You need €20 000 in 6 years. An account pays 2% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €20 0002% / 12n = 6
PV = €
2.Time value of moneyHard
You deposit €5 000 at the start of every year for 30 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 30? Find FV.
A = €5 000i = 5%n = 30
FV = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 4: €6 000, €8 000, €4 000, €9 000. The cost of capital is 12%. Find the net present value NPV.
A project costs €50 000 and returns €35 500, €18 500, €27 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 35 500CF2 = 18 500CF3 = 27 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €1 154 when the yield to maturity is 2.5%. Find the coupon rate c in percent.
F = €1 000n = 12P = €1 154YTM = 2.5%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 6 years. It sells for €1 028. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 3% (2×)n = 6P = €1 028
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €3 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 4% and the tax rate is 25.8%. Find the WACC.
A company has 2 500 000 shares at €13.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 5% and the tax rate is 28%. The share’s beta is 0.9, the risk-free rate 1% and the market return 8%. Find the WACC.
β = 0.9rf = 1%rm = 8%rd = 5%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 15, 27, 58, 57, 51.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is above 40?
μ = 50σ = 20x = 40
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.2, 0.1, 0.2, 0.35, 0.15
Var(X) =
12.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 4 correct answers?
n = 6p = 0.4k = 4
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0749. Find z to two decimals.
A = 0.0749
z =
14.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 101 h, known standard deviation 15 h. Model 2: 100 phones, mean 97 h, known standard deviation 15 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 16. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 0.5?
σ = 16E = 0.595%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 12 students, mean score 86.5, standard deviation 10. Group 2: 9 students, mean score 98.8, standard deviation 9. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 12x̄₁ = 86.5s₁ = 10n₂ = 9x̄₂ = 98.8s₂ = 9
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 501.1. The population standard deviation is 5. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 501.1σ = 5n = 8195%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.
n = 5x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 17740 €
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 = 0.02/12 = 0.001667
N = 12·6 = 72
Present value
PV = FV/(1 + i)N = 20 000/1.127 = €17 740.18
Answer
PV = €17 740.18
2.Time value of money
Answer: FV = 348804 €
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)30 − 1)/0.05 = 66.44
Annuity due
Each deposit is invested one year longer.
FV = 5 000·66.44·1.05 = €348 803.95
Answer
FV = €348 803.95
3.NPV & IRR
Answer: NPV = 301.5 €
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 each cash flow
CF0 = −20 000
CF1/(1 + 0.12)1 = 6 000/1.12 = 5 357.14
CF2/(1 + 0.12)2 = 8 000/1.254 = 6 377.55
CF3/(1 + 0.12)3 = 4 000/1.405 = 2 847.12
CF4/(1 + 0.12)4 = 9 000/1.574 = 5 719.66
Add them up
NPV = €301.48
Answer
NPV = €301.48
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 30.83 %
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(30%) = 543.92
NPV(35%) = −2 578.88
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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 + 0.025)−12 = €743.56
Coupon
C = (1154 − 743.56)/10.26 = €40.01
Coupon rate
c = 40.01/1000 = 4.001%
Answer
c = 4.001%
6.Bond pricing & yield
Answer: YTM = 2.495 %
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 = €15.00, N = 12
Solve numerically
i = 1.247% per half-year
YTM
YTM = 2·1.247% = 2.495%
Answer
YTM = 2.495%
7.CAPM & WACC
Answer: WACC = 9.272 %
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).
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
11.Probability
Answer: Var(X) = 1.828
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.
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) = 15·0.44·0.62 = 15·0.0256·0.36 = 0.1382
Answer
P = 0.1382
13.Normal distribution
Answer: z = 1.44
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 − 0.0749 = 0.9251
Read the z-table backwards
Find 0.9251 inside the table and read off z.
z = 1.44
Answer
z = 1.44
14.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 difference
μ = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
15.Confidence intervals
Answer: lower limit = −0.2738 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 = 101 − 97 = 4
Standard error
SE = √(15²/50 + 15²/100) = 2.598
Margin of error
E = 1.645·2.598 = 4.274
Interval
4 ± 4.274 → [−0.2738, 8.274]
Answer
lower limit = −0.2738 h
The interval contains 0, so the data do not show a significant difference.
16.Confidence intervals
Answer: n = 3934
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 = (1.96·16/0.5)² = 3934
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 3934
Answer
n = 3934
17.Hypothesis tests & confidence intervals
Answer: t = −2.908
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.
Pooled variance
sp² = (11·10² + 8·9²)/19 = 92
sp = 9.592
Standard error
SE = 9.592·√(1/12 + 1/9) = 4.23
Test statistic
t = (86.5 − 98.8)/4.23 = −2.908
Answer
t = −2.908
With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: U = 502.2
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).
Margin of error
E = 1.96·5/√81 = 1.089
Interval limit
U = 501.1 + 1.089 = 502.2
Answer
U = 502.2
19.Linear regression & correlation
Answer: r² = 0.8685
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 = 132/6 = 22
ȳ = 361/6 = 60.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 183
Syy = Σ(y − ȳ)² = 550.8
Coefficient of determination
r² = 183²/(70·550.8) = 0.8685
Answer
r² = 0.8685
About 86.85% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 53.8
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 = 25/5 = 5
ȳ = 277/5 = 55.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 32
Regression line
b = 32/10 = 3.2
a = 55.4 − 3.2·5 = 39.4
ŷ = 39.4 + 3.2·x
Prediction
ŷ = 39.4 + 3.2·4.5 = 53.8
Answer
ŷ = 53.8
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