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: 38, 52, 32, 35, 15.

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

A quantity is normally distributed with mean 70 and standard deviation 10. What proportion P of values is below 65?

μ = 70σ = 10x = 65
P =
3.ProbabilityMedium
xp(x)100.35200.25300.3400.1μ = 21.5

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

x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.3, 0.1
Var(X) =
4.ProbabilityHard
€p(x)−50.84100.15450.01

A game costs €5 to play. You win €15 with probability 0.15 and €50 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€15: p = 0.15€50: p = 0.01
E(G) = €
5.Normal distributionMedium
85130.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85 and 130.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 85x2 = 130.75
P =
6.Normal distributionHard
380420460500540580620hbottom 85%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 85% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h85%
x = h
7.Confidence intervalsMedium
x̄E₁ = 10.390%, n = 25x̄E₂ = ?99%, n = 50

A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 10.3. What is the margin of error E₂ of a 99% interval based on 50 observations?

90%n₁ = 25E₁ = 10.399%n₂ = 50
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(19) = 1.7290LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 11 plots, mean yield 79 kg, standard deviation 7.5 kg. Plot group 2: 10 plots, mean 82 kg, standard deviation 4.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 11x̄₁ = 79s₁ = 7.5n₂ = 10x̄₂ = 82s₂ = 4.590%t(19) = 1.729
E = kg
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 501.8 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 501.8s = 12n = 64
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.01βσ = 11, power 80%, 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 80%. The standard deviation is 11. Find the minimum sample size n.

α = 0.01power = 80%δ = 2.5σ = 11zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
xy360574787910311115xy5106080100120

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy2641061682010231226xy5101020x = 11

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 11.

n = 6x = 11
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.28

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 = 3.6² + 17.6² + (−2.4)² + 0.6² + (−19.4)² = 705.2

Standard deviation

s = √(705.2/4) = 13.28

Answer

s = 13.28
2.Descriptive statistics

Answer: P = 0.3085

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 = 0.3085

Answer

P = 0.3085
3.Probability

Answer: Var(X) = 102.8

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 = 10·0.35 + 20·0.25 + 30·0.3 + 40·0.1 = 21.5

Mean of the squares

E(X²) = 100·0.35 + 400·0.25 + 900·0.3 + 1600·0.1 = 565

Variance

Var(X) = 565 − 21.5² = 102.8

Answer

Var(X) = 102.8
4.Probability

Answer: E(G) = −2.25 €

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: 10 (p = 0.15), 45 (p = 0.01), −5 (p = 0.84)

Expected value

E(G) = 10·0.15 + 45·0.01 − 5·0.84 = −2.25 €

Answer

E(G) = −2.25 €

A negative value means you lose money on average.

5.Normal distribution

Answer: P = 0.8212

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 = (85 − 100)/15 = −1
z₂ = (130.75 − 100)/15 = 2.05

Area between

P = Φ(2.05) − Φ(−1) = 0.9798 − 0.1587 = 0.8212

Answer

P = 0.8212
6.Normal distribution

Answer: x = 542 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) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 500 + 1.04·40 = 542 h

Answer

x = 542 h
7.Confidence intervals

Answer: E₂ = 11.41

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.

Ratio of margins

E₂/E₁ = (2.576/1.645)·√(25/50) = 1.107

New margin

E₂ = 10.3·1.107 = 11.41

Answer

E₂ = 11.41
8.Confidence intervals

Answer: E = 4.73 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 = (10·7.5² + 9·4.5²)/19 = 39.2
sp = 6.261

Margin of error

E = 1.729·6.261·√(1/11 + 1/10) = 4.73

Answer

E = 4.73 kg
9.Hypothesis tests & confidence intervals

Answer: t = 1.2

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/√64 = 12/8 = 1.5

Test statistic

t = (501.8 − 500)/1.5 = 1.2

Answer

t = 1.2
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 195

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 + 0.842)·11/2.5)² = 194.3

Round up

n = 195

Answer

n = 195
11.Linear regression & correlation

Answer: r² = 0.9986

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 = 439/5 = 87.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 278
Syy = Σ(y − ȳ)² = 1935

Coefficient of determination

r² = 278²/(40·1935) = 0.9986

Answer

r² = 0.9986

About 99.86% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 25

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 = 42/6 = 7
ȳ = 101/6 = 16.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 143

Regression line

b = 143/70 = 2.043
a = 16.83 − 2.043·7 = 2.533
ŷ = 2.533 + 2.043·x

Prediction

ŷ = 2.533 + 2.043·11 = 25

Answer

ŷ = 25

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