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
Introductory Statistics: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
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
- Convert the value to a z-score: z = (x − μ)/σ.
- Look up Φ(z), the area to the left of z.
- For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
- 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
- Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
- Find the critical value for the confidence level.
- Margin of error: E = critical value × standard deviation/√n.
- 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
- State H₀ and H₁ and decide whether the test is one- or two-sided.
- Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
- Compare with the critical value, or find the p-value.
- 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
Find the sample standard deviation s of the data: 59, 13, 41, 58, 37.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 160.4?
μ = 170σ = 12x = 160.4
P =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.05, 0.35, 0.05, 0.2, 0.35
σ =
4.ProbabilityHard
X has the distribution shown, and Y = 3X + 20. Find the standard deviation σY of Y.
x: 10, 20, 30, 40p(x): 0.3, 0.35, 0.1, 0.25Y = 3X + 20
σY =
5.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 5 of them exceeds 398.2 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 5> 398.2 kg
P =
6.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 25% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g25%
x = g
7.Confidence intervalsMedium
A sample of 15 items has mean 19.5 and standard deviation s = 17.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 15x̄ = 19.5s = 17.595%t(14) = 2.145
lower limit =
8.Confidence intervalsHard
The population standard deviation is 35. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1?
σ = 35E = 199%
n =
9.Hypothesis tests & confidence intervalsMedium
16 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 0.91 and the standard deviation of the differences is 1.4. Find the paired test statistic t for H₀: μd = 0.
n = 16d̄ = 0.91sd = 1.4
t =
10.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 27. Find the minimum sample size n.
α = 0.05power = 90%δ = 5σ = 27zα = 1.645zβ = 1.282
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 6
r =
12.Linear regression & correlationHard
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 = 25?
n = 5ŷ = 25
x =
3Solutions
1.Descriptive statistics
Answer: s = 18.78
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 = 208/5 = 41.6
Squared deviations
Σ(x − x̄)² = 17.4² + (−28.6)² + (−0.6)² + 16.4² + (−4.6)² = 1411
Standard deviation
s = √(1411/4) = 18.78
Answer
s = 18.78
2.Descriptive statistics
Answer: P = 0.2119
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 0.2119
Answer
P = 0.2119
3.Probability
Answer: σ = 1.396
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 = 1·0.05 + 2·0.35 + 3·0.05 + 4·0.2 + 5·0.35 = 3.45
Mean of the squares
E(X²) = 1·0.05 + 4·0.35 + 9·0.05 + 16·0.2 + 25·0.35 = 13.85
Variance
Var(X) = 13.85 − 3.45² = 1.947
σ = √1.947 = 1.396
Answer
σ = 1.396
4.Probability
Answer: σY = 34.34
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 = 23, Var(X) = 131, σX = 11.45
Linear change
σY = |3|·11.45 = 34.34
Answer
σY = 34.34
The mean does move: E(Y) = 3·23 + 20 = 89.
5.Normal distribution
Answer: P = 0.8413
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 = 425 kg, σT = 12·√5 = 26.83 kg
Standardise
z = (398.2 − 425)/26.83 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
6.Normal distribution
Answer: x = 246.7 g
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) = 0.25
z from the table
z ≈ −0.67
Back to the original units
x = 250 − 0.67·5 = 246.7 g
Answer
x = 246.7 g
7.Confidence intervals
Answer: lower limit = 9.808
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 = 14, t = 2.145
Margin of error
E = 2.145·17.5/√15 = 9.692
Interval
19.5 ± 9.692 → [9.808, 29.19]
Answer
lower limit = 9.808
8.Confidence intervals
Answer: n = 8129
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 = (2.576·35/1)² = 8129
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 8129
Answer
n = 8129
9.Hypothesis tests & confidence intervals
Answer: t = 2.6
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 of the mean difference
SE = 1.4/√16 = 0.35
Test statistic
t = 0.91/0.35 = 2.6
Answer
t = 2.6
With df = 15, the two-sided 5% critical value is 2.131; |t| is larger, so H₀ is rejected.
10.Hypothesis tests & confidence intervals
Answer: n = 250
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 + 1.282)·27/5)² = 249.8
Round up
n = 250
Answer
n = 250
11.Linear regression & correlation
Answer: r = 0.9954
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 = 185/6 = 30.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 283
Syy = Σ(y − ȳ)² = 1155
Correlation coefficient
r = 283/√(70·1155) = 0.9954
Answer
r = 0.9954
12.Linear regression & correlation
Answer: x = 1.424
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 = 99/5 = 19.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −33
Regression line
b = −33/10 = −3.3
a = 19.8 − (−3.3)·3 = 29.7
ŷ = 29.7 − 3.3·x
Solve for x
x = (25 − 29.7)/(−3.3) = 1.424
Answer
x = 1.424
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