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
15202530354045

Find the sample standard deviation s of the data: 23, 24, 32, 46, 15, 19, 17.

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

A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is below 440?

μ = 500σ = 40x = 440
P =
3.ProbabilityMedium
xp(x)10.3520.2530.3540.05μ = 2.1

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4p(x): 0.35, 0.25, 0.35, 0.05
Var(X) =
4.ProbabilityHard
n = 10, p = 0.6: P(X = 2) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 2 correct answers?

n = 10p = 0.6k = 2
P =
5.Normal distributionMedium
6077P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 60 and 77 points? Give P as a decimal.

μ = 65σ = 10x1 = 60x2 = 77
P =
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
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 80 phones, mean 67.5 h, known standard deviation 3 h. Model 2: 40 phones, mean 70.5 h, known standard deviation 13 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 67.5 hσ₁ = 3 hn₂ = 40x̄₂ = 70.5 hσ₂ = 13 h90%
lower limit = h
8.Confidence intervalsHard
μ₁ − μ₂, t(10) = 1.8120L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 96 kg, standard deviation 4 kg. Plot group 2: 6 plots, mean 107 kg, standard deviation 8.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 96s₁ = 4n₂ = 6x̄₂ = 107s₂ = 8.590%t(10) = 1.812
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 52.04 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 52.04s = 12n = 100
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.01βσ = 16, power 95%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 2.5 units with a power of 95%. The standard deviation is 16. Find the minimum sample size n.

α = 0.01power = 95%δ = 2.5σ = 16zα = 2.326zβ = 1.645
n =
11.Linear regression & correlationMedium
xy175420722389268829111xy2025306080100

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 115Σy = 414Σx² = 2735Σy² = 36086Σxy = 9912
r =
12.Linear regression & correlationHard
xy22486108111017xy51051015ŷ = 5

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. For which value x does the line predict ŷ = 5?

n = 5ŷ = 5
x =

3Solutions

1.Descriptive statistics

Answer: s = 10.76

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 = 176/7 = 25.14

Squared deviations

Σ(x − x̄)² = (−2.143)² + (−1.143)² + 6.857² + 20.86² + (−10.14)² + (−6.143)² + (−8.143)² = 694.9

Standard deviation

s = √(694.9/6) = 10.76

Answer

s = 10.76
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) = 0.06681
P = 0.06681

Answer

P = 0.06681
3.Probability

Answer: Var(X) = 0.89

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 = 1·0.35 + 2·0.25 + 3·0.35 + 4·0.05 = 2.1

Mean of the squares

E(X²) = 1·0.35 + 4·0.25 + 9·0.35 + 16·0.05 = 5.3

Variance

Var(X) = 5.3 − 2.1² = 0.89

Answer

Var(X) = 0.89
4.Probability

Answer: P = 0.01062

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 = 2) = 45·0.62·0.48 = 45·0.36·0.0006554 = 0.01062

Answer

P = 0.01062
5.Normal distribution

Answer: P = 0.5764

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 = (60 − 65)/10 = −0.5
z₂ = (77 − 65)/10 = 1.2

Area between

P = Φ(1.2) − Φ(−0.5) = 0.8849 − 0.3085 = 0.5764

Answer

P = 0.5764
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: lower limit = −6.426 h

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.

Point estimate

x̄₁ − x̄₂ = 67.5 − 70.5 = −3

Standard error

SE = √(3²/80 + 13²/40) = 2.083

Margin of error

E = 1.645·2.083 = 3.426

Interval

−3 ± 3.426 → [−6.426, 0.426]

Answer

lower limit = −6.426 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: lower limit = −17.95 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 = (5·4² + 5·8.5²)/10 = 44.13
sp = 6.643

Margin of error

E = 1.812·6.643·√(1/6 + 1/6) = 6.949

Interval

−11 ± 6.949 → [−17.95, −4.051]

Answer

lower limit = −17.95 kg
9.Hypothesis tests & confidence intervals

Answer: t = 1.7

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/√100 = 12/10 = 1.2

Test statistic

t = (52.04 − 50)/1.2 = 1.7

Answer

t = 1.7
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 646

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 + 1.645)·16/2.5)² = 645.9

Round up

n = 646

Answer

n = 646
11.Linear regression & correlation

Answer: r = 0.9671

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 = 9912 − 115·414/5 = 390
Sxx = 2735 − 115²/5 = 90
Syy = 36086 − 414²/5 = 1807

Correlation coefficient

r = 390/√(90·1807) = 0.9671

Answer

r = 0.9671
12.Linear regression & correlation

Answer: x = 3.212

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 = 48/5 = 9.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 66

Regression line

b = 66/40 = 1.65
a = 9.6 − 1.65·6 = −0.3
ŷ = −0.3 + 1.65·x

Solve for x

x = (5 − (−0.3))/1.65 = 3.212

Answer

x = 3.212

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