Flashcards · 9709 · Topic 6

Probability & Statistics 2 flashcards

30 cards on Topic 6, subtopics 6.1 to 6.5. Pick a subtopic or study them all.

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Poisson Po(λ): P(X = r), mean and variance: P(X = r) = e^(−λ) λʳ/r!, r = 0, 1, 2, … Mean = variance = λ.
Conditions for a Poisson model: Events occur singly, independently, at random and at a constant mean rate.
Faults occur at a mean rate of 0.8 per metre. What is the distribution for 3 metres?: Scale λ with the interval: λ = 3 × 0.8 = 2.4, so X ~ Po(2.4).
When can B(n, p) be approximated by a Poisson distribution?: When n is large and np is small: the syllabus guide is n > 50 and np < 5. Use Po(np), with no continuity correction.
When can Po(λ) be approximated by a normal distribution?: When λ > 15 (approximately). Use N(λ, λ) with a continuity correction.
X ~ Po(λ₁) and Y ~ Po(λ₂) are independent. What is the distribution of X + Y?: X + Y ~ Po(λ₁ + λ₂). This holds only for sums: 2X is not Poisson.

E(aX + b) and Var(aX + b): E(aX + b) = aE(X) + b Var(aX + b) = a²Var(X) Adding a constant does not change the variance.
E(aX ± bY) and Var(aX ± bY) for independent X and Y: E(aX ± bY) = aE(X) ± bE(Y) Var(aX ± bY) = a²Var(X) + b²Var(Y) Variances always add, even for a difference.
X₁ + X₂ + X₃ (three separate items) vs 3X (one item tripled): Both have mean 3μ. X₁ + X₂ + X₃ has variance 3σ². 3X has variance 9σ².
Var(X) = 3. Find Var(5 − 2X).: (−2)² × 3 = 12. The constant 5 does not change the variance.
What is the distribution of a linear combination of independent normal variables?: It is also normal, with mean and variance found from the rules for linear combinations.

Properties of a probability density function f(x): f(x) ≥ 0 for all x, and the total area under f is 1: ∫ f(x) dx = 1 over the whole range. P(a < X < b) = ∫ₐᵇ f(x) dx.
E(X) and Var(X) for a continuous random variable: E(X) = ∫ x f(x) dx Var(X) = ∫ x² f(x) dx − {E(X)}²
How do you find the median of a continuous random variable?: Solve ∫ f(x) dx = ½, integrating from the lower end of the range up to m. Quartiles use ¼ and ¾.
f(x) = (3/8)x² for 0 ≤ x ≤ 2. Find E(X) and Var(X).: E(X) = (3/8)∫₀² x³ dx = (3/8)(4) = 1.5. E(X²) = (3/8)∫₀² x⁴ dx = (3/8)(32/5) = 2.4. Var(X) = 2.4 − 1.5² = 0.15.
For a continuous random variable, what is P(X = a)?: 0, because a single value has no area. So P(X < a) = P(X ≤ a).

Unbiased estimate of the population variance: s² = (1/(n − 1))(Σx² − (Σx)²/n). Dividing by n would, on average, underestimate the population variance.
Why must a sample be random?: A random sample gives every member of the population an equal chance of selection, so the results are not biased. A method that leaves out part of the population (one time, one place, volunteers only) is unsatisfactory.
Distribution of the sample mean X̄: Normal population: X̄ ~ N(μ, σ²/n) exactly, for any n. Other or unknown population, large n: X̄ ~ N(μ, σ²/n) approximately, by the Central Limit Theorem.
Confidence interval for a population mean: x̄ ± z × σ/√n (or s/√n for a large sample). Interpretation: about 95% of intervals built this way contain the true mean (for a 95% interval).
z-values for 90%, 95%, 98% and 99% confidence intervals: 90%: 1.645 95%: 1.960 98%: 2.326 99%: 2.576
Approximate confidence interval for a population proportion: p̂ ± z √(p̂(1 − p̂)/n), where p̂ is the sample proportion.
σ = 5. What is the smallest n for which a 95% confidence interval for μ has width less than 2?: 2 × 1.96 × 5/√n < 2, so √n > 9.8 and n > 96.04. Smallest n = 97.

Steps of a hypothesis test: 1. State H₀ and H₁ in terms of the parameter (p, λ or μ). 2. State the distribution under H₀. 3. Find the tail probability or test statistic. 4. Compare with the significance level. 5. Conclude in context, without overstating.
Type I and Type II errors: Type I: rejecting H₀ when it is true. Type II: accepting H₀ when it is false.
P(Type I error) for a normal test and for a binomial or Poisson test: Normal test: equals the significance level. Binomial or Poisson test: the actual probability of the rejection region under H₀, usually less than the significance level.
How do you compare in a two-tailed test?: Compare each tail probability with half the significance level (for example 2.5% in each tail at the 5% level).
H₀: p = 0.4, H₁: p > 0.4, n = 15, 5% level. P(X ≥ 9) = 0.0950 and P(X ≥ 10) = 0.0338. Find the rejection region.: The first upper tail below 0.05 starts at 10, so the rejection region is X ≥ 10. P(Type I error) = 0.0338.
How do you find P(Type II error)?: Find the probability of landing in the acceptance region, using the stated alternative value of the parameter (not the H₀ value).
Test statistic for a population mean with σ known (normal population): z = (x̄ − μ)/(σ/√n), using the H₀ value of μ. For a large sample with σ unknown, use s in place of σ.