Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

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.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 46, 46, 24, 54, 48, 13, 45.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is above 65?

μ = 70σ = 5x = 65
P =
3.ProbabilityMedium
xp(x)00.0510.3520.3530.25μ = 1.8

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3p(x): 0.05, 0.35, 0.35, 0.25
σ =
4.ProbabilityHard
n = 10, p = 0.2: P(X = 2) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 2 correct answers?

n = 10p = 0.2k = 2
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8165

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8165. Find z to two decimals.

A = 0.8165
z =
6.Normal distributionHard
157164171178185192199cmbottom 15%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 15% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm15%
x = cm
7.Confidence intervalsMedium
n = 15, sd​ = 5.60L = ?Ud̄ = −8.1E95%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −8.1 and standard deviation 5.6. Find the lower limit of the 95% confidence interval for the mean difference.

n = 15d̄ = −8.1sd = 5.695%t(14) = 2.145
lower limit = mmHg
8.Confidence intervalsHard
p ≈ 0.6, n = ?p̂ − Ep̂ + Ep̂E = 0.0395%

A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.03?

p ≈ 0.6E = 0.0395%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 102.6 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 102.6s = 10n = 64
t =
10.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 190%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 190%
n =
11.Linear regression & correlationMedium
xy140243353454558xy24405060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy1874207622962410026110xy202580100x = 19

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (19, 74) is made. Find its residual e = y − ŷ.

n = 5(19, 74)
e =

3Solutions

1.Descriptive statistics

Answer: s = 14.94

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 = 276/7 = 39.43

Squared deviations

Σ(x − x̄)² = 6.571² + 6.571² + (−15.43)² + 14.57² + 8.571² + (−26.43)² + 5.571² = 1340

Standard deviation

s = √(1340/6) = 14.94

Answer

s = 14.94
2.Descriptive statistics

Answer: P = 0.8413

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) = 0.1587
P = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
3.Probability

Answer: σ = 0.8718

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 + 1·0.35 + 2·0.35 + 3·0.25 = 1.8

Mean of the squares

E(X²) = 0·0.05 + 1·0.35 + 4·0.35 + 9·0.25 = 4

Variance

Var(X) = 4 − 1.8² = 0.76
σ = √0.76 = 0.8718

Answer

σ = 0.8718
4.Probability

Answer: P = 0.302

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 = 2) = 45·0.22·0.88 = 45·0.04·0.1678 = 0.302

Answer

P = 0.302
5.Normal distribution

Answer: z = 1.33

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 + 0.8165)/2 = 0.9082

Read the z-table backwards

Find 0.9082 inside the table and read off z.

z = 1.33

Answer

z = 1.33
6.Normal distribution

Answer: x = 170.7 cm

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) = 0.15

z from the table

z ≈ −1.04

Back to the original units

x = 178 − 1.04·7 = 170.7 cm

Answer

x = 170.7 cm
7.Confidence intervals

Answer: lower limit = −11.2 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 = 14, t = 2.145

Margin of error

E = 2.145·5.6/√15 = 3.101

Interval

−8.1 ± 3.101 → [−11.2, −4.999]

Answer

lower limit = −11.2 mmHg
8.Confidence intervals

Answer: n = 1025

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.96²·0.6·0.4/0.03² = 1024

Round up

n = 1025

Answer

n = 1025
9.Hypothesis tests & confidence intervals

Answer: t = 2.08

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

SE = 10/√64 = 10/8 = 1.25

Test statistic

t = (102.6 − 100)/1.25 = 2.08

Answer

t = 2.08
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 390

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).

Solve for n

n = (1.645·12/1)² = 389.7

Round up

n = 390

Answer

n = 390
11.Linear regression & correlation

Answer: r² = 0.9313

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 = 248/5 = 49.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 47
Syy = Σ(y − ȳ)² = 237.2

Coefficient of determination

r² = 47²/(10·237.2) = 0.9313

Answer

r² = 0.9313

About 93.13% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = −2.8

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 = 110/5 = 22
ȳ = 456/5 = 91.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 192

Regression line

b = 192/40 = 4.8
a = 91.2 − 4.8·22 = −14.4
ŷ = −14.4 + 4.8·x

Predicted value

ŷ = −14.4 + 4.8·19 = 76.8

Residual

e = 74 − 76.8 = −2.8

Answer

e = −2.8

The point lies below the line.

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