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: 39, 28, 20, 58, 47, 49, 31.

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

Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 115.

μ = 100σ = 15x = 115
z =
3.ProbabilityMedium
red: 5blue: 5

A bag holds 5 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 5
P =
4.ProbabilityHard
n = 10, p = 0.6: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 10p = 0.6
P =
5.Normal distributionMedium
640 g642.4 gT = X₁ + … + X₁₆?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 16 of them is below 642.4 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 16< 642.4 g
P =
6.Normal distributionHard
35455565758595pointstop 75%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 75% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points75%
x = points
7.Confidence intervalsMedium
n = 11, s = 4LUx̄ = 112.5E = ?90%

A sample of 11 items has mean 112.5 and standard deviation s = 4. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 11x̄ = 112.5s = 490%t(10) = 1.812
E =
8.Confidence intervalsHard
p ≈ 0.8, n = ?p̂ − Ep̂ + Ep̂E = 0.0499%

A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.8E = 0.0499%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −0.78 and the standard deviation of the differences is 1.8. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = −0.78sd = 1.8
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 83 of 150 patients on treatment A recover, and 102 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 83n₁ = 150x₂ = 102n₂ = 200
z =
11.Linear regression & correlationMedium
xy2741261681610241228xy510102030

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 =
12.Linear regression & correlationHard
xy2541161682410271234xy510102030x = 7

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 7.

n = 6x = 7
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.36

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 = 272/7 = 38.86

Squared deviations

Σ(x − x̄)² = 0.1429² + (−10.86)² + (−18.86)² + 19.14² + 8.143² + 10.14² + (−7.857)² = 1071

Standard deviation

s = √(1071/6) = 13.36

Answer

s = 13.36
2.Descriptive statistics

Answer: z = 1

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 = (115 − 100)/15 = 1

Answer

z = 1
3.Probability

Answer: P = 0.2222

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.

First and second draw

P = 5/10 · 4/9 = 20/90 = 0.2222

Answer

P = 0.2222
4.Probability

Answer: P = 0.001678

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.

Zero and one failure

P(X = 0) = 0.410 = 0.0001049
P(X = 1) = 10·0.6·0.49 = 0.001573

Add them

P = 0.0001049 + 0.001573 = 0.001678

Answer

P = 0.001678
5.Normal distribution

Answer: P = 0.6915

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 total

μT = 640 g, σT = 1.2·√16 = 4.8 g

Standardise

z = (642.4 − 640)/4.8 = 0.5

Probability

P = Φ(0.5) = 0.6915

Answer

P = 0.6915
6.Normal distribution

Answer: x = 58.3 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) = 0.25

z from the table

z ≈ −0.67

Back to the original units

x = 65 − 0.67·10 = 58.3 points

Answer

x = 58.3 points
7.Confidence intervals

Answer: E = 2.185

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 = 10, t = 1.812

Margin of error

E = 1.812·4/√11 = 2.185

Answer

E = 2.185
8.Confidence intervals

Answer: n = 664

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 = 2.576²·0.8·0.2/0.04² = 663.6

Round up

n = 664

Answer

n = 664
9.Hypothesis tests & confidence intervals

Answer: t = −1.501

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 of the mean difference

SE = 1.8/√12 = 0.5196

Test statistic

t = −0.78/0.5196 = −1.501

Answer

t = −1.501

With df = 11, the two-sided 5% critical value is 2.201; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: z = 0.8037

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 = 83/150 = 0.5533
p̂₂ = 102/200 = 0.51
p̂ = 185/350 = 0.5286

Standard error under H₀

SE = √(0.5286·0.4714·(1/150 + 1/200)) = 0.05392

Test statistic

z = (0.5533 − 0.51)/0.05392 = 0.8037

Answer

z = 0.8037

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r² = 0.9568

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 = 103/6 = 17.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 141
Syy = Σ(y − ȳ)² = 296.8

Coefficient of determination

r² = 141²/(70·296.8) = 0.9568

Answer

r² = 0.9568

About 95.68% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 19.5

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 = 117/6 = 19.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 201

Regression line

b = 201/70 = 2.871
a = 19.5 − 2.871·7 = −0.6
ŷ = −0.6 + 2.871·x

Prediction

ŷ = −0.6 + 2.871·7 = 19.5

Answer

ŷ = 19.5

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