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: 45, 42, 60, 45, 41.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is below 85?
μ = 100σ = 15x = 85
P =
3.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
4.ProbabilityHard
A game costs €10 to play. You win €20 with probability 0.15 and €500 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€500: p = 0.02
E(G) = €
5.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 454 and 510 h? Give P as a decimal.
μ = 500σ = 40x1 = 454x2 = 510
P =
6.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 80% of them lie (the 80th percentile).
μ = 32σ = 6P = 80%
x = min
7.Confidence intervalsMedium
A sample of 8 items has mean 80.5 and standard deviation s = 7. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 8x̄ = 80.5s = 790%t(7) = 1.895
E =
8.Confidence intervalsHard
The population standard deviation is 36. 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?
σ = 36E = 195%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 500.9 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 500.9s = 5n = 81
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 4. Find the smallest sample size n.
σ = 4E = 195%
n =
11.Linear regression & correlationMedium
The table gives 6 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 = 6
r² =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (6.5, 70) is made. Find its residual e = y − ŷ.
n = 6(6.5, 70)
e =
3Solutions
1.Descriptive statistics
Answer: s = 7.701
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 = 233/5 = 46.6
Squared deviations
Σ(x − x̄)² = (−1.6)² + (−4.6)² + 13.4² + (−1.6)² + (−5.6)² = 237.2
Standard deviation
s = √(237.2/4) = 7.701
Answer
s = 7.701
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) = 0.1587
P = 0.1587
Answer
P = 0.1587
3.Probability
Answer: P = 0.3056
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Complement
P(no six) = (5/6)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
4.Probability
Answer: E(G) = 3 €
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), 490 (p = 0.02), −10 (p = 0.83)
Expected value
E(G) = 10·0.15 + 490·0.02 − 10·0.83 = 3 €
Answer
E(G) = 3 €
A positive value means you win money on average.
5.Normal distribution
Answer: P = 0.4736
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 = (454 − 500)/40 = −1.15
z₂ = (510 − 500)/40 = 0.25
Area between
P = Φ(0.25) − Φ(−1.15) = 0.5987 − 0.1251 = 0.4736
Answer
P = 0.4736
6.Normal distribution
Answer: x = 37.05 min
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) = 0.8 → z = 0.8416
Back to x
x = 32 + 0.8416·6 = 37.05 min
Answer
x = 37.05 min
7.Confidence intervals
Answer: E = 4.69
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 = 1.895·7/√8 = 4.69
Answer
E = 4.69
8.Confidence intervals
Answer: n = 4979
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 = (1.96·36/1)² = 4979
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 4979
Answer
n = 4979
9.Hypothesis tests & confidence intervals
Answer: t = 1.62
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 = 5/√81 = 5/9 = 0.5556
Test statistic
t = (500.9 − 500)/0.5556 = 1.62
Answer
t = 1.62
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 62
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).
Solve for n
n = (1.96·4/1)² = 61.47
Round up
n = 62
Answer
n = 62
11.Linear regression & correlation
Answer: r² = 0.9893
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 = 48/6 = 8
ȳ = 465/6 = 77.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 437
Syy = Σ(y − ȳ)² = 2758
Coefficient of determination
r² = 437²/(70·2758) = 0.9893
Answer
r² = 0.9893
About 98.93% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = −2.124
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 = 27/6 = 4.5
ȳ = 382/6 = 63.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 74
Regression line
b = 74/17.5 = 4.229
a = 63.67 − 4.229·4.5 = 44.64
ŷ = 44.64 + 4.229·x
Predicted value
ŷ = 44.64 + 4.229·6.5 = 72.12
Residual
e = 70 − 72.12 = −2.124
Answer
e = −2.124
The point lies below the line.
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