Flashcards · 9660 · Topic 6

Unit S2: Statistics (International A-level) flashcards

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S2.1 Poisson distribution

Poisson distribution Po(λ): probability, mean and variance: P(X = r) = e^(−λ) λʳ/r!, r = 0, 1, 2, … Mean λ, variance λ, standard deviation √λ.
Conditions for a Poisson model: Events occur singly, at random, independently of each other, and at a constant average rate. Link each condition to the context.
Faults occur at 12 per hour. What distribution models faults in 30 minutes and in 5 minutes?: New λ = rate × length: 30 minutes gives Po(6); 5 minutes gives Po(1).
Poisson approximation to the binomial: B(n, p) ≈ Po(np) when n is large and p is small. Example: B(80, 0.025) ≈ Po(2), so P(X = 3) ≈ e⁻² × 8/6 = 0.1804 (exact 0.1827).
Sum of independent Poisson variables: If X ~ Po(a) and Y ~ Po(b) are independent, X + Y ~ Po(a + b). Add the rates, not the standard deviations. A difference of Poisson variables is not Poisson.
How can data suggest that a Poisson model is a poor fit?: For a Poisson model the mean and variance should be close. A variance much larger than the mean suggests clustering or a changing rate. Comment in context.

S2.2 Continuous random variables

Conditions for a valid pdf, and the probability that X equals a single value: f(x) ≥ 0 everywhere and the total area is 1. f(x) is a height, not a probability, and can exceed 1. P(X = a) = 0, so P(X < a) = P(X ≤ a).
Link between the pdf f and the cdf F: F(x) = P(X ≤ x) = ∫ from −∞ to x of f(t) dt; f(x) = dF/dx. P(a < X < b) = F(b) − F(a); P(X > a) = 1 − F(a).
Median and quartiles of a continuous random variable: Median m: F(m) = 1/2. Lower quartile: F = 1/4. Upper quartile: F = 3/4. For a piecewise pdf, compare the target with F at each break point to choose the piece.
E(X), E(g(X)) and Var(X) for a continuous random variable: E(X) = ∫ x f(x) dx. E(g(X)) = ∫ g(x) f(x) dx. Var(X) = E(X²) − μ². Integrate over the range where f is non-zero.
f(x) = 3(1 − x)² for 0 ≤ x ≤ 1. Find E(X) and Var(X).: E(X) = 3 ∫ (x − 2x² + x³) dx = 3(1/2 − 2/3 + 1/4) = 1/4. E(X²) = 3(1/3 − 1/2 + 1/5) = 1/10. Var(X) = 1/10 − 1/16 = 3/80.
f(x) = 2(1 − x) for 0 ≤ x ≤ 1. Find the median.: F(x) = 2x − x². Solve 2m − m² = 1/2: m² − 2m + 1/2 = 0, m = 1 ± 1/√2. Only m = 1 − 1/√2 ≈ 0.293 lies in [0, 1].

S2.3 Exponential distribution

When does an exponential distribution arise?: When events occur singly, independently, at random and at a constant average rate λ (a Poisson process), the length of the interval between events has an exponential distribution Exp(λ).
Exponential Exp(λ): pdf, cdf and upper tail (x ≥ 0): f(x) = λe^(−λx) F(x) = 1 − e^(−λx) P(X > x) = e^(−λx) P(a < X < b) = e^(−λa) − e^(−λb)
Mean, variance and median of Exp(λ): Mean 1/λ. Variance 1/λ² (so the standard deviation is also 1/λ). Median ln 2/λ. If the mean m is given, λ = 1/m.
X ~ Exp(0.25). Find P(X > 3), the mean and E(X²).: P(X > 3) = e^(−0.75) = 0.472. Mean 4 (and standard deviation 4). E(X²) = Var(X) + (E(X))² = 16 + 16 = 32.
Link between Exp(λ) and the Poisson distribution: The number of events in time t is Po(λt), so P(no events in time t) = e^(−λt) = P(X > t). Both routes give the same value.

S2.4 Normal distribution

Normal distribution: notation, standardising and the proportions within 1, 2 and 3 standard deviations: X ~ N(μ, σ²); the second parameter is the variance. Z = (X − μ)/σ ~ N(0, 1). About 0.6827 within μ ± σ, 0.9545 within μ ± 2σ, 0.9973 within μ ± 3σ.
Distribution of X + Y, X − Y and aX + bY for independent normal X and Y: X + Y ~ N(μ₁ + μ₂, σ₁² + σ₂²) X − Y ~ N(μ₁ − μ₂, σ₁² + σ₂²) aX + bY ~ N(aμ₁ + bμ₂, a²σ₁² + b²σ₂²) Variances always add.
X ~ N(μ, σ²), P(X < 10) = 0.05 and P(X > 22) = 0.20. Find μ and σ (z-values 1.6449 and 0.8416).: μ − 1.6449σ = 10 and μ + 0.8416σ = 22. Subtract: 2.4865σ = 12, σ = 4.83. μ = 10 + 1.6449 × 4.826 = 17.9 (3 s.f.).
X ~ N(5, 1²) and Y ~ N(8, 2²) are independent. Find P(2X + Y > 21), given Φ(1.06) = 0.8554.: 2X + Y ~ N(18, 4 × 1 + 4) = N(18, 8). z = 3/√8 = 1.06, so P = 1 − 0.8554 = 0.1446.
Total of three independent copies of X compared with 3X: X₁ + X₂ + X₃: mean 3μ, variance 3σ². 3X: mean 3μ, variance 9σ².
How do you find the probability that one normal quantity exceeds another, such as a book fitting a gap?: Use the difference: it fits when G − B > 0. G − B is normal with mean μ_G − μ_B and variance σ_G² + σ_B² (independent). Standardise 0 and find the probability.

S2.5 Estimation

Parameter compared with statistic: Parameter: a fixed (usually unknown) number describing the population, such as μ or σ². Statistic: a number calculated from a sample, such as x̄ or s²; it varies from sample to sample.
Unbiased estimator, and the unbiased estimate of σ²: T is an unbiased estimator of θ if E(T) = θ. s² = Σ(x − x̄)²/(n − 1) = (Σx² − n x̄²)/(n − 1). Dividing by n − 1, not n, makes it unbiased.
Distribution of the sample mean from a normal population, and the standard error: X̄ ~ N(μ, σ²/n), exactly, for any n. Standard error = σ/√n, estimated by s/√n.
The Central Limit Theorem: For a large random sample from any population with mean μ and variance σ², X̄ is approximately N(μ, σ²/n). The total is approximately N(nμ, nσ²). It does not make single values normal.
n = 20, Σx = 164, Σx² = 1420. Find x̄ and s².: x̄ = 164/20 = 8.2. s² = (1420 − 20 × 8.2²)/19 = 75.2/19 = 3.958 (3 d.p.).
σ = 10. Find the smallest sample size for a standard error below 1.5.: 10/√n < 1.5, so √n > 6.667 and n > 44.4. Round up: n = 45.

S2.6 Hypothesis testing

H₀, H₁, and one-tailed compared with two-tailed tests: H₀ gives the parameter one value (p = p₀, λ = λ₀ or μ = μ₀). H₁ says it is greater, less or different. > or <: one-tailed. ≠: two-tailed, with the level shared between the tails.
Critical region and actual significance level: Critical region: values of the test statistic that lead to rejecting H₀. Take the biggest region that does not exceed the stated level (half of it per tail if two-tailed). Actual significance level: the true probability of the critical region under H₀.
Type I and Type II errors: Type I: rejecting H₀ when H₀ is true. P(Type I error) = the actual significance level. Type II: not rejecting H₀ when H₀ is false. Describe each in the context of the question.
X ~ B(15, 0.6) under H₀, H₁: p < 0.6, observed 5, 5% level. P(X ≤ 5) = 0.0338. Conclusion?: 0.0338 ≤ 0.05, so reject H₀. There is sufficient evidence at the 5% level that the proportion is less than 0.6.
Test statistics for a population mean: z or t?: Normal population, σ known (or large n): z = (x̄ − μ₀)/(σ/√n), compared with N(0, 1); for large n with σ unknown, use s. Normal population, σ unknown, small n: t = (x̄ − μ₀)/(s/√n) with n − 1 degrees of freedom.
z critical values for one-tailed 5%, one-tailed 1% and two-tailed 5% tests: One-tailed 5%: 1.645. One-tailed 1%: 2.326. Two-tailed 5%: ±1.960.