A credit card charges a nominal interest rate of 19.5% per year, compounded quarterly. Find the effective annual rate EAR in percent.
r = 19.5%m = 4
EAR = %
2.Time value of moneyHard
You save €150 at the end of every month for 14 years. The account pays 3.6% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €1503.6% / 12N = 168
FV = €
3.NPV & IRRMedium
A machine costs €80 000 today, lasts 4 years and costs €5 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.
C0 = €80 000M = €5 000n = 4r = 12%
EAC = €
4.NPV & IRRHard
A project costs €150 000 today. Its cash flow is €27 500 at the end of year 1 and grows by 5% per year until the last cash flow at the end of year 6. The cost of capital is 10%. Find the net present value NPV.
C0 = €150 000CF1 = €27 500g = 5%n = 6r = 10%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €1 188 when the yield to maturity is 4%. Find the coupon rate c in percent.
F = €1 000n = 12P = €1 188YTM = 4%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 9 years. It sells for €757. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 5% (2×)n = 9P = €757
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 0.6. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.
D/E = 0.6re = 12%rd = 3%T = 25.8%
WACC = %
8.CAPM & WACCHard
A company has 5 000 000 shares at €49.00 each and 5 000 bonds trading at €960 each. The pre-tax cost of debt is 4.5% and the tax rate is 28%. The share’s beta is 1.2, the risk-free rate 1% and the market return 5%. Find the WACC.
β = 1.2rf = 1%rm = 5%rd = 4.5%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 60, 32, 17, 56, 33.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is below 175?
μ = 170σ = 5x = 175
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.05, 0.3, 0.35, 0.3
Var(X) =
12.ProbabilityHard
A game costs €3 to play. You win €6 with probability 0.3 and €30 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€6: p = 0.3€30: p = 0.02
E(G) = €
13.Normal distributionMedium
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 0.01 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 = 0.01 mm
P =
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 114 h, known standard deviation 13 h. Model 2: 30 phones, mean 105 h, known standard deviation 4 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 17. 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 = 17E = 299%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 5 students, mean score 24.5, standard deviation 3. Group 2: 11 students, mean score 20.4, standard deviation 5. 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 = 5x̄₁ = 24.5s₁ = 3n₂ = 11x̄₂ = 20.4s₂ = 5
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 20 of 100 patients on treatment A recover, and 60 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 20n₁ = 100x₂ = 60n₂ = 250
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (6, 76) is made. Find its residual e = y − ŷ.
n = 5(6, 76)
e =
3Solutions
1.Time value of money
Answer: EAR = 20.97 %
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 per period
r/m = 0.195/4 = 0.04875
Compound over one year
EAR = (1 + 0.04875)4 − 1 = 1.21 − 1 = 20.97%
Answer
EAR = 20.97%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: FV = 32704 €
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 €25 200 yourself; the rest is interest.
3.NPV & IRR
Answer: EAC = 31339 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.12−4)/0.12 = 3.037
Annualised purchase price
80 000/3.037 = 26 338.75
EAC
EAC = 26 338.75 + 5 000 = €31 338.75
Answer
EAC = €31 338.75
4.NPV & IRR
Answer: NPV = −16047 €
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.
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: c = 6.003 %
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 + 0.04)−12 = €624.60
Coupon
C = (1188 − 624.60)/9.385 = €60.03
Coupon rate
c = 60.03/1000 = 6.003%
Answer
c = 6.003%
6.Bond pricing & yield
Answer: YTM = 8.996 %
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 = €25.00, N = 18
Solve numerically
i = 4.498% per half-year
YTM
YTM = 2·4.498% = 8.996%
Answer
YTM = 8.996%
7.CAPM & WACC
Answer: WACC = 8.335 %
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 + 0.6) = 0.625
D/V = 0.6/(1 + 0.6) = 0.375
WACC
WACC = 0.625·12% + 0.375·3%·(1 − 0.258) = 8.335%
Answer
WACC = 8.335%
8.CAPM & WACC
Answer: WACC = 5.751 %
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 = 1
Standard normal table
Φ(1) = 0.8413
P = 0.8413
Answer
P = 0.8413
11.Probability
Answer: Var(X) = 0.79
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.
Mean
μ = 0·0.05 + 1·0.3 + 2·0.35 + 3·0.3 = 1.9
Mean of the squares
E(X²) = 0·0.05 + 1·0.3 + 4·0.35 + 9·0.3 = 4.4
Variance
Var(X) = 4.4 − 1.9² = 0.79
Answer
Var(X) = 0.79
12.Probability
Answer: E(G) = −0.6 €
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.
Net outcomes
G: 3 (p = 0.3), 27 (p = 0.02), −3 (p = 0.68)
Expected value
E(G) = 3·0.3 + 27·0.02 − 3·0.68 = −0.6 €
Answer
E(G) = −0.6 €
A negative value means you lose money on average.
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 = 0.9207
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 = 0.01414 mm
Standardise
z = (0 − 0.02)/0.01414 = −1.41
Probability
P = 1 − Φ(−1.41) = 0.9207
Answer
P = 0.9207
15.Confidence intervals
Answer: E = 3.254 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 = 114 − 105 = 9
Standard error
SE = √(13²/50 + 4²/30) = 1.978
Margin of error
E = 1.645·1.978 = 3.254
Answer
E = 3.254 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 480
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 = (2.576·17/2)² = 479.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 480
Answer
n = 480
17.Hypothesis tests & confidence intervals
Answer: t = 1.682
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 = (4·3² + 10·5²)/14 = 20.43
sp = 4.52
Standard error
SE = 4.52·√(1/5 + 1/11) = 2.438
Test statistic
t = (24.5 − 20.4)/2.438 = 1.682
Answer
t = 1.682
With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: z = −0.8051
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 = 20/100 = 0.2
p̂₂ = 60/250 = 0.24
p̂ = 80/350 = 0.2286
Standard error under H₀
SE = √(0.2286·0.7714·(1/100 + 1/250)) = 0.04968
Test statistic
z = (0.2 − 0.24)/0.04968 = −0.8051
Answer
z = −0.8051
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
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.
Centred sums
Sxy = 2823 − 36·429/6 = 249
Sxx = 286 − 36²/6 = 70
Syy = 31581 − 429²/6 = 907.5
Correlation coefficient
r = 249/√(70·907.5) = 0.9879
Answer
r = 0.9879
20.Linear regression & correlation
Answer: e = 2.95
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 = 35/5 = 7
ȳ = 393/5 = 78.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 222
Regression line
b = 222/40 = 5.55
a = 78.6 − 5.55·7 = 39.75
ŷ = 39.75 + 5.55·x
Predicted value
ŷ = 39.75 + 5.55·6 = 73.05
Residual
e = 76 − 73.05 = 2.95
Answer
e = 2.95
The point lies above the line.
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