CFA Level I quantitative methods and fixed income practice: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Time value of money

Future value
FV = PV·(1 + i)n
Present value
PV = FV/(1 + i)n
Several periods per year
i = r/m, N = m·n
Annuity, present value
PV = PMT·[1 − (1 + i)−n]/i
Annuity, future value
FV = PMT·[(1 + i)n − 1]/i

Method

  1. Draw a timeline with every cash flow and its timing.
  2. Match the rate to the period: a monthly rate for monthly payments.
  3. Use FV = PV·(1 + i)ⁿ for single amounts and the annuity formulas for equal payments.
  4. Check the sign and the order of magnitude of the answer.

NPV & IRR

Net present value
NPV = −C0 + Σ CFt/(1 + r)t
Equal cash flows
NPV = −C0 + CF·[1 − (1 + r)−n]/r
Internal rate of return
the r at which NPV = 0
Decision
accept when NPV > 0 (or IRR > required return)

Method

  1. List all cash flows with their year; the investment at year 0 is negative.
  2. Discount each cash flow: CFt/(1 + r)t.
  3. Add them up to get the NPV. Positive means the project earns more than r.
  4. For the IRR, find the rate that makes NPV zero by trial or interpolation.

Bond pricing & yield

Bond price
P = C·(1 − (1 + y)−n)/y + F/(1 + y)n
Semi-annual
C/2, y/2 and 2n periods
Current yield
C/P
Price and yield
yield up → price down

Method

  1. Find the coupon per period and the yield per period.
  2. Price = present value of the coupons (an annuity) + present value of the face value.
  3. For semi-annual coupons use half the coupon, half the yield and twice the periods.
  4. Check: if the yield is above the coupon rate, the price is below face value.

CAPM & WACC

CAPM
re = rf + β·(rm − rf)
WACC
E/V·re + D/V·rd·(1 − T)
Firm value
V = E + D

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

2Practice problems

1.Time value of moneyMedium
years0123PV = ?€15 000

You need €15 000 in 3 years. An account pays 4.8% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €15 0004.8% / 4n = 3
PV = €
2.Time value of moneyHard
months01…204…360€300 000AB = ?A4.2% / 12

A mortgage of €300 000 is repaid with equal monthly payments over 30 years at 4.2% per year, compounded monthly. How much is still owed just after the payment at the end of year 17? Find the outstanding balance B.

P = €300 0004.2% / 12N = 360k = 204
B = €
3.NPV & IRRMedium
years01234−45 00019 00013 00013 00010 000

An investment of €45 000 gives these cash flows at the end of years 1 to 4: €19 000, €13 000, €13 000, €10 000. The cost of capital is 15%. Find the net present value NPV.

r = 15%CF0 = −45 000CF1 = 19 000CF2 = 13 000CF3 = 13 000CF4 = 10 000
NPV = €
4.NPV & IRRHard
years0123−10 0007 0007 5005 500

A project costs €10 000 and returns €7 000, €7 500, €5 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 7 000CF2 = 7 500CF3 = 5 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…24€22.50€22.50€22.50 + €1 000P = ?

A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 6% per year. Find the price P of the bond.

F = €1 000coupon = 4.5%YTM = 6%n = 12
P = €
6.Bond pricing & yieldHard
years0123456P€40€40€40 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 6 years. The yield to maturity is 3.5%. Find the Macaulay duration D in years.

F = €1 000coupon = 4%n = 6YTM = 3.5%
D = years
7.CAPM & WACCMedium
WACC = ?ordinary shares€10 000 000re​ = 13%preferred shares€500 000rp​ = 7%debt€5 000 000rd​ = 4%T = 25.8%

A company is financed with ordinary shares worth €10 000 000, preferred shares worth €500 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 7% and the pre-tax cost of debt 4%. The tax rate is 25.8%. Find the WACC.

E = €10 000 000Ps = €500 000D = €5 000 000re = 13%rp = 7%rd = 4%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €30.50D0​ = €0.90g = 5%debt€18 300 000rd​ = 6%T = 20%

A company has 1 000 000 shares trading at €30.50; the book value of its equity is €15 300 000. It just paid a dividend of €0.90 per share, and dividends grow by 5% per year. Its debt has a market value of €18 300 000 and a pre-tax cost of 6%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.

N = 1 000 000P0 = €30.50book equity = €15 300 000D0 = €0.90g = 5%D = €18 300 000rd = 6%T = 20%
WACC = %
9.Descriptive statisticsMedium
10152025303540

Find the sample standard deviation s of the data: 31, 33, 43, 10, 39.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is above 32?

μ = 50σ = 15x = 32
P =
11.ProbabilityMedium
xp(x)00.1510.3520.330.2μ = 1.55

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2, 3p(x): 0.15, 0.35, 0.3, 0.2
Var(X) =
12.ProbabilityHard
n = 9, p = 0.4: P(X = 1) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?

n = 9p = 0.4k = 1
P =
13.Normal distributionMedium
2335.3P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23 and 35.3 min? Give P as a decimal.

μ = 32σ = 6x1 = 23x2 = 35.3
P =
14.Normal distributionHard
4 s0X − Y?

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
n = 8, s = 12L = ?Ux̄ = 109.5E90%

A sample of 8 items has mean 109.5 and standard deviation s = 12. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 8x̄ = 109.5s = 1290%t(7) = 1.895
lower limit =
16.Confidence intervalsHard
σ = 21, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 21. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 3?

σ = 21E = 390%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.0452.045t = ?df = 29

Two teaching methods are compared. Group 1: 18 students, mean score 112, standard deviation 10. Group 2: 13 students, mean score 104.1, standard deviation 2. 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 = 18x̄₁ = 112s₁ = 10n₂ = 13x̄₂ = 104.1s₂ = 2
t =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 64 measurements has mean 201.5. The population standard deviation is 20. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 201.5σ = 20n = 6495%
L =
19.Linear regression & correlationMedium
xy2641061582110291232xy510102030

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy167019832297251222813531157xy15202530100150x = 26.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (26.5, 126) is made. Find its residual e = y − ŷ.

n = 6(26.5, 126)
e =

3Solutions

1.Time value of money

Answer: PV = 12999 €

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 = 0.048/4 = 0.012
N = 4·3 = 12

Present value

PV = FV/(1 + i)N = 15 000/1.154 = €12 999.45

Answer

PV = €12 999.45
2.Time value of money

Answer: B = 176124 €

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 payment

i = 0.0035, N = 360
A = 300 000·0.0035/(1 − (1 + 0.0035)−360) = €1 467.05

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 360 − 204 = 156
B = 1467·(1 − (1 + 0.0035)−156)/0.0035 = €176 123.86

Answer

B = €176 123.86
3.NPV & IRR

Answer: NPV = −4383 €

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 = −45 000
CF1/(1 + 0.15)1 = 19 000/1.15 = 16 521.74
CF2/(1 + 0.15)2 = 13 000/1.322 = 9 829.87
CF3/(1 + 0.15)3 = 13 000/1.521 = 8 547.71
CF4/(1 + 0.15)4 = 10 000/1.749 = 5 717.53

Add them up

NPV = €−4 383.15

Answer

NPV = €−4 383.15

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: IRR = 46.69 %

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(45%) = 198.86
NPV(50%) = −370.37

Narrow down

Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.

IRR ≈ 45 + 5·198.86/(198.86 − (−370.37)) = 46.75%
exact: NPV(46.69%) = 0 → IRR = 46.69%

Answer

IRR = 46.69%
5.Bond pricing & yield

Answer: P = 873 €

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 = €22.50, i = 0.03, N = 24

Present value of the coupons

€22.50·(1 − 0.4919)/0.03 = €381.05

Present value of the face value

€1 000·0.4919 = €491.93

Price

P = €872.98

Answer

P = €872.98

The yield is above the coupon rate, so the bond sells at a discount.

6.Bond pricing & yield

Answer: D = 5.46 years

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Present value of each cash flow

t = 1: 40/1.035 = 38.65, t·PV = 38.65
t = 2: 40/1.071 = 37.34, t·PV = 74.68
t = 3: 40/1.109 = 36.08, t·PV = 108.23
t = 4: 40/1.148 = 34.86, t·PV = 139.43
t = 5: 40/1.188 = 33.68, t·PV = 168.39
t = 6: 1040/1.229 = 846.04, t·PV = 5 076.24

Duration

P = 1 026.64
D = 5 605.63/1 026.64 = 5.46

Answer

D = 5.46 years
7.CAPM & WACC

Answer: WACC = 9.57 %

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 + 500 000 + 5 000 000 = €15 500 000
E/V = 0.6452, Ps/V = 0.03226, D/V = 0.3226

WACC

WACC = 0.6452·13% + 0.03226·7% + 0.3226·4%·(1 − 0.258) = 9.57%

Answer

WACC = 9.57%
8.CAPM & WACC

Answer: WACC = 6.861 %

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 equity

D1 = 0.9·1.05 = €0.95
re = 0.945/30.5 + 0.05 = 8.098%

Market values

E = 1 000 000·€30.50 = €30 500 000
V = €48 800 000, E/V = 0.625, D/V = 0.375

WACC

WACC = 0.625·8.098% + 0.375·6%·(1 − 0.2) = 6.861%

Answer

WACC = 6.861%
9.Descriptive statistics

Answer: s = 12.77

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.

Mean

x̄ = 156/5 = 31.2

Squared deviations

Σ(x − x̄)² = (−0.2)² + 1.8² + 11.8² + (−21.2)² + 7.8² = 652.8

Standard deviation

s = √(652.8/4) = 12.77

Answer

s = 12.77
10.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
11.Probability

Answer: Var(X) = 0.9475

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 = 0·0.15 + 1·0.35 + 2·0.3 + 3·0.2 = 1.55

Mean of the squares

E(X²) = 0·0.15 + 1·0.35 + 4·0.3 + 9·0.2 = 3.35

Variance

Var(X) = 3.35 − 1.55² = 0.9475

Answer

Var(X) = 0.9475
12.Probability

Answer: P = 0.06047

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 = 1) = 9·0.41·0.68 = 9·0.4·0.0168 = 0.06047

Answer

P = 0.06047
13.Normal distribution

Answer: P = 0.642

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 = (23 − 32)/6 = −1.5
z₂ = (35.3 − 32)/6 = 0.55

Area between

P = Φ(0.55) − Φ(−1.5) = 0.7088 − 0.06681 = 0.642

Answer

P = 0.642
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 = 101.5

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.

Critical value

df = n − 1 = 7, t = 1.895

Margin of error

E = 1.895·12/√8 = 8.04

Interval

109.5 ± 8.04 → [101.5, 117.5]

Answer

lower limit = 101.5
16.Confidence intervals

Answer: n = 133

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 = (1.645·21/3)² = 132.6

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 133

Answer

n = 133
17.Hypothesis tests & confidence intervals

Answer: t = 2.796

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 = (17·10² + 12·2²)/29 = 60.28
sp = 7.764

Standard error

SE = 7.764·√(1/18 + 1/13) = 2.826

Test statistic

t = (112 − 104.1)/2.826 = 2.796

Answer

t = 2.796

With df = 29, the two-sided 5% critical value is 2.045; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: L = 196.6

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 = 1.96·20/√64 = 4.9

Interval limit

L = 201.5 − 4.9 = 196.6

Answer

L = 196.6
19.Linear regression & correlation

Answer: r² = 0.9876

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 = 113/6 = 18.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 193
Syy = Σ(y − ȳ)² = 538.8

Coefficient of determination

r² = 193²/(70·538.8) = 0.9876

Answer

r² = 0.9876

About 98.76% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = −2.267

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 = 141/6 = 23.5
ȳ = 664/6 = 110.7

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 924

Regression line

b = 924/157.5 = 5.867
a = 110.7 − 5.867·23.5 = −27.2
ŷ = −27.2 + 5.867·x

Predicted value

ŷ = −27.2 + 5.867·26.5 = 128.3

Residual

e = 126 − 128.3 = −2.267

Answer

e = −2.267

The point lies below the line.

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