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
152025303540

Find the sample standard deviation s of the data: 27, 16, 44, 14, 39, 28.

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

Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 112.

μ = 100σ = 20x = 112
z =
3.ProbabilityMedium
xp(x)100.3200.1300.35400.25μ = 25.5

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

x: 10, 20, 30, 40p(x): 0.3, 0.1, 0.35, 0.25
Var(X) =
4.ProbabilityHard
€p(x)−30.8930.1570.01

A game costs €3 to play. You win €6 with probability 0.1 and €60 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€6: p = 0.1€60: p = 0.01
E(G) = €
5.Normal distributionMedium
200 kg200.7 kgT = X₁ + … + X₈?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 8 of them exceeds 200.7 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 8> 200.7 kg
P =
6.Normal distributionHard
μ = 250x = ?25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).

μ = 250σ = 5P = 25%
x = g
7.Confidence intervalsMedium
n = 6, s = 15.5LU = ?x̄ = 81E95%

A sample of 6 items has mean 81 and standard deviation s = 15.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 6x̄ = 81s = 15.595%t(5) = 2.571
upper limit =
8.Confidence intervalsHard
p ≈ 0.3, n = ?p̂ − Ep̂ + Ep̂E = 0.0499%

A pilot study suggests that about 0.3 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.3E = 0.0499%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 200.9 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 200.9s = 5n = 81
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 16 of 80 patients on treatment A recover, and 31 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 16n₁ = 80x₂ = 31n₂ = 120
z =
11.Linear regression & correlationMedium
xy274156218271033xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 103Σx² = 220Σy² = 2533Σxy = 746
r =
12.Linear regression & correlationHard
xy130226321421516616xy24615202530x = 3.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 26) is made. Find its residual e = y − ŷ.

n = 6(3.5, 26)
e =

3Solutions

1.Descriptive statistics

Answer: s = 11.98

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 = 168/6 = 28

Squared deviations

Σ(x − x̄)² = (−1)² + (−12)² + 16² + (−14)² + 11² + 0² = 718

Standard deviation

s = √(718/5) = 11.98

Answer

s = 11.98
2.Descriptive statistics

Answer: z = 0.6

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 = (112 − 100)/20 = 0.6

Answer

z = 0.6
3.Probability

Answer: Var(X) = 134.8

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 = 10·0.3 + 20·0.1 + 30·0.35 + 40·0.25 = 25.5

Mean of the squares

E(X²) = 100·0.3 + 400·0.1 + 900·0.35 + 1600·0.25 = 785

Variance

Var(X) = 785 − 25.5² = 134.8

Answer

Var(X) = 134.8
4.Probability

Answer: E(G) = −1.8 €

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: 3 (p = 0.1), 57 (p = 0.01), −3 (p = 0.89)

Expected value

E(G) = 3·0.1 + 57·0.01 − 3·0.89 = −1.8 €

Answer

E(G) = −1.8 €

A negative value means you lose money on average.

5.Normal distribution

Answer: P = 0.3121

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 total

μT = 200 kg, σT = 0.5·√8 = 1.414 kg

Standardise

z = (200.7 − 200)/1.414 = 0.49

Probability

P = 1 − Φ(0.49) = 0.3121

Answer

P = 0.3121
6.Normal distribution

Answer: x = 246.6 g

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 = 250 + −0.6745·5 = 246.63 g

Answer

x = 246.63 g
7.Confidence intervals

Answer: upper limit = 97.27

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 = 5, t = 2.571

Margin of error

E = 2.571·15.5/√6 = 16.27

Interval

81 ± 16.27 → [64.73, 97.27]

Answer

upper limit = 97.27
8.Confidence intervals

Answer: n = 871

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 = 2.576²·0.3·0.7/0.04² = 870.9

Round up

n = 871

Answer

n = 871
9.Hypothesis tests & confidence intervals

Answer: t = 1.62

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 = 5/√81 = 5/9 = 0.5556

Test statistic

t = (200.9 − 200)/0.5556 = 1.62

Answer

t = 1.62
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = −0.9532

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.

Sample proportions

p̂₁ = 16/80 = 0.2
p̂₂ = 31/120 = 0.2583
p̂ = 47/200 = 0.235

Standard error under H₀

SE = √(0.235·0.765·(1/80 + 1/120)) = 0.0612

Test statistic

z = (0.2 − 0.2583)/0.0612 = −0.9532

Answer

z = −0.9532

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.9981

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 = 746 − 30·103/5 = 128
Sxx = 220 − 30²/5 = 40
Syy = 2533 − 103²/5 = 411.2

Correlation coefficient

r = 128/√(40·411.2) = 0.9981

Answer

r = 0.9981
12.Linear regression & correlation

Answer: e = 4.333

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 = 21/6 = 3.5
ȳ = 130/6 = 21.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −50

Regression line

b = −50/17.5 = −2.857
a = 21.67 − (−2.857)·3.5 = 31.67
ŷ = 31.67 − 2.857·x

Predicted value

ŷ = 31.67 + (−2.857)·3.5 = 21.67

Residual

e = 26 − 21.67 = 4.333

Answer

e = 4.333

The point lies above the line.

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