S3 · Statistics 3

Statistics 3 (S3, paper WST03/01, 1 hour 30 minutes, 75 marks, first assessment June 2020) is the final statistics unit: S1–S2 are assumed. The booklet carries the Normal and chi-squared tables plus the Spearman critical values. This reference covers combinations of Normal variables, sampling methods, estimators and the sample mean's distribution, confidence intervals and hypothesis tests 假设检验 for means (one and two samples), goodness of fit and contingency tables, and rank correlation with its test.

Must-know for S3 (not in the booklet):

Independent Normal variables combine to a Normal: add means, add squared coefficients times variances. For $3S$ where $S \sim N(18, 0.03^2)$: $3S \sim N(54, 9 \times 0.0009)$ — the mean triples but the variance multiplies by nine. Differences work the same way: $C_1 - C_2 \sim N(0, 2\sigma^2)$, and $P(|C_1 - C_2| > c)$ is twice the one-tail probability by symmetry.

Worked example (WST03 June 2025 Q8). $S \sim N(18, 0.03^2)$, $C \sim N(25, 0.03^2)$, $P \sim N(15, 0.2^2)$ independently (kg). (a) $S_1+S_2+S_3 \sim N(54, 0.0027)$ (sd 0.05196): $P(>54.1) = 1 - \Phi(1.924) = 0.0274$. (b) $C_1 - C_2 \sim N(0, 0.0018)$: $P(|C_1-C_2| > 0.02) = 2(1 - \Phi(0.4714)) = 2 \times 0.3192 = 0.638$. (c) With $T = \sum_1^{28} S_i + \sum_1^5 C_i + P_1$, the event $T < 30P_1 + 190$ rearranges to $\sum S_i + \sum C_i - 29P_1 < 190$ — a Normal with mean $504 + 125 - 435 = 194$ and variance $28(0.0009) + 5(0.0009) + 29^2(0.04) = 33.6697$, so $P = \Phi\left(\tfrac{190-194}{5.803}\right) = 1 - 0.7549 = 0.245$.

The simple random sample 简单随机抽样 gives every member (and every sample of size $n$) an equal chance; use random numbers on a sampling frame. Systematic sampling 系统抽样 takes every $x$th member from a random start — for 20 from 360, $x = 18$; it avoids choosing individuals by personal preference, but periodic patterns in the list can distort the sample, but members in the same stretch can never appear together (spacings are locked). Stratified sampling 分层抽样 splits by strata and samples each proportionally — often useful when groups differ from each other and values within each group are relatively similar. Quota sampling 配额抽样 fills interviewer-set quotas without a frame — cheap but vulnerable to interviewer bias. Know one advantage and one disadvantage of each.

A statistic 统计量 contains no unknown population parameters. An estimator is unbiased 无偏 if its expectation equals the parameter; its bias 偏差 is the gap: $\bar X$ for $\mu$, $s^2 = \dfrac{\sum x_i^2 - n\bar x^2}{n-1}$ for $\sigma^2$ (the $n-1$ is the correction). Bias is $\mathrm{E}(\hat\theta) - \theta$ in sign and size. For independent observations from a Normal population, the sample mean satisfies $\bar X \sim N\left(\mu, \tfrac{\sigma^2}n\right)$: bigger samples shrink the standard error 标准误 $\sigma/\sqrt n$.

Worked example (WST03 June 2025 Q6). $R = \tfrac12X_1 + \tfrac34X_{20}$: $\mathrm{E}(R) = (\tfrac12 + \tfrac34)\mu = \tfrac54\mu \ne \mu$, so $R$ is biased with bias $\tfrac\mu4$. Given $S = \tfrac{aX_1 - bX_2}2$ and $T = \tfrac{a\sum_1^{14}X_i + b\sum_{15}^{20}X_i}{20}$ both unbiased: the expectations force $a - b = 2$ and $14a + 6b = 20$, so $a = 1.6$, $b = -0.4$.

Worked example — recovering parameters (WST03 June 2025 Q7). Given $P(\bar X_n < 45.2) = 0.3446$ and $P(\bar X_n > 75.2) = 0.0179$: the z-values are $-0.4$ and $2.1$, so $\mu - 0.4\tfrac\sigma{\sqrt n} = 45.2$ and $\mu + 2.1\tfrac\sigma{\sqrt n} = 75.2$. Subtracting: $2.5\tfrac\sigma{\sqrt n} = 30$, giving $\sigma = 12\sqrt n$ (SE 12) and $\mu = 45.2 + 4.8 = 50$; then $P(\bar X_n > 59) = 1 - \Phi(0.75) = 0.2266$.

A confidence interval 置信区间 for $\mu$ (variance known) is $\bar x \pm z\tfrac{\sigma}{\sqrt n}$ with $z = 1.96$ (95%), $1.6449$ (one-tailed 95% critical), $2.5758$ (99%). Interpretation: the procedure captures $\mu$ in 95% of samples — not "the probability that μ lies inside is 0.95". The test statistic $z = \dfrac{\bar x - \mu}{\sigma/\sqrt n} \sim N(0, 1)$; when the variance is unknown and $n$ is large, substitute $s$ (the Central Limit Theorem 中心极限定理 also licenses non-Normal populations under independent sampling from the same population with finite, nonzero variance when the sample is sufficiently large). Conclusions state the evidence in context.

Worked example — two samples with a gap (WST03 June 2025 Q4). Grey squirrels ($n = 80$, $\bar x = 500$, $s_x^2 = 1156$) and red ($n = 80$; $\sum y = 23200$, $\sum y^2 = 6741351$, giving $\bar y = 290$ and $s_y^2 = \tfrac{6741351 - 80(290)^2}{79} = 169$). Test $\text{H}_0\colon \mu_x - \mu_y = 200$ against $> 200$ at 5%:

so reject H₀ — the grey mean is more than 200 g above the red. The independent samples are large, so sample variances estimate the unknown population variances in this approximate test; this does not assert exact equality $\sigma^2=s^2$.

Goodness of fit 拟合优度检验 compares observed counts with a model's expected ones: $X^2 = \sum \frac{(O-E)^2}{E}$. Degrees of freedom = (cells − 1) − (parameters estimated from the data): a binomial with p̂ estimated loses one more. Combine tail cells whose expected frequency 期望频数 is below 5 before anything else. Contingency tables 列联表 test association between two classifications with $E = \frac{\text{row} \times \text{column}}{\text{total}}$ and $\mathrm{df} = (r-1)(c-1)$.

Worked example — binomial fit (WST03 June 2025 Q5). Males in 250 litters of 5 ($O = 2, 40, 90, 85, 30, 3$): $\hat p = \tfrac{610}{1250} = 0.488$, giving expected frequencies $8.80, 41.92, 79.91, 76.16, 36.30, 6.92$. The $r$-class contributes $\tfrac{(2-8.796)^2}{8.796} = 5.25$, so with the given $\sum = 5.70$ for the rest, $X^2 = 10.95$ on degrees of freedom 自由度 $6 - 1 - 1 = 4$; the 5% critical value 9.488 is exceeded — reject: the binomial is not a suitable model.

Worked example — association (WST03 June 2025 Q3). Engine type × first-problem timing (250 cars, 3×3): the diesel/month 1–6 cell has $E = \tfrac{35\times36}{250} = 5.04$ and contributes $\tfrac{(10-5.04)^2}{5.04} = 4.88$; with 7.444 for the other eight cells, $X^2 = 12.325 > 9.488$ on 4 df — evidence of an association between engine type and when problems start.

Spearman's rank correlation coefficient measures monotonic agreement between two rankings; test $\text{H}_0\colon\rho = 0$ against the table and conclude in words about the rankings.

Spearman's rank correlation coefficient 斯皮尔曼等级相关系数 $r_s = 1 - \tfrac{6\sum d^2}{n(n^2-1)}$ measures monotonic agreement between two rankings (no ties examined). Test $\text{H}_0\colon \rho = 0$ against one- or two-tailed alternatives using the table of critical values; conclude in context ("insufficient evidence of positive agreement between the judges").

Worked example (WST03 June 2025 Q2). Two judges rank 10 performances: the rank differences give $\sum d^2 = 66$, so $r_s = 1 - \tfrac{6(66)}{990} = 0.6$. At the 1% one-tailed level the critical value 临界值 is 0.7455: $0.6 < 0.7455$, so do not reject H₀ — insufficient evidence of positive correlation between the judges' rankings.

• Combine independent Normals (squared coefficients on variances); handle differences and |differences| by symmetry.
• Name and justify the four sampling methods; compute systematic spacing and its impossibility properties.
• Check statistics for parameters, compute expectations for bias, solve paired unbiased-estimator equations.
• Use SE σ/√n; recover (μ, σ) from two sample-mean probabilities; compute probabilities for X̄.
• Build and interpret z-based confidence intervals; run one- and two-sample tests with variance substitution and CLT stated.
• Compute expected frequencies (model or contingency), combine low cells, subtract estimated parameters from the df, compare with the χ² table.
• Compute Spearman's rₛ from ranks and test against the table with hypotheses stated in symbols and the conclusion in words.