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
303540455055

Find the sample standard deviation s of the data: 53, 48, 54, 55, 27.

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

Exam scores are normally distributed with mean 50 and standard deviation 20. Find the z-score z of a score of 70.

μ = 50σ = 20x = 70
z =
3.ProbabilityMedium
xp(x)00.2510.0520.3530.35μ = 1.8

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

x: 0, 1, 2, 3p(x): 0.25, 0.05, 0.35, 0.35
σ =
4.ProbabilityHard
€p(x)−100.83100.15900.02

A game costs €10 to play. You win €20 with probability 0.15 and €100 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€20: p = 0.15€100: p = 0.02
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9649

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9649. Find z to two decimals.

A = 0.9649
z =
6.Normal distributionHard
μ = 500x = ?5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 5% of them lie (the 5th percentile).

μ = 500σ = 40P = 5%
x = h
7.Confidence intervalsMedium
x̄E₁ = 490%, n = 50x̄E₂ = ?95%, n = 50

A 90% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 4. What is the margin of error E₂ of a 95% interval from the same data?

90%n₁ = 50E₁ = 495%
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.0450L = ?Ux̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 20 plots, mean yield 19 kg, standard deviation 1.5 kg. Plot group 2: 11 plots, mean 23 kg, standard deviation 3 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 20x̄₁ = 19s₁ = 1.5n₂ = 11x̄₂ = 23s₂ = 395%t(29) = 2.045
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 100 parts has mean length 113.5 mm, and the line's known standard deviation is 8 mm. Line 2: a sample of 30 parts has mean 111.9 mm, known standard deviation 10 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 100x̄₁ = 113.5 mmσ₁ = 8 mmn₂ = 30x̄₂ = 111.9 mmσ₂ = 10 mm
z =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 81 measurements has mean 120.4. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 120.4σ = 4n = 8199%
L =
11.Linear regression & correlationMedium
xy136347555769979xy5406080

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy16301840203722512449xy20304050ŷ = 44

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 44?

n = 5ŷ = 44
x =

3Solutions

1.Descriptive statistics

Answer: s = 11.72

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 = 237/5 = 47.4

Squared deviations

Σ(x − x̄)² = 5.6² + 0.6² + 6.6² + 7.6² + (−20.4)² = 549.2

Standard deviation

s = √(549.2/4) = 11.72

Answer

s = 11.72
2.Descriptive statistics

Answer: z = 1

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 = (70 − 50)/20 = 1

Answer

z = 1
3.Probability

Answer: σ = 1.166

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 = 0·0.25 + 1·0.05 + 2·0.35 + 3·0.35 = 1.8

Mean of the squares

E(X²) = 0·0.25 + 1·0.05 + 4·0.35 + 9·0.35 = 4.6

Variance

Var(X) = 4.6 − 1.8² = 1.36
σ = √1.36 = 1.166

Answer

σ = 1.166
4.Probability

Answer: E(G) = −5 €

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), 90 (p = 0.02), −10 (p = 0.83)

Expected value

E(G) = 10·0.15 + 90·0.02 − 10·0.83 = −5 €

Answer

E(G) = −5 €

A negative value means you lose money on average.

5.Normal distribution

Answer: z = −1.81

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 − 0.9649 = 0.0351

Read the z-table backwards

Find 0.0351 inside the table and read off z.

z = −1.81

Answer

z = −1.81
6.Normal distribution

Answer: x = 434.2 h

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) = 0.05 → z = −1.645

Back to x

x = 500 + −1.645·40 = 434.2 h

Answer

x = 434.2 h
7.Confidence intervals

Answer: E₂ = 4.766

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 = (1.96/1.645)·√(50/50) = 1.191

New margin

E₂ = 4·1.191 = 4.766

Answer

E₂ = 4.766
8.Confidence intervals

Answer: lower limit = −5.642 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 = (19·1.5² + 10·3²)/29 = 4.578
sp = 2.14

Margin of error

E = 2.045·2.14·√(1/20 + 1/11) = 1.642

Interval

−4 ± 1.642 → [−5.642, −2.358]

Answer

lower limit = −5.642 kg
9.Hypothesis tests & confidence intervals

Answer: z = 0.8027

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 = √(8²/100 + 10²/30) = √3.973 = 1.993

Test statistic

z = (113.5 − 111.9)/1.993 = 0.8027

Answer

z = 0.8027

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

10.Hypothesis tests & confidence intervals

Answer: L = 119.3

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).

Margin of error

E = 2.576·4/√81 = 1.145

Interval limit

L = 120.4 − 1.145 = 119.3

Answer

L = 119.3
11.Linear regression & correlation

Answer: r² = 0.9945

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 = 286/5 = 57.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 216
Syy = Σ(y − ȳ)² = 1173

Coefficient of determination

r² = 216²/(40·1173) = 0.9945

Answer

r² = 0.9945

About 99.45% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 21.06

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 = 100/5 = 20
ȳ = 207/5 = 41.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 98

Regression line

b = 98/40 = 2.45
a = 41.4 − 2.45·20 = −7.6
ŷ = −7.6 + 2.45·x

Solve for x

x = (44 − (−7.6))/2.45 = 21.06

Answer

x = 21.06

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