📚 Year 10 CAIE Statistics: Cross-curricular Integrated Problem Drills | Year 10 CAIE 统计：跨学科综合题型训练
In the CAIE Year 10 Statistics course, you are expected not only to master isolated statistical techniques but also to apply them in realistic, multi-subject contexts. This article presents a series of cross-curricular integrated problems that blend statistics with physics, biology, economics, geography, sports science, psychology, business, and environmental studies. Each scenario is designed to strengthen your ability to choose appropriate statistical tools, interpret data meaningfully, and communicate findings clearly – exactly what the exam demands.
在 CAIE Year 10 统计课程中，你不仅要掌握单一的统计方法，还要能够在真实的多学科情境中加以应用。本文提供一系列跨学科综合题型，将统计与物理、生物、经济、地理、体育科学、心理学、商业和环境研究结合起来。每个情境都旨在提升你选择合适的统计工具、有意义地解读数据以及清晰地表达结论的能力，这正是考试所要求的。
1. Physics & Statistics: Measurement Uncertainty and Repeated Trials | 物理与统计：测量误差与重复试验
A student measures the period of a pendulum 10 times using a stopwatch. The results (in seconds) are: 1.81, 1.78, 1.83, 1.79, 1.82, 1.80, 1.78, 1.81, 1.79, 1.82. The physics teacher asks for the best estimate of the period and an assessment of measurement precision. From a statistical perspective, this is a classic dataset for calculating the mean and standard deviation. The mean period is easily found as the sum divided by 10. The sample standard deviation quantifies the spread, and the standard error of the mean gives the precision of the estimate. However, the student must also consider systematic error (e.g., reaction time) which statistics alone cannot detect. In the exam, you might be asked to calculate the 95% confidence interval for the true period, assuming the measurements are normally distributed.
一名学生用秒表测量单摆周期 10 次，结果（秒）为：1.81, 1.78, 1.83, 1.79, 1.82, 1.80, 1.78, 1.81, 1.79, 1.82。物理老师要求给出周期的最佳估计值和测量精度的评估。从统计角度看，这是一个计算均值和标准差的经典数据集。平均周期很容易求得，总和除以 10。样本标准差表示数据的离散程度，均值的标准误反映估计的精确度。然而，学生还需考虑系统误差（例如反应时间），这单靠统计是检测不出来的。在考试中，你可能会被要求计算真实周期的 95% 置信区间，前提是测量值服从正态分布。
For this data set: Σx = 18.03, so x̄ = 1.803 s. ∑(x – x̄)² = 0.00281, sample standard deviation s = √[0.00281/(10−1)] ≈ 0.0177 s. Standard error = s/√10 ≈ 0.0056 s. A 95% confidence interval using t₉,₀.₀₂₅ ≈ 2.262 gives 1.803 ± 2.262×0.0056 = (1.790, 1.816) seconds. This interval suggests the true period is likely between 1.79 s and 1.82 s. Always link such calculations back to the physics context: the small spread indicates good precision, but the student should also compare with the known value for that pendulum length to check accuracy.
对这个数据集：Σx = 18.03，故 x̄ = 1.803 秒。∑(x – x̄)² = 0.00281，样本标准差 s = √[0.00281/9] ≈ 0.0177 秒。标准误 = s/√10 ≈ 0.0056 秒。使用 t₉,₀.₀₂₅ ≈ 2.262 计算的 95% 置信区间为 1.803 ± 2.262×0.0056 = (1.790, 1.816) 秒。这个区间表明真实周期很可能介于 1.79 秒与 1.82 秒之间。始终要将这些计算与物理背景联系起来：较小的离散程度说明精度良好，但该生还应与已知的该摆长下的周期值对比，以检验准确度。
2. Biology & Statistics: Modelling Bacterial Growth | 生物与统计：细菌生长建模
A biologist records the number of bacterial colonies in a petri dish every hour for 8 hours. The counts are: 2, 5, 11, 23, 48, 95, 190, 385. This suggests exponential growth. In statistics, we often linearise such data by taking logarithms. The transformed data (ln of count) should follow a linear trend. You can then use least-squares linear regression to find the growth rate. The slope of the regression line represents the hourly growth rate constant. The intercept gives the logarithm of the initial colony count. This interdisciplinary problem requires you to justify why logarithmic transformation is appropriate, perform the regression, and interpret the parameters biologically.
一位生物学家每小时记录培养皿中细菌菌落数量，共 8 小时。计数为：2, 5, 11, 23, 48, 95, 190, 385。这提示呈指数增长。在统计中，我们常通过对数变换将这类数据线性化。变换后的数据（计数的自然对数）应呈线性趋势。然后你可以使用最小二乘线性回归求出生长速率。回归线的斜率代表每小时生长速率常数，截距给出初始菌落数的对数值。这道跨学科题目要求你论证为何对数变换是合适的，进行回归分析，并从生物学角度解释参数。
Let y = ln(count). Calculating: ln(2)=0.693, ln(5)=1.609, ln(11)=2.398, ln(23)=3.135, ln(48)=3.871, ln(95)=4.554, ln(190)=5.247, ln(385)=5.953. Time t = 0,1,2,…,7. Using a calculator or formulae, Σt=28, Σy=27.46, Σt²=140, Σty=126.33. The slope b = (nΣty − ΣtΣy) / (nΣt² − (Σt)²) = (8×126.33 − 28×27.46) / (8×140 − 28²) = (1010.64 − 768.88) / (1120 − 784) = 241.76 / 336 ≈ 0.720. Hence the hourly growth rate constant is 0.720, meaning the population multiplies by e⁰·⁷² ≈ 2.05 each hour. The initial colony count estimate = e^(intercept). Intercept a = (Σy − bΣt)/n = (27.46 − 0.720×28)/8 = (27.46 − 20.16)/8 = 7.3/8 = 0.9125, so estimated initial count = e⁰·⁹¹²⁵ ≈ 2.49, close to 2. The model fits well; you could assess goodness-of-fit using R².
设 y = ln(计数)。计算结果：ln(2)=0.693, ln(5)=1.609, ln(11)=2.398, ln(23)=3.135, ln(48)=3.871, ln(95)=4.554, ln(190)=5.247, ln(385)=5.953。时间 t = 0,1,2,…,7。利用计算器或公式，Σt=28, Σy=27.46, Σt²=140, Σty=126.33。斜率 b = (8×126.33 − 28×27.46) / (8×140 − 28²) = (1010.64 − 768.88) / (1120 − 784) = 241.76 / 336 ≈ 0.720。因此每小时生长速率常数为 0.720，意味着菌落数每小时乘以 e⁰·⁷² ≈ 2.05 倍。初始菌落数估计值 = e^(截距)。截距 a = (27.46 − 0.720×28)/8 = 0.9125，故估计初始数 = e⁰·⁹¹²⁵ ≈ 2.49，与实际 2 接近。模型拟合良好；你可以用 R² 评估拟合优度。
3. Economics & Statistics: Constructing a Weighted Price Index | 经济与统计：构建加权价格指数
An economics student collects price and quantity data for three goods in a small town for 2020 (base year) and 2023. Good A: price $2 (2020), $2.50 (2023), quantity 100 units. Good B: price $5, $6, quantity 50. Good C: price $10, $12, quantity 20. The task is to compute a Laspeyres price index and a Paasche price index, then explain why they differ. This combines statistics with economic theory. The Laspeyres index uses base-year quantities as weights, reflecting the cost of the base-period basket at current prices. The Paasche index uses current quantities, showing the cost of the current basket. The difference illustrates substitution bias. In Year 10, you need to know how to compute weighted means and understand the concept of weighting.
一名经济学学生收集了某小镇三种商品在 2020 年（基年）和 2023 年的价格和数量数据。商品 A：价格 $2 (2020)，$2.50 (2023)，数量 100 单位。商品 B：价格 $5，$6，数量 50。商品 C：价格 $10，$12，数量 20。任务是计算拉氏价格指数和帕氏价格指数，然后解释两者差异。这道题将统计与经济理论结合。拉氏指数以基年数量为权重，反映基期篮子按现价计算的成本。帕氏指数使用现期数量，反映现期篮子的成本。两者差异说明了替代偏差。在 Year 10，你需要知道如何计算加权平均数并理解加权的概念。
Laspeyres Index = (Σpₙq₀ / Σp₀q₀) × 100. p₀q₀: A 2×100=200, B 5×50=250, C 10×20=200; Σ = 650. pₙq₀: A 2.5×100=250, B 6×50=300, C 12×20=240; Σ = 790. Index = (790/650)×100 ≈ 121.5. Paasche Index = (Σpₙqₙ / Σp₀qₙ) × 100. Assuming quantities remained the same (for simplicity, often they change; but let’s assume no quantity change so index would be the same; to illustrate difference, usually quantities change). If instead in 2023 quantity of A rose to 120, B dropped to 40, C unchanged: p₀qₙ: A 2×120=240, B 5×40=200, C 10×20=200; Σ=640. pₙqₙ: A 2.5×120=300, B 6×40=240, C 12×20=240; Σ=780. Paasche index = (780/640)×100 = 121.9. Slightly lower because consumers substituted away from B. This demonstrates the weighting effect.
拉氏指数 = (Σpₙq₀ / Σp₀q₀) × 100。p₀q₀：A 200，B 250，C 200；总和 650。pₙq₀：A 250，B 300，C 240；总和 790。指数 = (790/650)×100 ≈ 121.5。帕氏指数 = (Σpₙqₙ / Σp₀qₙ) × 100。假设 2023 年 A 数量增至 120，B 降至 40，C 不变：p₀qₙ：A 240，B 200，C 200；总和 640。pₙqₙ：A 300，B 240，C 240；总和 780。帕氏指数 = (780/640)×100 = 121.9。略低是因为消费者减少了对 B 的购买，从而体现了替代效应。这展示了权重的作用。
4. Geography & Statistics: Analysing Population Pyramids and Dependency Ratios | 地理与统计：人口金字塔与抚养比分析
A geography class examines the age distribution of Country X. The data: 0–14 years: 23% of population, 15–64: 65%, 65+: 12%. The total population is 8.5 million. Students must calculate the youth dependency ratio, old-age dependency ratio, and total dependency ratio. This is a straightforward application of ratios and percentages. The youth dependency ratio = (population aged 0–14) / (population aged 15–64) × 100. The old-age dependency ratio = (65+) / (15–64) × 100. These statistics help compare the economic burden on the working-age population across different countries. A histogram or population pyramid can also be constructed from such data, and measures of central tendency like median age can be estimated from a cumulative frequency graph.
地理课分析 X 国的人口年龄结构。数据：0–14 岁占 23%，15–64 岁占 65%，65 岁以上占 12%。总人口 850 万。学生须计算少儿抚养比、老年抚养比和总抚养比。这是比率和百分比的直接应用。少儿抚养比 = (0–14 岁人口) / (15–64 岁人口) × 100。老年抚养比 = (65 岁以上) / (15–64 岁) × 100。这些统计指标有助于比较不同国家劳动年龄人口所承受的经济负担。根据这类数据还可以绘制直方图或人口金字塔，并且中位年龄等集中趋势指标可通过累积频率图估算。
Calculations: working-age population = 0.65 × 8.5m = 5.525m; youth = 0.23 × 8.5m = 1.955m; elderly = 0.12 × 8.5m = 1.02m. Youth dependency ratio = (1.955/5.525)×100 ≈ 35.4%. Old-age dependency ratio = (1.02/5.525)×100 ≈ 18.5%. Total dependency ratio = 35.4 + 18.5 = 53.9%. This means for every 100 working-age people, there are about 54 dependants. If the total population grows to 10 million with the same percentages, the ratios remain unchanged because they are proportions. In a scatter plot, you might compare these ratios against GDP per capita for several countries to explore correlation.
计算：劳动年龄人口 = 0.65 × 850 万 = 552.5 万；少儿人口 = 0.23 × 850 万 = 195.5 万；老年人口 = 0.12 × 850 万 = 102 万。少儿抚养比 = (195.5/552.5)×100 ≈ 35.4%。老年抚养比 = (102/552.5)×100 ≈ 18.5%。总抚养比 = 35.4 + 18.5 = 53.9%。这意味着每 100 名劳动年龄人口要抚养约 54 人。如果总人口增长至 1000 万且比例不变，由于是比例数据，这些比值保持不变。在散点图中，你可以比较多个国家的抚养比与人均 GDP，探索其相关关系。
5. Sports Science & Statistics: Comparing Athlete Performance | 体育科学与统计：运动员成绩比较
A coach records the 100 m sprint times (in seconds) for two athletes over six races. Athlete P: 11.2, 11.0, 11.3, 10.9, 11.1, 11.2. Athlete Q: 11.0, 11.5, 10.8, 11.7, 10.9, 11.3. Which athlete is more consistent? The question requires comparing measures of dispersion, such as the range, interquartile range, and standard deviation. While the means may be similar, consistency is judged by variability. A box-and-whisker plot would visually compare medians and spreads. The coach might also want to test whether the difference in means is statistically significant, introducing the t-test. For Year 10, emphasis is on calculating and interpreting standard deviation.
一位教练记录了六场比赛中两名运动员的 100 米短跑时间（秒）。运动员 P：11.2, 11.0, 11.3, 10.9, 11.1, 11.2。运动员 Q：11.0, 11.5, 10.8, 11.7, 10.9, 11.3。哪位运动员表现更稳定？该问题需要比较离散程度指标，例如全距、四分位距和标准差。虽然均值可能相近，但稳定性通过变异程度来判断。箱线图可以直观地比较中位数和分散程度。教练可能还想检验均值差异是否显著，这便引入了 t 检验。Year 10 的重点是计算和解释标准差。
For Athlete P: sorted times 10.9, 11.0, 11.1, 11.2, 11.2, 11.3. Mean = (11.2+11.0+11.3+10.9+11.1+11.2)/6 = 66.7/6 = 11.117 s. Median = (11.1+11.2)/2 = 11.15 s. Sum of squared deviations = (11.2−11.117)² + (11.0−11.117)² + … = 0.00694 + 0.0136 + 0.0336 + 0.0469 + 0.00028 + 0.00694 = 0.1083. Sample s = √(0.1083/5) ≈ √0.02166 = 0.147 s. For Q: sorted 10.8, 10.9, 11.0, 11.3, 11.5, 11.7. Mean = (11.0+11.5+10.8+11.7+10.9+11.3)/6 = 67.2/6 = 11.2 s. Median = (11.0+11.3)/2 = 11.15 s. Sum of squared deviations = 0.04 + 0.09 + 0.16 + 0.25 + 0.09 + 0.01 = 0.64. s = √(0.64/5) = √0.128 = 0.358 s. Athlete P has a much smaller standard deviation (0.147 vs 0.358), indicating greater consistency. The ranges confirm this: P’s range 0.4 s, Q’s range 0.9 s.
运动员 P：排序 10.9, 11.0, 11.1, 11.2, 11.2, 11.3。均值 = 66.7/6 = 11.117 秒。中位数 = 11.15 秒。离差平方和 = 0.1083。样本标准差 s = √(0.1083/5) ≈ 0.147 秒。运动员 Q：排序 10.8, 10.9, 11.0, 11.3, 11.5, 11.7。均值 = 67.2/6 = 11.2 秒。中位数 = 11.15 秒。离差平方和 = 0.64，s = √(0.64/5) = 0.358 秒。运动员 P 的标准差远小于 Q（0.147 对 0.358），表明更稳定。全距也证实了这一点：P 全距 0.4 秒，Q 全距 0.9 秒。
6. Psychology & Statistics: Experimental Design and Control Groups | 心理学与统计：实验设计与对照组
A psychology student investigates whether background music affects concentration. She gives a memory test to 20 participants: 10 with music (experimental group) and 10 without (control group). The scores (out of 30) are: Music group: 14, 18, 12, 20, 15, 17, 13, 19, 16, 14. No-music group: 22, 25, 20, 27, 24, 23, 26, 21, 28, 22. The task: calculate summary statistics for each group, produce side-by-side box plots, and discuss whether the difference could be due to chance. This requires calculating quartiles and constructing box plots. Later, a hypothesis test (Mann–Whitney U or t-test) could be used, but at Year 10 level, descriptive comparison and recognising sampling variability are key.
一位心理学学生探究背景音乐是否影响注意力。她让 20 名参与者完成记忆测试：10 人听音乐（实验组），10 人不听（对照组）。得分（满分 30）：音乐组：14, 18, 12, 20, 15, 17, 13, 19, 16, 14。无音乐组：22, 25, 20, 27, 24, 23, 26, 21, 28, 22。任务：计算每组的汇总统计量，绘制并排箱线图，讨论差异是否可能由偶然因素造成。这需要计算四分位数并绘制箱线图。后续可用假设检验（曼–惠特尼 U 或 t 检验），但 Year 10 的重点是描述性比较和认识抽样变异性。
Music group sorted: 12,13,14,14,15,16,17,18,19,20. n=10. Q1 position = (10+1)/4 = 2.75 → Q1 = 13.25 (interpolation: 13 + 0.75×(14−13) = 13.25). Q2 median = (15+16)/2 = 15.5. Q3 position = 3(11)/4 = 8.25 → Q3 = 18 + 0.25×(19−18) = 18.25. IQR = 5.0. No-music sorted: 20,21,22,22,23,24,25,26,27,28. Q1 pos 2.75 → 21 + 0.75×(22−21) = 21.75. Q2 = (23+24)/2 = 23.5. Q3 pos 8.25 → 26 + 0.25×(27−26) = 26.25. IQR = 4.5. The box plots would show non-overlapping notches if drawn, indicating a likely significant difference. The student should note that with small samples, the difference might arise from sampling variability, but the consistency of the separation suggests a real effect. Confounding variables (e.g., fatigue, musical taste) should also be discussed.
音乐组排序：12,13,14,14,15,16,17,18,19,20。n=10。Q1 位置 = 2.75 → Q1 = 13.25。Q2 中位数 = 15.5。Q3 位置 = 8.25 → Q3 = 18.25。IQR = 5.0。无音乐组排序：20,21,22,22,23,24,25,26,27,28。Q1 = 21.75，Q2 = 23.5，Q3 = 26.25，IQR = 4.5。若绘制带有凹槽的箱线图，两组凹槽不重叠，表明差异很可能显著。学生应指出，样本量较小时，该差异可能只是抽样波动，但数据分离的一致性暗示存在真实效应。还应讨论混杂变量（如疲劳、音乐品味）。
7. Business & Statistics: Sampling Methods for Market Research | 商业与统计：市场调研的抽样方法
A new café chain wants to estimate the average amount customers are willing to spend per visit. The target population is all adults in a city of 500,000. The marketing team proposes an online survey distributed via social media, but a statistician recommends a stratified random sample based on age groups and income levels. Explain why the online survey may lead to biased results, and how stratified sampling can improve representativeness. This problem evaluates understanding of sampling bias, sampling frames, and the purpose of stratification. You might be asked to calculate the required sample size per stratum if the overall budget allows surveying 600 people, given the population proportions.
一家新连锁咖啡馆想要估计顾客每次愿意消费的平均金额。目标总体是某城市 50 万成年人。营销团队提议通过社交媒体发布在线调查，但统计学家建议按年龄组和收入水平进行分层随机抽样。解释为什么在线调查可能导致偏差结果，以及分层抽样如何提高代表性。这道题考查对抽样偏差、抽样框以及分层目的的理解。如果总体预算允许调查 600 人，你可能会被要求根据各层的人口比例计算每层所需的样本量。
Possible population strata: 18–30 years (30%), 31–50 (40%), 51+ (30%). With a budget of 600 surveys, a proportional stratified sample would allocate: 18–30: 0.30×600 = 180; 31–50: 0.40×600 = 240; 51+: 0.30×600 = 180. Within each stratum, individuals should be randomly selected (e.g., from electoral rolls or a customer database). The online survey suffers from voluntary response bias: only those who see the post and feel strongly might respond, and it over-represents younger, tech-savvy people. Statistics offers tools to assess bias by comparing sample demographics to known population demographics. Post-stratification weighting could be applied, but it’s better to design the sample properly from the start.
可能的人口分层：18–30 岁（30%），31–50 岁（40%），51 岁以上（30%）。在 600 份调查预算下，比例分层抽样分配为：18–30 岁 180 人，31–50 岁 240 人，51 岁以上 180 人。各层内应采用随机选择（如从选民名册或客户数据库中抽取）。在线调查存在自愿回答偏差：只有看到帖子且感受强烈的人才会回应，且过度代表年轻、精通技术的人群。统计学提供工具，通过比较样本人口结构与已知总体人口结构来评估偏差。事后可进行分层加权调整，但更好的做法是从一开始就妥善设计样本。
8. Environmental Science & Statistics: Monitoring Air Quality | 环境科学与统计：监测空气质量
The daily PM₂.₅ concentration (µg/m³) is recorded in an urban area for 30 consecutive days. The environmental scientist suspects that values are higher on weekdays due to traffic. The data is summarised by day type. Weekdays (n=22): mean = 35.2, s = 8.1. Weekends (n=8): mean = 28.6, s = 6.5. Is the difference statistically meaningful? Although Year 10 doesn’t require formal hypothesis testing, you can explore this by calculating a 95% confidence interval for the difference of two means, or simply comparing box plots. This integrates environmental awareness with statistical inference. Also, you can discuss the assumptions required (approximate normality, independence).
在一个城区连续 30 天记录每日 PM₂.₅ 浓度（µg/m³）。环境科学家怀疑周中数值因交通而偏高。数据按日期类型汇总。工作日（n=22）：均值 = 35.2，s = 8.1。周末（n=8）：均值 = 28.6，s = 6.5。差异有统计意义吗？虽然 Year 10 不要求正式假设检验，但你可以通过计算两均值之差的 95% 置信区间或简单比较箱线图来探究。这道题将环保意识与统计推断结合起来。此外，还可以讨论所需的假设（近似正态、独立性）。
The estimated difference in means is 35.2 − 28.6 = 6.6 µg/m³. The standard error of the difference (assuming unequal variances) is √(s₁²/n₁ + s₂²/n₂) = √(8.1²/22 + 6.5²/8) = √(65.61/22 + 42.25/8) = √(2.982 + 5.281) = √8.263 = 2.875. Using t-distribution with conservative df = min(21,7) = 7, t₇,₀.₀₂₅ = 2.365 (approx). 95% CI: 6.6 ± 2.365×2.875 = 6.6 ± 6.80 = (−0.2, 13.4). The interval includes zero, so at the 5% significance level, we cannot rule out that the true difference could be zero. However, the interval is wide due to small weekend sample. This exercise highlights the importance of sample size and the role of confidence intervals in decision-making.
均值估计差值为 35.2 − 28.6 = 6.6 µg/m³。差值标准误（假设方差不相等）为 √(8.1²/22 + 6.5²/8) = √(2.982 + 5.281) = 2.875。采用保守自由度 7 的 t 分布，t₇,₀.₀₂₅ ≈ 2.365。95% 置信区间：6.6 ± 2.365×2.875 = 6.6 ± 6.80 = (−0.2, 13.4)。区间包含零，因此在 5% 显著性水平下，不能排除真实差值为零的可能。然而，由于周末样本量较小，区间较宽。这道练习突显了样本量的重要性以及置信区间在决策中的作用。
9. General Problem-Solving Strategy: Interpreting Scenarios and Choosing Tools | 通用解题策略：解读情境与工具选择
In a typical cross-curricular exam question, the scenario is given, data provided, and then a series of sub-questions requiring calculations and interpretations. Read the scenario carefully, identify the variables and their types (categorical, discrete, continuous). Determine the appropriate summary statistics: for symmetric data, mean and standard deviation; for skewed data, median and IQR. Know when to use a bar chart vs a histogram vs a scatter graph. In correlation questions, distinguish between dependent and independent variables. Always relate your statistical findings back to the context – a calculation is incomplete without a contextual sentence. Practice extracting assumptions and evaluating limitations, such as sample size or potential biases.
在典型的跨学科考题中，会先给出情境和数据，然后有一系列子问题要求计算和解读。仔细阅读情境，识别变量及其类型（类别、离散、连续）。确定合适的汇总统计量：对称数据用均值和标准差，偏态数据用中位数和 IQR。知道何时用条形图、直方图或散点图。在相关性问题中，区分因变量和自变量。始终将统计发现联系回情境——没有结合上下文句子的计算是不完整的。练习提炼假设和评估局限性，如样本量或潜在偏差。
For instance, if a question combines geography and statistics by asking you to compare earthquake magnitudes across two regions, you should first note that magnitude scales are continuous and often right-skewed. Thus, medians are better for central tendency, and box plots for comparison. If given latitude and magnitude, a scatter plot might reveal a relationship. The key is flexible thinking. The CAIE exam rewards students who can justify their choice of statistical technique, not just perform it. Build a mental checklist: type of data, objective (compare, predict, describe), distribution shape, and assumptions.
例如，如果一道题目结合地理与统计，要求比较两个地区的地震震级，你应首先注意到震级是连续数据且常右偏。因此，中位数更适合衡量集中趋势，箱线图适合进行比较。如果还给出纬度数据，散点图可能揭示关系。关键在于灵活思维。CAIE 考试奖励那些能够证明所选统计方法合理性的学生，而不仅仅是会计算。建立一个心理检查清单：数据类型、目标（比较、预测、描述）、分布形状和假设。
10. Practice Integrated Task: A Cross-curricular Data Analysis | 综合练习任务：跨学科数据分析
Here is a combined exercise to test your skills. A health researcher records the following for 12 participants: daily screen time (hours) and quality of sleep rating (1-10, higher is better). Data: (3,9), (5,7), (2,10), (7,5), (4,8), (6,6), (8,4), (1,10), (5,6), (3,8), (6,5), (4,7). Construct a scatter plot, calculate the product-moment correlation coefficient, find the equation of the regression line of sleep rating on screen time, and then interpret the slope. Discuss whether reducing screen time could improve sleep.
Published by TutorHao | Year 10 统计 Revision Series | aleveler.com
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