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: 22, 43, 45, 11, 46, 24, 55.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 50.
μ = 70σ = 10x = 50
z =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.05, 0.3, 0.3, 0.35
σ =
4.ProbabilityHard
A game costs €10 to play. You win €50 with probability 0.15 and €200 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€50: p = 0.15€200: p = 0.02
E(G) = €
5.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 45.5 and 70.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 45.5x2 = 70.5
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 top 80% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g80%
x = g
7.Confidence intervalsMedium
A sample of 8 items has mean 20 and standard deviation s = 2.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 8x̄ = 20s = 2.595%t(7) = 2.365
lower limit =
8.Confidence intervalsHard
The population standard deviation is 19. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1.5?
σ = 19E = 1.595%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 90.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 90.7s = 12n = 16
t =
10.Hypothesis tests & confidence intervalsHard
In a trial, 40 of 100 patients on treatment A recover, and 162 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 40n₁ = 100x₂ = 162n₂ = 300
z =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 110Σx² = 55Σy² = 2508Σxy = 302
r =
12.Linear regression & correlationHard
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 = 41?
n = 5ŷ = 41
x =
3Solutions
1.Descriptive statistics
Answer: s = 16.08
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 = 246/7 = 35.14
Squared deviations
Σ(x − x̄)² = (−13.14)² + 7.857² + 9.857² + (−24.14)² + 10.86² + (−11.14)² + 19.86² = 1551
Standard deviation
s = √(1551/6) = 16.08
Answer
s = 16.08
2.Descriptive statistics
Answer: z = −2
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 = (50 − 70)/10 = −2
Answer
z = −2
3.Probability
Answer: σ = 9.206
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 = 10·0.05 + 20·0.3 + 30·0.3 + 40·0.35 = 29.5
Mean of the squares
E(X²) = 100·0.05 + 400·0.3 + 900·0.3 + 1600·0.35 = 955
Variance
Var(X) = 955 − 29.5² = 84.75
σ = √84.75 = 9.206
Answer
σ = 9.206
4.Probability
Answer: E(G) = 1.5 €
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: 40 (p = 0.15), 190 (p = 0.02), −10 (p = 0.83)
Expected value
E(G) = 40·0.15 + 190·0.02 − 10·0.83 = 1.5 €
Answer
E(G) = 1.5 €
A positive value means you win money on average.
5.Normal distribution
Answer: P = 0.6833
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 = (45.5 − 65)/10 = −1.95
z₂ = (70.5 − 65)/10 = 0.55
Area between
P = Φ(0.55) − Φ(−1.95) = 0.7088 − 0.02559 = 0.6833
Answer
P = 0.6833
6.Normal distribution
Answer: x = 245.8 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) = 0.2
z from the table
z ≈ −0.84
Back to the original units
x = 250 − 0.84·5 = 245.8 g
Answer
x = 245.8 g
7.Confidence intervals
Answer: lower limit = 17.91
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 = 2.365
Margin of error
E = 2.365·2.5/√8 = 2.09
Interval
20 ± 2.09 → [17.91, 22.09]
Answer
lower limit = 17.91
8.Confidence intervals
Answer: n = 617
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 = (1.96·19/1.5)² = 616.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 617
Answer
n = 617
9.Hypothesis tests & confidence intervals
Answer: t = −3.1
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/√16 = 12/4 = 3
Test statistic
t = (90.7 − 100)/3 = −3.1
Answer
t = −3.1
t0.025; df = 15 = 2.131
|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: z = −2.425
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.
Sample proportions
p̂₁ = 40/100 = 0.4
p̂₂ = 162/300 = 0.54
p̂ = 202/400 = 0.505
Standard error under H₀
SE = √(0.505·0.495·(1/100 + 1/300)) = 0.05773
Test statistic
z = (0.4 − 0.54)/0.05773 = −2.425
Answer
z = −2.425
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
11.Linear regression & correlation
Answer: r = −0.9439
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 = 302 − 15·110/5 = −28
Sxx = 55 − 15²/5 = 10
Syy = 2508 − 110²/5 = 88
Correlation coefficient
r = −28/√(10·88) = −0.9439
Answer
r = −0.9439
12.Linear regression & correlation
Answer: x = 1.96
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 = 262/5 = 52.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 150
Regression line
b = 150/40 = 3.75
a = 52.4 − 3.75·5 = 33.65
ŷ = 33.65 + 3.75·x
Solve for x
x = (41 − 33.65)/3.75 = 1.96
Answer
x = 1.96
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