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
2025303540455055

Find the sample standard deviation s of the data: 56, 18, 39, 27, 22.

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

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is above 180?

μ = 170σ = 10x = 180
P =
3.ProbabilityMedium
xp(x)00.3510.320.230.140.05μ = 1.2

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

x: 0, 1, 2, 3, 4p(x): 0.35, 0.3, 0.2, 0.1, 0.05
Var(X) =
4.ProbabilityHard
xp(x)10.1520.0530.2540.3550.2μ

X has the distribution shown, and Y = 5X + 1. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.15, 0.05, 0.25, 0.35, 0.2Y = 5X + 1
σY =
5.Normal distributionMedium
50.585.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 50.5 and 85.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 50.5x2 = 85.5
P =
6.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 8, s = 3L = ?Ux̄ = 87.5E90%

A sample of 8 items has mean 87.5 and standard deviation s = 3. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 8x̄ = 87.5s = 390%t(7) = 1.895
lower limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(24) = 1.7110L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 19 plots, mean yield 71.5 kg, standard deviation 2 kg. Plot group 2: 7 plots, mean 62 kg, standard deviation 2 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 19x̄₁ = 71.5s₁ = 2n₂ = 7x̄₂ = 62s₂ = 290%t(24) = 1.711
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 59.67 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 59.67s = 20n = 36
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 16 measurements has mean 497.1. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 497.1σ = 4n = 1699%
L =
11.Linear regression & correlationMedium
xy129334549752967xy54060

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
xy336447549652756xy464050ŷ = 46

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 46?

n = 5ŷ = 46
x =

3Solutions

1.Descriptive statistics

Answer: s = 15.37

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 = 162/5 = 32.4

Squared deviations

Σ(x − x̄)² = 23.6² + (−14.4)² + 6.6² + (−5.4)² + (−10.4)² = 945.2

Standard deviation

s = √(945.2/4) = 15.37

Answer

s = 15.37
2.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
3.Probability

Answer: Var(X) = 1.36

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 + 1·0.3 + 2·0.2 + 3·0.1 + 4·0.05 = 1.2

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.2 + 9·0.1 + 16·0.05 = 2.8

Variance

Var(X) = 2.8 − 1.2² = 1.36

Answer

Var(X) = 1.36
4.Probability

Answer: σY = 6.403

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.

Spread of X

μ = 3.4, Var(X) = 1.64, σX = 1.281

Linear change

σY = |5|·1.281 = 6.403

Answer

σY = 6.403

The mean does move: E(Y) = 5·3.4 + 1 = 18.

5.Normal distribution

Answer: P = 0.9063

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 = (50.5 − 65)/10 = −1.45
z₂ = (85.5 − 65)/10 = 2.05

Area between

P = Φ(2.05) − Φ(−1.45) = 0.9798 − 0.07353 = 0.9063

Answer

P = 0.9063
6.Normal distribution

Answer: P = 0.9207

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 difference

μ = 25.02 − 25 = 0.02 mm, σ = 0.01414 mm

Standardise

z = (0 − 0.02)/0.01414 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207
7.Confidence intervals

Answer: lower limit = 85.49

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 = 7, t = 1.895

Margin of error

E = 1.895·3/√8 = 2.01

Interval

87.5 ± 2.01 → [85.49, 89.51]

Answer

lower limit = 85.49
8.Confidence intervals

Answer: lower limit = 7.987 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 = (18·2² + 6·2²)/24 = 4
sp = 2

Margin of error

E = 1.711·2·√(1/19 + 1/7) = 1.513

Interval

9.5 ± 1.513 → [7.987, 11.01]

Answer

lower limit = 7.987 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.901

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 = 20/√36 = 20/6 = 3.333

Test statistic

t = (59.67 − 50)/3.333 = 2.901

Answer

t = 2.901
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 494.5

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/√16 = 2.576

Interval limit

L = 497.1 − 2.576 = 494.5

Answer

L = 494.5
11.Linear regression & correlation

Answer: r² = 0.9617

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 = 231/5 = 46.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 188
Syy = Σ(y − ȳ)² = 918.8

Coefficient of determination

r² = 188²/(40·918.8) = 0.9617

Answer

r² = 0.9617

About 96.17% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 4.556

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 = 240/5 = 48

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 45

Regression line

b = 45/10 = 4.5
a = 48 − 4.5·5 = 25.5
ŷ = 25.5 + 4.5·x

Solve for x

x = (46 − 25.5)/4.5 = 4.556

Answer

x = 4.556

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