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
25303540455055

Find the sample standard deviation s of the data: 54, 44, 39, 50, 55, 47, 24.

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

A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is above 36?

μ = 60σ = 20x = 36
P =
3.ProbabilityMedium
red: 8blue: 7

A bag holds 8 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 7
P =
4.ProbabilityHard
€p(x)−10.720.25190.05

A game costs €1 to play. You win €3 with probability 0.25 and €20 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€3: p = 0.25€20: p = 0.05
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.4313

Drag z so that the shaded area between −z and z under the standard normal curve is 0.4313. Find z to two decimals.

A = 0.4313
z =
6.Normal distributionHard
μ = 178x = ?25%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 25% of them lie (the 25th percentile).

μ = 178σ = 7P = 25%
x = cm
7.Confidence intervalsMedium
n = 10, sd​ = 7.70LU = ?d̄ = −4.9E95%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −4.9 and standard deviation 7.7. Find the upper limit of the 95% confidence interval for the mean difference.

n = 10d̄ = −4.9sd = 7.795%t(9) = 2.262
upper limit = mmHg
8.Confidence intervalsHard
p ≈ 0.8, n = ?p̂ − Ep̂ + Ep̂E = 0.0490%

A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.8E = 0.0490%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 50 parts has mean length 124 mm, and the line's known standard deviation is 11 mm. Line 2: a sample of 30 parts has mean 129.8 mm, known standard deviation 7 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 124 mmσ₁ = 11 mmn₂ = 30x̄₂ = 129.8 mmσ₂ = 7 mm
z =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.05βσ = 25, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 80%. The standard deviation is 25. Find the minimum sample size n.

α = 0.05power = 80%δ = 3σ = 25zα = 1.645zβ = 0.842
n =
11.Linear regression & correlationMedium
xy3635727759891110213107xy5106080100

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy351453561665765xy4650556065x = 4.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 5x = 4.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 10.7

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 = 313/7 = 44.71

Squared deviations

Σ(x − x̄)² = 9.286² + (−0.7143)² + (−5.714)² + 5.286² + 10.29² + 2.286² + (−20.71)² = 687.4

Standard deviation

s = √(687.4/6) = 10.7

Answer

s = 10.7
2.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
3.Probability

Answer: P = 0.2667

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 = 8/15 · 7/14 = 56/210 = 0.2667

Answer

P = 0.2667
4.Probability

Answer: E(G) = 0.75 €

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: 2 (p = 0.25), 19 (p = 0.05), −1 (p = 0.7)

Expected value

E(G) = 2·0.25 + 19·0.05 − 1·0.7 = 0.75 €

Answer

E(G) = 0.75 €

A positive value means you win money on average.

5.Normal distribution

Answer: z = 0.57

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 + 0.4313)/2 = 0.7157

Read the z-table backwards

Find 0.7157 inside the table and read off z.

z = 0.57

Answer

z = 0.57
6.Normal distribution

Answer: x = 173.3 cm

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) = 0.25 → z = −0.6745

Back to x

x = 178 + −0.6745·7 = 173.28 cm

Answer

x = 173.28 cm
7.Confidence intervals

Answer: upper limit = 0.6079 mmHg

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.

Critical value

df = 9, t = 2.262

Margin of error

E = 2.262·7.7/√10 = 5.508

Interval

−4.9 ± 5.508 → [−10.41, 0.6079]

Answer

upper limit = 0.6079 mmHg
8.Confidence intervals

Answer: n = 271

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 = 1.645²·0.8·0.2/0.04² = 270.6

Round up

n = 271

Answer

n = 271
9.Hypothesis tests & confidence intervals

Answer: z = −2.881

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.

Standard error of the difference

SE = √(11²/50 + 7²/30) = √4.053 = 2.013

Test statistic

z = (124 − 129.8)/2.013 = −2.881

Answer

z = −2.881

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 430

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 + 0.842)·25/3)² = 429.5

Round up

n = 430

Answer

n = 430
11.Linear regression & correlation

Answer: r² = 0.973

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 = 48/6 = 8
ȳ = 508/6 = 84.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 324
Syy = Σ(y − ȳ)² = 1541

Coefficient of determination

r² = 324²/(70·1541) = 0.973

Answer

r² = 0.973

About 97.3% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 57

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 = 25/5 = 5
ȳ = 295/5 = 59

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 40

Regression line

b = 40/10 = 4
a = 59 − 4·5 = 39
ŷ = 39 + 4·x

Prediction

ŷ = 39 + 4·4.5 = 57

Answer

ŷ = 57

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