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
1520253035404550

Find the sample standard deviation s of the data: 52, 41, 11, 42, 26.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is below 85?

μ = 100σ = 15x = 85
P =
3.ProbabilityMedium
n = 7k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 7? The order does not matter.

n = 7k = 2
N =
4.ProbabilityHard
€p(x)−50.83200.15450.02

A game costs €5 to play. You win €25 with probability 0.15 and €50 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€25: p = 0.15€50: p = 0.02
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8461

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8461. Find z to two decimals.

A = 0.8461
z =
6.Normal distributionHard
μ = 100x = ?99%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 99% of them lie (the 99th percentile).

μ = 100σ = 15P = 99%
x = points
7.Confidence intervalsMedium
n = 21, s = 13L = ?Ux̄ = 10.5E95%

A sample of 21 items has mean 10.5 and standard deviation s = 13. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 21x̄ = 10.5s = 1395%t(20) = 2.086
lower limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(19) = 2.0930LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 5 plots, mean yield 69.5 kg, standard deviation 8 kg. Plot group 2: 16 plots, mean 77.5 kg, standard deviation 3.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 5x̄₁ = 69.5s₁ = 8n₂ = 16x̄₂ = 77.5s₂ = 3.595%t(19) = 2.093
E = kg
9.Hypothesis tests & confidence intervalsMedium
−2.0862.086t = ?df = 20

Two teaching methods are compared. Group 1: 7 students, mean score 110, standard deviation 3. Group 2: 15 students, mean score 98, standard deviation 11. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 7x̄₁ = 110s₁ = 3n₂ = 15x̄₂ = 98s₂ = 11
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.05βσ = 6, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 95%. The standard deviation is 6. Find the minimum sample size n.

α = 0.05power = 95%δ = 3σ = 6zα = 1.645zβ = 1.645
n =
11.Linear regression & correlationMedium
xy167019772285251012812431126xy1520253080100120

The table gives 6 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 = 6Σx = 141Σy = 583Σx² = 3471Σy² = 59507Σxy = 14356
r =
12.Linear regression & correlationHard
xy3475617719801188xy5106080x = 6

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (6, 67) is made. Find its residual e = y − ŷ.

n = 5(6, 67)
e =

3Solutions

1.Descriptive statistics

Answer: s = 16.04

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 = 172/5 = 34.4

Squared deviations

Σ(x − x̄)² = 17.6² + 6.6² + (−23.4)² + 7.6² + (−8.4)² = 1029

Standard deviation

s = √(1029/4) = 16.04

Answer

s = 16.04
2.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(−1) = 0.1587
P = 0.1587

Answer

P = 0.1587
3.Probability

Answer: N = 21

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(7, 2) = (7·6)/(2·1) = 21

Answer

N = 21
4.Probability

Answer: E(G) = −0.25 €

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: 20 (p = 0.15), 45 (p = 0.02), −5 (p = 0.83)

Expected value

E(G) = 20·0.15 + 45·0.02 − 5·0.83 = −0.25 €

Answer

E(G) = −0.25 €

A negative value means you lose money on average.

5.Normal distribution

Answer: z = −1.02

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 − 0.8461 = 0.1539

Read the z-table backwards

Find 0.1539 inside the table and read off z.

z = −1.02

Answer

z = −1.02
6.Normal distribution

Answer: x = 134.9 points

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 for the percentile

Φ(z) = 0.99 → z = 2.326

Back to x

x = 100 + 2.326·15 = 134.89 points

Answer

x = 134.89 points
7.Confidence intervals

Answer: lower limit = 4.582

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 = 20, t = 2.086

Margin of error

E = 2.086·13/√21 = 5.918

Interval

10.5 ± 5.918 → [4.582, 16.42]

Answer

lower limit = 4.582
8.Confidence intervals

Answer: E = 5.159 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 = (4·8² + 15·3.5²)/19 = 23.14
sp = 4.811

Margin of error

E = 2.093·4.811·√(1/5 + 1/16) = 5.159

Answer

E = 5.159 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.804

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.

Pooled variance

sp² = (6·3² + 14·11²)/20 = 87.4
sp = 9.349

Standard error

SE = 9.349·√(1/7 + 1/15) = 4.279

Test statistic

t = (110 − 98)/4.279 = 2.804

Answer

t = 2.804

With df = 20, the two-sided 5% critical value is 2.086; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 44

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 = ((1.645 + 1.645)·6/3)² = 43.3

Round up

n = 44

Answer

n = 44
11.Linear regression & correlation

Answer: r = 0.9769

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 = 14356 − 141·583/6 = 655.5
Sxx = 3471 − 141²/6 = 157.5
Syy = 59507 − 583²/6 = 2859

Correlation coefficient

r = 655.5/√(157.5·2859) = 0.9769

Answer

r = 0.9769
12.Linear regression & correlation

Answer: e = 2.65

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 = 347/5 = 69.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 202

Regression line

b = 202/40 = 5.05
a = 69.4 − 5.05·7 = 34.05
ŷ = 34.05 + 5.05·x

Predicted value

ŷ = 34.05 + 5.05·6 = 64.35

Residual

e = 67 − 64.35 = 2.65

Answer

e = 2.65

The point lies above the line.

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