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
253035404550

Find the sample standard deviation s of the data: 52, 51, 29, 21, 48, 34.

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

A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is above 47.5?

μ = 50σ = 5x = 47.5
P =
3.ProbabilityMedium
xp(x)00.310.320.1530.240.05μ = 1.4

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3, 4p(x): 0.3, 0.3, 0.15, 0.2, 0.05
σ =
4.ProbabilityHard
n = 10, p = 0.1: P(X = 4) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 4 correct answers?

n = 10p = 0.1k = 4
P =
5.Normal distributionMedium
105 s108.2 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 108.2 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 108.2 s
P =
6.Normal distributionHard
380420460500540580620hbottom 75%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h75%
x = h
7.Confidence intervalsMedium
n = 15, s = 6L = ?Ux̄ = 39E99%

A sample of 15 items has mean 39 and standard deviation s = 6. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 15x̄ = 39s = 699%t(14) = 2.977
lower limit =
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0399%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.03?

p = 0.5E = 0.0399%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 81 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 50 parts has mean 89.4 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 = 40x̄₁ = 81 mmσ₁ = 18 mmn₂ = 50x̄₂ = 89.4 mmσ₂ = 7 mm
z =
10.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 1.595%
n =
11.Linear regression & correlationMedium
xy126223320421518xy242025

The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy237334445548649752xy2464050x = 5.5

The table gives 6 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 = 5.5.

n = 6x = 5.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 12.98

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 = 235/6 = 39.17

Squared deviations

Σ(x − x̄)² = 12.83² + 11.83² + (−10.17)² + (−18.17)² + 8.833² + (−5.167)² = 842.8

Standard deviation

s = √(842.8/5) = 12.98

Answer

s = 12.98
2.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
3.Probability

Answer: σ = 1.241

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 = 0·0.3 + 1·0.3 + 2·0.15 + 3·0.2 + 4·0.05 = 1.4

Mean of the squares

E(X²) = 0·0.3 + 1·0.3 + 4·0.15 + 9·0.2 + 16·0.05 = 3.5

Variance

Var(X) = 3.5 − 1.4² = 1.54
σ = √1.54 = 1.241

Answer

σ = 1.241
4.Probability

Answer: P = 0.01116

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 = 4) = 210·0.14·0.96 = 210·0.0001·0.5314 = 0.01116

Answer

P = 0.01116
5.Normal distribution

Answer: P = 0.3085

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 sum

μ = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (108.2 − 105)/6.403 = 0.5

Upper tail

P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085

Answer

P = 0.3085
6.Normal distribution

Answer: x = 527 h

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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 500 + 0.67·40 = 527 h

Answer

x = 527 h
7.Confidence intervals

Answer: lower limit = 34.39

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 = 14, t = 2.977

Margin of error

E = 2.977·6/√15 = 4.612

Interval

39 ± 4.612 → [34.39, 43.61]

Answer

lower limit = 34.39
8.Confidence intervals

Answer: n = 1844

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 = 2.576²·0.5·0.5/0.03² = 1843

Round up

n = 1844

Answer

n = 1844
9.Hypothesis tests & confidence intervals

Answer: z = −2.788

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 = √(18²/40 + 7²/50) = √9.08 = 3.013

Test statistic

z = (81 − 89.4)/3.013 = −2.788

Answer

z = −2.788

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

10.Hypothesis tests & confidence intervals

Answer: n = 246

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).

Solve for n

n = (1.96·12/1.5)² = 245.9

Round up

n = 246

Answer

n = 246
11.Linear regression & correlation

Answer: r = −0.9333

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 = 15/5 = 3
ȳ = 108/5 = 21.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −18
Syy = Σ(y − ȳ)² = 37.2

Correlation coefficient

r = −18/√(10·37.2) = −0.9333

Answer

r = −0.9333
12.Linear regression & correlation

Answer: ŷ = 47.68

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 = 27/6 = 4.5
ȳ = 265/6 = 44.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 61.5

Regression line

b = 61.5/17.5 = 3.514
a = 44.17 − 3.514·4.5 = 28.35
ŷ = 28.35 + 3.514·x

Prediction

ŷ = 28.35 + 3.514·5.5 = 47.68

Answer

ŷ = 47.68

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