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
2025303540455055

Find the sample standard deviation s of the data: 59, 17, 43, 44, 21, 40, 30.

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

Exam scores are normally distributed with mean 170 and standard deviation 15. Find the z-score z of a score of 147.5.

μ = 170σ = 15x = 147.5
z =
3.ProbabilityMedium
xp(x)00.110.120.1530.340.35μ = 2.7

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

x: 0, 1, 2, 3, 4p(x): 0.1, 0.1, 0.15, 0.3, 0.35
σ =
4.ProbabilityHard
€p(x)−30.7590.21470.05

A game costs €3 to play. You win €12 with probability 0.2 and €150 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€12: p = 0.2€150: p = 0.05
E(G) = €
5.Normal distributionMedium
246.25259.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 246.25 and 259.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 246.25x2 = 259.25
P =
6.Normal distributionHard
557085100115130145pointsbottom 30%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points30%
x = points
7.Confidence intervalsMedium
x̄E₁ = 9.590%, n = 25x̄E₂ = ?90%, n = 100

A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 9.5. What is the margin of error E₂ if the same study uses 100 observations instead?

90%n₁ = 25E₁ = 9.5n₂ = 100
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(24) = 2.7970L = ?Ux̄₁ − x̄₂E99%

Two fertilisers are tested. Plot group 1: 16 plots, mean yield 87 kg, standard deviation 4.5 kg. Plot group 2: 10 plots, mean 94 kg, standard deviation 3 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 16x̄₁ = 87s₁ = 4.5n₂ = 10x̄₂ = 94s₂ = 399%t(24) = 2.797
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 51.61 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 51.61s = 5n = 81
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.01βσ = 25, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 5 units with a power of 80%. The standard deviation is 25. Find the minimum sample size n.

α = 0.01power = 80%δ = 5σ = 25zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
xy148258368470585684xy2466080

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy231335441543643754xy246304050ŷ = 45

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 45?

n = 6ŷ = 45
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.6

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 = 254/7 = 36.29

Squared deviations

Σ(x − x̄)² = 22.71² + (−19.29)² + 6.714² + 7.714² + (−15.29)² + 3.714² + (−6.286)² = 1279

Standard deviation

s = √(1279/6) = 14.6

Answer

s = 14.6
2.Descriptive statistics

Answer: z = −1.5

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 = (147.5 − 170)/15 = −1.5

Answer

z = −1.5
3.Probability

Answer: σ = 1.308

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 = 0·0.1 + 1·0.1 + 2·0.15 + 3·0.3 + 4·0.35 = 2.7

Mean of the squares

E(X²) = 0·0.1 + 1·0.1 + 4·0.15 + 9·0.3 + 16·0.35 = 9

Variance

Var(X) = 9 − 2.7² = 1.71
σ = √1.71 = 1.308

Answer

σ = 1.308
4.Probability

Answer: E(G) = 6.9 €

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: 9 (p = 0.2), 147 (p = 0.05), −3 (p = 0.75)

Expected value

E(G) = 9·0.2 + 147·0.05 − 3·0.75 = 6.9 €

Answer

E(G) = 6.9 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.7412

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 = (246.25 − 250)/5 = −0.75
z₂ = (259.25 − 250)/5 = 1.85

Area between

P = Φ(1.85) − Φ(−0.75) = 0.9678 − 0.2266 = 0.7412

Answer

P = 0.7412
6.Normal distribution

Answer: x = 92.2 points

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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 100 − 0.52·15 = 92.2 points

Answer

x = 92.2 points
7.Confidence intervals

Answer: E₂ = 4.75

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.

Ratio of margins

E₂/E₁ = (1.645/1.645)·√(25/100) = 0.5

New margin

E₂ = 9.5·0.5 = 4.75

Answer

E₂ = 4.75
8.Confidence intervals

Answer: lower limit = −11.51 kg

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.

Pooled standard deviation

sp² = (15·4.5² + 9·3²)/24 = 16.03
sp = 4.004

Margin of error

E = 2.797·4.004·√(1/16 + 1/10) = 4.514

Interval

−7 ± 4.514 → [−11.51, −2.486]

Answer

lower limit = −11.51 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.898

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 = 5/√81 = 5/9 = 0.5556

Test statistic

t = (51.61 − 50)/0.5556 = 2.898

Answer

t = 2.898
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 251

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 + 0.842)·25/5)² = 250.9

Round up

n = 251

Answer

n = 251
11.Linear regression & correlation

Answer: r² = 0.9457

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 = 21/6 = 3.5
ȳ = 413/6 = 68.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 131.5
Syy = Σ(y − ȳ)² = 1045

Coefficient of determination

r² = 131.5²/(17.5·1045) = 0.9457

Answer

r² = 0.9457

About 94.57% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 5.452

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 = 27/6 = 4.5
ȳ = 247/6 = 41.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 70.5

Regression line

b = 70.5/17.5 = 4.029
a = 41.17 − 4.029·4.5 = 23.04
ŷ = 23.04 + 4.029·x

Solve for x

x = (45 − 23.04)/4.029 = 5.452

Answer

x = 5.452

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