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
15202530354045505560

Find the sample standard deviation s of the data: 59, 24, 60, 18, 11, 14, 43.

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

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is above 125?

μ = 100σ = 20x = 125
P =
3.ProbabilityMedium
xp(x)00.3510.1520.330.2μ = 1.35

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

x: 0, 1, 2, 3p(x): 0.35, 0.15, 0.3, 0.2
σ =
4.ProbabilityHard
€p(x)−30.73120.25570.02

A game costs €3 to play. You win €15 with probability 0.25 and €60 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€15: p = 0.25€60: p = 0.02
E(G) = €
5.Normal distributionMedium
400 kg403 kgT = X₁ + … + X₁₆?

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 16 of them is below 403 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 16< 403 kg
P =
6.Normal distributionHard
μ = 500x = ?75%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 75% of them lie (the 75th percentile).

μ = 500σ = 40P = 75%
x = h
7.Confidence intervalsMedium
n = 16, s = 2LUx̄ = 100.5E = ?95%

A sample of 16 items has mean 100.5 and standard deviation s = 2. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 16x̄ = 100.5s = 295%t(15) = 2.131
E =
8.Confidence intervalsHard
σ = 32, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 32. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 0.5?

σ = 32E = 0.590%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.521.52p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 4. A random sample of 64 observations has a mean of 249.24. Test H₀: μ = 250 against H₁: μ ≠ 250. Find the p-value p.

μ₀ = 250σ = 4n = 64x̄ = 249.24H₁: μ ≠ μ₀
p =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 29, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 29. Find the minimum sample size n.

α = 0.05power = 95%δ = 4σ = 29zα = 1.645zβ = 1.645
n =
11.Linear regression & correlationMedium
xy3365407469471153xy5104050

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 35Σy = 222Σx² = 285Σy² = 10030Σxy = 1636
r =
12.Linear regression & correlationHard
xy12221831541659xy24101520ŷ = 12

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 12?

n = 5ŷ = 12
x =

3Solutions

1.Descriptive statistics

Answer: s = 21.04

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 = 229/7 = 32.71

Squared deviations

Σ(x − x̄)² = 26.29² + (−8.714)² + 27.29² + (−14.71)² + (−21.71)² + (−18.71)² + 10.29² = 2655

Standard deviation

s = √(2655/6) = 21.04

Answer

s = 21.04
2.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
3.Probability

Answer: σ = 1.152

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 = 0·0.35 + 1·0.15 + 2·0.3 + 3·0.2 = 1.35

Mean of the squares

E(X²) = 0·0.35 + 1·0.15 + 4·0.3 + 9·0.2 = 3.15

Variance

Var(X) = 3.15 − 1.35² = 1.327
σ = √1.327 = 1.152

Answer

σ = 1.152
4.Probability

Answer: E(G) = 1.95 €

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: 12 (p = 0.25), 57 (p = 0.02), −3 (p = 0.73)

Expected value

E(G) = 12·0.25 + 57·0.02 − 3·0.73 = 1.95 €

Answer

E(G) = 1.95 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.9332

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 = 400 kg, σT = 0.5·√16 = 2 kg

Standardise

z = (403 − 400)/2 = 1.5

Probability

P = Φ(1.5) = 0.9332

Answer

P = 0.9332
6.Normal distribution

Answer: x = 527 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) = 0.75 → z = 0.6745

Back to x

x = 500 + 0.6745·40 = 526.98 h

Answer

x = 526.98 h
7.Confidence intervals

Answer: E = 1.065

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 = 15, t = 2.131

Margin of error

E = 2.131·2/√16 = 1.065

Answer

E = 1.065
8.Confidence intervals

Answer: n = 11084

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.

Solve for n

n = (1.645·32/0.5)² = 11084

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 11084

Answer

n = 11084
9.Hypothesis tests & confidence intervals

Answer: p = 0.1285

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.

Test statistic

SE = σ/√n = 4/√64 = 0.5
z = (249.24 − 250)/0.5 = −1.52

Tail area

Φ(1.52) = 0.9357
1 − Φ(1.52) = 0.0643

p-value

p = 2·0.0643 = 0.1285

Answer

p = 0.1285

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 569

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 + 1.645)·29/4)² = 568.9

Round up

n = 569

Answer

n = 569
11.Linear regression & correlation

Answer: r = 0.9852

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 = 1636 − 35·222/5 = 82
Sxx = 285 − 35²/5 = 40
Syy = 10030 − 222²/5 = 173.2

Correlation coefficient

r = 82/√(40·173.2) = 0.9852

Answer

r = 0.9852
12.Linear regression & correlation

Answer: x = 4.429

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 = 80/5 = 16

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −28

Regression line

b = −28/10 = −2.8
a = 16 − (−2.8)·3 = 24.4
ŷ = 24.4 − 2.8·x

Solve for x

x = (12 − 24.4)/(−2.8) = 4.429

Answer

x = 4.429

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