📚 Year 13 CAIE Statistics: Case Study Practice | Year 13 CAIE 统计：案例分析实战演练
Welcome to this intensive case study revision session tailored for Year 13 CAIE Statistics. Through a series of real-world style problems, we will apply key techniques from Probability & Statistics 2, including confidence intervals, hypothesis tests for means and proportions, the Poisson distribution, and the use of normal and t-distributions. Each case breaks down the thought process, calculations, and interpretation, helping you build confidence for the exam.
欢迎来到为 Year 13 CAIE 统计量身定制的案例实战复习课。通过一系列贴近真实情境的问题，我们将运用概率与统计2中的关键技巧，包括置信区间、均值和比例的假设检验、泊松分布，以及正态分布与 t 分布的应用。每个案例都拆解了思考过程、计算和结果解读，帮你建立应考信心。
1. Confidence Interval for Mean Lifetime | 平均寿命的置信区间
A manufacturer claims that its LED bulbs have a mean lifetime of 15000 hours. A consumer group tests 50 bulbs and finds a sample mean of 14800 hours. Assume the population standard deviation is known to be 800 hours. Construct a 95% confidence interval for the true mean lifetime and comment on the claim.
某制造商声称其 LED 灯泡的平均寿命为 15000 小时。一家消费者组织测试了 50 个灯泡，得到样本均值为 14800 小时。假设已知总体标准差为 800 小时。构建总体平均寿命的 95% 置信区间，并对该声称加以评论。
Since the population standard deviation σ is known and the sample size n = 50 is large, we use a z-interval. The standard error is SE = σ/√n = 800/√50 ≈ 113.14.
由于总体标准差 σ 已知且样本量 n = 50 足够大，我们采用 z 区间。标准误为 SE = σ/√n = 800/√50 ≈ 113.14。
x̄ ± z × SE → 14800 ± 1.96 × 113.14 = (14578.2, 15021.8)
The 95% confidence interval is (14578, 15022) hours. Since the claimed value of 15000 lies inside this interval, there is insufficient evidence to reject the manufacturer’s claim at the 5% significance level.
95% 置信区间为 (14578, 15022) 小时。由于声称值 15000 位于该区间内，没有足够证据在 5% 显著性水平下拒绝制造商的说法。
2. t-Test for Small Sample Mean | 小样本均值的 t 检验
A new energy bar is supposed to contain 20 g of protein. A random sample of 10 bars has a mean protein content of 18.9 g with a sample standard deviation of 1.5 g. Test at the 5% significance level whether the mean protein content is less than 20 g.
一款新的能量棒标称含有 20 g 蛋白质。随机抽取 10 根，测得蛋白质均值为 18.9 g，样本标准差为 1.5 g。在 5% 显著性水平下检验平均蛋白质含量是否低于 20 g。
Since the population variance is unknown and the sample size is small, we use a one-sample t‑test. Hypotheses: H₀: μ = 20, H₁: μ < 20. The test statistic is t = (x̄ − μ) / (s/√n) = (18.9 − 20) / (1.5/√10) = −2.32.
因为总体方差未知且样本量小，我们使用单样本 t 检验。假设：H₀: μ = 20，H₁: μ < 20。检验统计量为 t = (x̄ − μ) / (s/√n) = (18.9 − 20) / (1.5/√10) = −2.32。
Degrees of freedom df = 9. The critical value for a one‑tailed test at α = 0.05 is t₀.₀₅,₉ = 1.833. Since −2.32 < −1.833, we reject H₀. There is sufficient evidence that the mean protein content is less than 20 g.
自由度 df = 9。α = 0.05 的单尾临界值为 t₀.₀₅,₉ = 1.833。由于 −2.32 < −1.833，我们拒绝 H₀。有充分证据表明平均蛋白质含量低于 20 g。
3. Proportion Test Using Normal Approximation | 比例检验（正态近似）
A politician claims that 60% of voters support her. In a poll of 400 voters, 228 express support. Test at the 1% level whether the true support rate differs from 60%.
一位政客声称有 60% 的选民支持她。在一项 400 名选民的民意调查中，有 228 人表示支持。在 1% 显著性水平下检验真实支持率是否不同于 60%。
The number of supporters follows a binomial distribution. Here np = 240 and nq = 160, both greater than 5, so a normal approximation is appropriate. Sample proportion p̂ = 228/400 = 0.57.
支持人数服从二项分布。此处 np = 240，nq = 160，均大于 5，适合用正态近似。样本比例 p̂ = 228/400 = 0.57。
H₀: p = 0.6, H₁: p ≠ 0.6. Test statistic: z = (p̂ − p) / √[p(1−p)/n] = (0.57 − 0.6) / √(0.6×0.4/400) ≈ −1.225. The two‑tailed critical value at 1% is ±2.576. Since |z| < 2.576, we do not reject H₀. The data do not provide enough evidence of a change in support.
H₀: p = 0.6, H₁: p ≠ 0.6。检验统计量：z = (p̂ − p) / √[p(1−p)/n] = (0.57 − 0.6) / √(0.6×0.4/400) ≈ −1.225。1% 双尾临界值为 ±2.576。因为 |z| < 2.576，我们不拒绝 H₀。数据未能提供支持率变化的充分证据。
4. Poisson Mean Test | 泊松均值的检验
A call centre historically receives 4.2 calls per hour. After introducing a new automated system, a random 20‑hour period yields 65 calls. Test at the 5% significance level whether the call rate has decreased.
某呼叫中心历史上每小时接到 4.2 通电话。引进新的自动化系统后，随机抽取 20 个小时时段共接到 65 通电话。在 5% 显著性水平下检验通话率是否下降。
The number of calls in 20 hours under the null hypothesis follows a Poisson distribution with mean λ = 4.2 × 20 = 84. Since λ is large, we use a normal approximation. Test statistic: z = (x − λ) / √λ = (65 − 84) / √84 ≈ −2.07.
在原假设下，20 小时的通话次数服从均值 λ = 4.2 × 20 = 84 的泊松分布。由于 λ 较大，我们采用正态近似。检验统计量：z = (x − λ) / √λ = (65 − 84) / √84 ≈ −2.07。
This is a one‑tailed test: H₀: λ = 84, H₁: λ < 84. The critical z‑value at α = 0.05 is −1.645. Because −2.07 < −1.645, we reject H₀. There is evidence that the call rate has significantly decreased after the new system.
这是一个单尾检验：H₀: λ = 84，H₁: λ < 84。α = 0.05 的临界 z 值为 −1.645。由于 −2.07 < −1.645，我们拒绝 H₀。有证据表明引进新系统后通话率显著下降。
5. Two‑Sample t‑Test (Independent Samples) | 独立双样本 t 检验
Two groups of students follow different revision programmes. Group A (nA = 25): mean = 72, SD = 8. Group B (nB = 30): mean = 68, SD = 9. Assuming equal population variances, test at α = 0.05 whether programme A yields a higher average score.
两组学生采用不同的复习方案。A 组（nA = 25）：均值 = 72，标准差 = 8。B 组（nB = 30）：均值 = 68，标准差 = 9。假设总体方差相等，在 α = 0.05 下检验 A 方案的平均成绩是否更高。
H₀: μA = μB, H₁: μA > μB. We first compute the pooled variance: Sp² = [(24)(64) + (29)(81)] / (25+30−2) = (1536 + 2349)/53 ≈ 73.30. Pooled SD Sp ≈ 8.56.
H₀: μA = μB, H₁: μA > μB。先计算合并方差：Sp² = [(24)(64) + (29)(81)] / (25+30−2) = (1536 + 2349)/53 ≈ 73.30。合并标准差 Sp ≈ 8.56。
The standard error of the difference is SE = Sp√(1/25 + 1/30) ≈ 8.56 × 0.270 ≈ 2.31. The t‑statistic is t = (72 − 68) / 2.31 ≈ 1.73. Degrees of freedom = 53, one‑tailed critical value t₀.₀₅,₅₃ ≈ 1.674. Since 1.73 > 1.674, we reject H₀. Programme A appears to produce a significantly higher mean score.
差异的标准误为 SE = Sp√(1/25 + 1/30) ≈ 8.56 × 0.270 ≈ 2.31。t 统计量 t = (72 − 68) / 2.31 ≈ 1.73。自由度 = 53，单尾临界值 t₀.₀₅,₅₃ ≈ 1.674。因为 1.73 > 1.674，拒绝 H₀。A 方案似乎带来显著更高的平均分。
6. Paired t‑Test | 配对 t 检验
Ten students take a test before and after a training course. The differences (after − before) are: 3, 5, −1, 4, 6, 2, 3, 4, 0, 5. Test at 5% significance whether the course has improved performance.
十名学生参加培训课程前后各进行一次测试。差值（后 − 前）为：3, 5, −1, 4, 6, 2, 3, 4, 0, 5。在 5% 显著性下检验培训是否提高了成绩。
The sample mean difference d̄ = 3.1 and the sample standard deviation of differences s_d ≈ 2.23. H₀: μ_d = 0, H₁: μ_d > 0. The test statistic is t = d̄ / (s_d/√n) = 3.1 / (2.23/√10) ≈ 4.40.
样本平均差值 d̄ = 3.1，差值的样本标准差 s_d ≈ 2.23。H₀: μ_d = 0, H₁: μ_d > 0。检验统计量 t = d̄ / (s_d/√n) = 3.1 / (2.23/√10) ≈ 4.40。
With df = 9, the one‑tailed critical value is t₀.₀₅,₉ = 1.833. Since 4.40 > 1.833, we reject H₀. There is strong evidence that the training course leads to improved scores.
自由度为 9，单尾临界值 t₀.₀₅,₉ = 1.833。由于 4.40 > 1.833，拒绝 H₀。有强证据表明培训课程提高了成绩。
7. Normal Distribution: Probabilities & Inverse Calculations | 正态分布：概率与逆运算
The weight of sugar bags from a filling machine is normally distributed with mean 500 g and standard deviation 10 g. (a) What proportion of bags weigh less than 490 g? (b) Find the weight that is exceeded by 95% of bags.
一台灌装机所装糖袋的重量服从均值为 500 g、标准差为 10 g 的正态分布。(a) 重量低于 490 g 的袋子占多大比例？(b) 求被 95% 的袋子超过的重量。
(a) z = (490 − 500) / 10 = −1. P(Z < −1) = 0.1587. So about 15.9% of bags weigh less than 490 g.
(a) z = (490 − 500) / 10 = −1。P(Z < −1) = 0.1587。因此约 15.9% 的袋子重量不足 490 g。
(b) “Exceeded by 95%” means the weight is the 5th percentile. The z‑value for a left‑tail probability of 0.05 is −1.645. Thus x = μ +
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