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)·p^k(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

Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
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

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

State H₀ and H₁ and decide whether the test is one- or two-sided.
Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
Compare with the critical value, or find the p-value.
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: S_xx = Σ(x − x̄)² = Σx² − (Σx)²/n, S_xy = Σxy − ΣxΣy/n
Slope: b = S_xy/S_xx
Intercept: a = ȳ − b·x̄
Correlation: r = S_xy/√(S_xxS_yy)
Residual: e = y − ŷ

2Practice problems

1.Descriptive statisticsMedium

Find the sample standard deviation s of the data: 20, 14, 24, 32, 60, 45.

n = 6

s =

2.Descriptive statisticsHard

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 30.

μ = 70σ = 20x = 30

z =

3.ProbabilityMedium

In how many ways N can a committee of 3 people be chosen from a group of 14? The order does not matter.

n = 14k = 3

N =

4.ProbabilityHard

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 4 correct answers?

n = 6p = 0.7k = 4

P =

5.Normal distributionMedium

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 8 of them exceeds 94.9 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 8> 94.9 kg

P =

6.Normal distributionHard

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).

μ = 100σ = 15P = 80%

x = points

7.Confidence intervalsMedium

A sample of 12 items has mean 28 and standard deviation s = 12.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 12x̄ = 28s = 12.595%t(11) = 2.201

lower limit =

8.Confidence intervalsHard

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.02?

p = 0.5E = 0.0299%

n =

9.Hypothesis tests & confidence intervalsMedium

A sample of 100 students has a mean score of 196.9 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 196.9s = 10n = 100

t =

10.Hypothesis tests & confidence intervalsHard

A one-sided z test at α = 0.01 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 30. Find the minimum sample size n.

α = 0.01power = 90%δ = 5σ = 30z_α = 2.326z_β = 1.282

n =

11.Linear regression & correlationMedium

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

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (4, 57) is made. Find its residual e = y − ŷ.

n = 6(4, 57)

e =

3Solutions

1.Descriptive statistics

Answer: s = 17.22

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 = 195/6 = 32.5

Squared deviations

Σ(x − x̄)² = (−12.5)² + (−18.5)² + (−8.5)² + (−0.5)² + 27.5² + 12.5² = 1484

Standard deviation

s = √(1484/5) = 17.22

Answer

s = 17.22

2.Descriptive statistics

Answer: z = −2

Standardise

z = (30 − 70)/20 = −2

Answer

z = −2

3.Probability

Answer: N = 364

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.

Combinations

C(14, 3) = (14·13·12)/(3·2·1) = 364

Answer

N = 364

4.Probability

Answer: P = 0.3241

Binomial formula

P(X = 4) = 15·0.7⁴·0.3² = 15·0.2401·0.09 = 0.3241

Answer

P = 0.3241

5.Normal distribution

Answer: P = 0.6879

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) = μ_X ± μ_Y. Variances always add, also for a difference: Var(X ± Y) = σ_X² + σ_Y², so σ = √(σ_X² + σ_Y²). 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 total

μ_T = 96 kg, σ_T = 0.8·√8 = 2.263 kg

Standardise

z = (94.9 − 96)/2.263 = −0.49

Probability

P = 1 − Φ(−0.49) = 0.6879

Answer

P = 0.6879

6.Normal distribution

Answer: x = 112.6 points

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 for the percentile

Φ(z) = 0.8 → z = 0.8416

Back to x

x = 100 + 0.8416·15 = 112.62 points

Answer

x = 112.62 points

7.Confidence intervals

Answer: lower limit = 20.06

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 = 11, t = 2.201

Margin of error

E = 2.201·12.5/√12 = 7.942

Interval

28 ± 7.942 → [20.06, 35.94]

Answer

lower limit = 20.06

8.Confidence intervals

Answer: n = 4148

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 = 2.576²·0.5·0.5/0.02² = 4147

Round up

n = 4148

Answer

n = 4148

9.Hypothesis tests & confidence intervals

Answer: t = −3.1

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/√100 = 10/10 = 1

Test statistic

t = (196.9 − 200)/1 = −3.1

Answer

t = −3.1

t_{0.025; df = 99} = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 469

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 1.282)·30/5)² = 468.6

Round up

n = 469

Answer

n = 469

11.Linear regression & correlation

Answer: r² = 0.8475

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 = 261/5 = 52.2

Sums of squares and products

S_xx = Σ(x − x̄)² = 10

S_xy = Σ(x − x̄)(y − ȳ) = 40

S_yy = Σ(y − ȳ)² = 188.8

Coefficient of determination

r² = 40²/(10·188.8) = 0.8475

Answer

r² = 0.8475

About 84.75% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = −1.662

Means

x̄ = 48/6 = 8

ȳ = 473/6 = 78.83

Sums of squares and products

S_xx = Σ(x − x̄)² = 70

S_xy = Σ(x − x̄)(y − ȳ) = 353

Regression line

b = 353/70 = 5.043

a = 78.83 − 5.043·8 = 38.49

ŷ = 38.49 + 5.043·x

Predicted value

ŷ = 38.49 + 5.043·4 = 58.66

Residual

e = 57 − 58.66 = −1.662

Answer

e = −1.662

The point lies below the line.

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