Introductory Statistics: cram pack

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

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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: 46, 51, 28, 20, 27, 37, 59.

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 162.5.

μ = 170σ = 15x = 162.5
z =
3.ProbabilityMedium
n = 6k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 6? The order does not matter.

n = 6k = 4
N =
4.ProbabilityHard
n = 7, p = 0.1: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 7p = 0.1
P =
5.Normal distributionMedium
96.25118P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 96.25 and 118 points? Give P as a decimal.

μ = 100σ = 15x1 = 96.25x2 = 118
P =
6.Normal distributionHard
157164171178185192199cmtop 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 top 15% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm15%
x = cm
7.Confidence intervalsMedium
n = 9, s = 6L = ?Ux̄ = 28E95%

A sample of 9 items has mean 28 and standard deviation s = 6. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 9x̄ = 28s = 695%t(8) = 2.306
lower limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(14) = 2.9770LU = ?x̄₁ − x̄₂E99%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 27 kg, standard deviation 7 kg. Plot group 2: 10 plots, mean 33 kg, standard deviation 2.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 27s₁ = 7n₂ = 10x̄₂ = 33s₂ = 2.599%t(14) = 2.977
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 52.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 52.4s = 8n = 64
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 21, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 90%. The standard deviation is 21. Find the minimum sample size n.

α = 0.01power = 90%δ = 4σ = 21zα = 2.326zβ = 1.282
n =
11.Linear regression & correlationMedium
xy2842062483310411248xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 174Σx² = 364Σy² = 6114Σxy = 1490
r =
12.Linear regression & correlationHard
xy2104146228291036xy510102030x = 7

The table gives 5 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 = 5x = 7
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 14.26

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 = 268/7 = 38.29

Squared deviations

Σ(x − x̄)² = 7.714² + 12.71² + (−10.29)² + (−18.29)² + (−11.29)² + (−1.286)² + 20.71² = 1219

Standard deviation

s = √(1219/6) = 14.26

Answer

s = 14.26
2.Descriptive statistics

Answer: z = −0.5

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 = (162.5 − 170)/15 = −0.5

Answer

z = −0.5
3.Probability

Answer: N = 15

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(6, 4) = (6·5·4·3)/(4·3·2·1) = 15

Answer

N = 15
4.Probability

Answer: P = 0.8503

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) = 0.97 = 0.4783
P(X = 1) = 7·0.1·0.96 = 0.372

Add them

P = 0.4783 + 0.372 = 0.8503

Answer

P = 0.8503
5.Normal distribution

Answer: P = 0.4836

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 = (96.25 − 100)/15 = −0.25
z₂ = (118 − 100)/15 = 1.2

Area between

P = Φ(1.2) − Φ(−0.25) = 0.8849 − 0.4013 = 0.4836

Answer

P = 0.4836
6.Normal distribution

Answer: x = 185.3 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) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 178 + 1.04·7 = 185.3 cm

Answer

x = 185.3 cm
7.Confidence intervals

Answer: lower limit = 23.39

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 = 8, t = 2.306

Margin of error

E = 2.306·6/√9 = 4.612

Interval

28 ± 4.612 → [23.39, 32.61]

Answer

lower limit = 23.39
8.Confidence intervals

Answer: upper limit = 1.131 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 = (5·7² + 9·2.5²)/14 = 21.52
sp = 4.639

Margin of error

E = 2.977·4.639·√(1/6 + 1/10) = 7.131

Interval

−6 ± 7.131 → [−13.13, 1.131]

Answer

upper limit = 1.131 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.4

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 = 8/√64 = 8/8 = 1

Test statistic

t = (52.4 − 50)/1 = 2.4

Answer

t = 2.4
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 = 359

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)·21/4)² = 358.8

Round up

n = 359

Answer

n = 359
11.Linear regression & correlation

Answer: r = 0.9948

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.

Centred sums

Sxy = 1490 − 42·174/6 = 272
Sxx = 364 − 42²/6 = 70
Syy = 6114 − 174²/6 = 1068

Correlation coefficient

r = 272/√(70·1068) = 0.9948

Answer

r = 0.9948
12.Linear regression & correlation

Answer: ŷ = 25.55

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 = 30/5 = 6
ȳ = 111/5 = 22.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 134

Regression line

b = 134/40 = 3.35
a = 22.2 − 3.35·6 = 2.1
ŷ = 2.1 + 3.35·x

Prediction

ŷ = 2.1 + 3.35·7 = 25.55

Answer

ŷ = 25.55

ExamProblems · examproblems.com/courses/statistics-1