Answer: A = 709.4 €
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 the yearly cash flows
8 500·3.17 = 26 943.86
Present value of the salvage value
4 500/1.464 = 3 073.56
NPV
NPV = −30 000 + 26 943.86 + 3 073.56 = €17.42
Answer
NPV = €17.42
4.NPV & IRR
Answer: IRR = 25.11 %
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.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(25%) = 32.00
NPV(30%) = −1 338.19
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 25 + 5·32.00/(32.00 − (−1 338.19)) = 25.12%
exact: NPV(25.11%) = 0 → IRR = 25.11%
Answer
IRR = 25.11%
5.Bond pricing & yield
Answer: HPR = 5.178 %
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.
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
€1 060 = €60·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(4%) = €1 120.04, P(5%) = €1 057.86
YTM = 4.964%
Answer
YTM = 4.964%
7.CAPM & WACC
Answer: WACC = 7.3 %
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 = €6 000 000, E/V = 0.5, D/V = 0.5
WACC
WACC = 0.5·11% + 0.5·4.5%·(1 − 0.2) = 7.3%
Answer
WACC = 7.3%
8.CAPM & WACC
Answer: WACC = 6.778 %
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 = (150 − 170)/10 = −2
Answer
z = −2
11.Probability
Answer: P = 0.6651
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.
Complement
P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
12.Probability
Answer: P = 0.02835
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.
Binomial formula
P(X = 4) = 5·0.34·0.71 = 5·0.0081·0.7 = 0.02835
Answer
P = 0.02835
13.Normal distribution
Answer: z = −0.63
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 of z
Φ(z) = 1 − 0.7357 = 0.2643
Read the z-table backwards
Find 0.2643 inside the table and read off z.
z = −0.63
Answer
z = −0.63
14.Normal distribution
Answer: x = 103.8 points
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) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 100 + 0.25·15 = 103.8 points
Answer
x = 103.8 points
15.Confidence intervals
Answer: E = 7.147 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 = 119 − 119.5 = −0.5
Standard error
SE = √(12²/50 + 17²/60) = 2.774
Margin of error
E = 2.576·2.774 = 7.147
Answer
E = 7.147 h
The interval contains 0, so the data do not show a significant difference.
16.Confidence intervals
Answer: n = 2835
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 = (2.576·31/1.5)² = 2834
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 2835
Answer
n = 2835
17.Hypothesis tests & confidence intervals
Answer: p = 0.06426
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 = 4/√64 = 0.5
z = (49.24 − 50)/0.5 = −1.52
Tail area
Φ(1.52) = 0.9357
1 − Φ(1.52) = 0.0643
p-value
p = 0.06426
Answer
p = 0.06426
p > 0.05, so H₀ is not rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: z = 1.862
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 = 113/250 = 0.452
p̂₂ = 42/120 = 0.35
p̂ = 155/370 = 0.4189
Standard error under H₀
SE = √(0.4189·0.5811·(1/250 + 1/120)) = 0.05479
Test statistic
z = (0.452 − 0.35)/0.05479 = 1.862
Answer
z = 1.862
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r² = 0.5749
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 = 279/5 = 55.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 27
Syy = Σ(y − ȳ)² = 126.8
Coefficient of determination
r² = 27²/(10·126.8) = 0.5749
Answer
r² = 0.5749
About 57.49% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 7.924
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 = 112/6 = 18.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 188
Regression line
b = 188/70 = 2.686
a = 18.67 − 2.686·7 = −0.1333
ŷ = −0.1333 + 2.686·x
Prediction
ŷ = −0.1333 + 2.686·3 = 7.924
Answer
ŷ = 7.924
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