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
262830323436

Find the sample standard deviation s of the data: 25, 36, 36, 36, 35, 26.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 68?

μ = 50σ = 12x = 68
P =
3.ProbabilityMedium
red: 8blue: 9

A bag holds 8 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 9
P =
4.ProbabilityHard
n = 9, p = 0.3: P(X = 5) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 5 correct answers?

n = 9p = 0.3k = 5
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9131

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9131. Find z to two decimals.

A = 0.9131
z =
6.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
7.Confidence intervalsMedium
x̄E₁ = 4.190%, n = 25x̄E₂ = ?90%, n = 50

A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 4.1. What is the margin of error E₂ if the same study uses 50 observations instead?

90%n₁ = 25E₁ = 4.1n₂ = 50
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(24) = 2.0640LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 20 plots, mean yield 16.5 kg, standard deviation 7.5 kg. Plot group 2: 6 plots, mean 19.5 kg, standard deviation 8.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 20x̄₁ = 16.5s₁ = 7.5n₂ = 6x̄₂ = 19.5s₂ = 8.595%t(24) = 2.064
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 116.1 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 116.1s = 12n = 64
t =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 100 measurements has mean 119.6. The population standard deviation is 4. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 119.6σ = 4n = 10090%
U =
11.Linear regression & correlationMedium
xy2404446528531058xy510405060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 247Σx² = 220Σy² = 12413Σxy = 1572
r =
12.Linear regression & correlationHard
xy3235297389351143xy5103040x = 6

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 6.

n = 5x = 6
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 5.317

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 = 194/6 = 32.33

Squared deviations

Σ(x − x̄)² = (−7.333)² + 3.667² + 3.667² + 3.667² + 2.667² + (−6.333)² = 141.3

Standard deviation

s = √(141.3/5) = 5.317

Answer

s = 5.317
2.Descriptive statistics

Answer: P = 0.06681

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
3.Probability

Answer: P = 0.2059

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 = 8/17 · 7/16 = 56/272 = 0.2059

Answer

P = 0.2059
4.Probability

Answer: P = 0.07351

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 = 5) = 126·0.35·0.74 = 126·0.00243·0.2401 = 0.07351

Answer

P = 0.07351
5.Normal distribution

Answer: z = −1.36

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 − 0.9131 = 0.0869

Read the z-table backwards

Find 0.0869 inside the table and read off z.

z = −1.36

Answer

z = −1.36
6.Normal distribution

Answer: P = 0.7389

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 difference

μ = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
7.Confidence intervals

Answer: E₂ = 2.899

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)·√(25/50) = 0.7071

New margin

E₂ = 4.1·0.7071 = 2.899

Answer

E₂ = 2.899
8.Confidence intervals

Answer: upper limit = 4.416 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 = (19·7.5² + 5·8.5²)/24 = 59.58
sp = 7.719

Margin of error

E = 2.064·7.719·√(1/20 + 1/6) = 7.416

Interval

−3 ± 7.416 → [−10.42, 4.416]

Answer

upper limit = 4.416 kg
9.Hypothesis tests & confidence intervals

Answer: t = −2.6

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 = 12/√64 = 12/8 = 1.5

Test statistic

t = (116.1 − 120)/1.5 = −2.6

Answer

t = −2.6
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: U = 120.3

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).

Margin of error

E = 1.645·4/√100 = 0.658

Interval limit

U = 119.6 + 0.658 = 120.3

Answer

U = 120.3
11.Linear regression & correlation

Answer: r = 0.9792

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 = 1572 − 30·247/5 = 90
Sxx = 220 − 30²/5 = 40
Syy = 12413 − 247²/5 = 211.2

Correlation coefficient

r = 90/√(40·211.2) = 0.9792

Answer

r = 0.9792
12.Linear regression & correlation

Answer: ŷ = 31.3

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 = 35/5 = 7
ȳ = 168/5 = 33.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 92

Regression line

b = 92/40 = 2.3
a = 33.6 − 2.3·7 = 17.5
ŷ = 17.5 + 2.3·x

Prediction

ŷ = 17.5 + 2.3·6 = 31.3

Answer

ŷ = 31.3

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