Cross-curricular Integrated Question Training for Year 10 Eduqas Statistics | 跨学科综合题型训练：Year 10 Eduqas 统计

📚 Cross-curricular Integrated Question Training for Year 10 Eduqas Statistics | 跨学科综合题型训练：Year 10 Eduqas 统计

Year 10 Eduqas Statistics encourages you to apply statistical thinking far beyond the maths classroom. Real problems in biology, geography, business and physics all demand the ability to collect, represent and interpret data. This article walks you through a series of integrated exercises drawn from different subjects, showing exactly how the statistical tools you are learning connect to the wider curriculum. Work through each section and you will build confidence in tackling the multi‑step, context‑rich questions that appear on your examination papers.

Year 10 Eduqas 统计课程鼓励你将统计思维延伸到数学教室之外。生物学、地理、商科和物理中的真实问题都需要数据收集、表示与解释的能力。本文带你遍历一系列来自不同学科的综合练习，向你展示正在学习的统计工具如何与更广泛的课程紧密相连。逐节完成这些内容，你将能够自信地应对考试中出现的多步骤、情境丰富的问题。

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1. Biology: Plant Growth Experiment | 生物学：植物生长实验

In a biology practical, two groups of 15 bean seedlings were grown under different light conditions: full daylight and partial shade. After three weeks, the height of each seedling was measured to the nearest millimetre. The raw data are shown in the table below. A biologist wants to compare the central tendency and spread of the two groups and decide whether light level affects growth. First, calculate the mean, median and range for each group. Which measure of average is more appropriate when there might be outliers, and why?

在一次生物实验课上，两组各15株豆苗分别在充分日照和半阴条件下生长。三周后，测量每株幼苗的高度，精确到毫米。原始数据如下表所示。生物学家希望比较两组的集中趋势和离散程度，并判断光照水平是否影响生长。首先，计算每组的平均数、中位数和极差。当可能存在异常值时，哪一种平均量数更为合适，为什么？

Full daylight (mm)	Partial shade (mm)
142, 156, 148, 163, 151, 149, 172, 147, 158, 155, 139, 161, 153, 145, 166	127, 131, 118, 155, 128, 129, 124, 133, 119, 126, 122, 130, 121, 200, 125

Solution approach: For full daylight, order the data: 139, 142, 145, 147, 148, 149, 151, 153, 155, 156, 158, 161, 163, 166, 172. The mean is (sum ÷ 15) = 2295 ÷ 15 = 153.0 mm. The median is the 8th value, 153 mm. The range is 172 − 139 = 33 mm. For partial shade, order: 118, 119, 121, 122, 124, 125, 126, 127, 128, 129, 130, 131, 133, 155, 200. The mean = 1998 ÷ 15 = 133.2 mm. The median is the 8th value, 127 mm. The range is 200 − 118 = 82 mm. The unusually tall seedling (200 mm) inflates the mean and range, so the median is a better summary of the typical growth in partial shade because it is resistant to the outlier.

解题思路：充分日照组排序：139, 142, 145, 147, 148, 149, 151, 153, 155, 156, 158, 161, 163, 166, 172。平均数 = 总和 ÷ 15 = 2295 ÷ 15 = 153.0 mm。中位数为第8个值，153 mm。极差 = 172 − 139 = 33 mm。半阴组排序：118, 119, 121, 122, 124, 125, 126, 127, 128, 129, 130, 131, 133, 155, 200。平均数 = 1998 ÷ 15 = 133.2 mm。中位数为第8个值，127 mm。极差 = 200 − 118 = 82 mm。那株异常高的幼苗（200 mm）夸大了平均数和极差，因此中位数能更好地概括半阴条件下的典型生长，因为它对异常值具有抵抗力。

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2. Geography: River Discharge Data | 地理：河流流量数据

A geography student recorded the discharge of a river (in cubic metres per second, m³/s) at the same gauging station on the first day of each month for one year. The values were: Jan 2.1, Feb 2.5, Mar 5.8, Apr 8.3, May 7.1, Jun 4.2, Jul 2.8, Aug 2.2, Sep 2.9, Oct 4.5, Nov 6.9, Dec 3.6. The student wants to display the seasonal pattern and identify the month with the peak flow. Suggest a suitable graph and describe how to construct it. Then calculate the three‑point moving averages to smooth out short‑term fluctuations and comment on the trend.

一位地理专业的学生在某测站记录了一条河流在一年中每月第一天的流量（单位：立方米每秒，m³/s）。数据为：1月 2.1, 2月 2.5, 3月 5.8, 4月 8.3, 5月 7.1, 6月 4.2, 7月 2.8, 8月 2.2, 9月 2.9, 10月 4.5, 11月 6.9, 12月 3.6。该学生希望展示季节性规律并找出峰值流量所在的月份。请建议一种合适的图表并说明其绘制方法。随后计算三点移动平均数以平滑短期波动，并评论其变化趋势。

A time‑series graph (line graph) with months on the horizontal axis and discharge on the vertical axis is ideal. Plot each point and join them with straight lines. The peak raw value is in April (8.3 m³/s). To smooth the series, compute moving averages: first average = (2.1 + 2.5 + 5.8) ÷ 3 = 3.47; second = (2.5 + 5.8 + 8.3) ÷ 3 = 5.53; third = (5.8 + 8.3 + 7.1) ÷ 3 = 7.07; fourth = (8.3 + 7.1 + 4.2) ÷ 3 = 6.53; fifth = (7.1 + 4.2 + 2.8) ÷ 3 = 4.70; sixth = (4.2 + 2.8 + 2.2) ÷ 3 = 3.07; seventh = (2.8 + 2.2 + 2.9) ÷ 3 = 2.63; eighth = (2.2 + 2.9 + 4.5) ÷ 3 = 3.20; ninth = (2.9 + 4.5 + 6.9) ÷ 3 = 4.77; tenth = (4.5 + 6.9 + 3.6) ÷ 3 = 5.00. The smoothed data reveal a gradual rise to April, a decline to August, and a secondary peak in November, clearly showing the seasonal rhythm of the river.

时间序列图（折线图）非常合适，横轴为月份，纵轴为流量。标出各数据点并用直线连接。原始数据的峰值出现在4月（8.3 m³/s）。为平滑序列，计算移动平均数：第一个平均值 = (2.1 + 2.5 + 5.8) ÷ 3 = 3.47；第二个 = (2.5 + 5.8 + 8.3) ÷ 3 = 5.53；第三个 = (5.8 + 8.3 + 7.1) ÷ 3 = 7.07；第四个 = (8.3 + 7.1 + 4.2) ÷ 3 = 6.53；第五个 = (7.1 + 4.2 + 2.8) ÷ 3 = 4.70；第六个 = (4.2 + 2.8 + 2.2) ÷ 3 = 3.07；第七个 = (2.8 + 2.2 + 2.9) ÷ 3 = 2.63；第八个 = (2.2 + 2.9 + 4.5) ÷ 3 = 3.20；第九个 = (2.9 + 4.5 + 6.9) ÷ 3 = 4.77；第十个 = (4.5 + 6.9 + 3.6) ÷ 3 = 5.00。平滑后的数据揭示了流量逐渐上升至4月，然后下降到8月，并在11月出现次级峰值，清晰地展现了河流的季节性节律。

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3. Business: Market Research Survey | 商业：市场调查

A small café surveyed 80 customers about their preferred hot drink. The responses were: coffee 34, tea 26, hot chocolate 14, other 6. The owner wants to present the findings to potential investors. Calculate the proportion of customers who prefer coffee and tea combined. Construct a pie chart by calculating the angle for each sector. Explain why a pie chart might be more effective than a bar chart for showing the market share of each drink.

一家小咖啡馆调查了80位顾客最喜欢的饮品。结果为：咖啡34人，茶26人，热巧克力14人，其他6人。店主希望向潜在投资者展示调查结果。计算喜欢咖啡和茶的顾客合起来所占的比例。通过计算每个扇形的角度，绘制一个饼图。解释在显示每种饮品的市场份额时，为什么饼图可能比条形图更有效。

The combined proportion for coffee and tea is (34 + 26) ÷ 80 = 60 ÷ 80 = 0.75, or 75%. For the pie chart, total angle 360°. Angle for coffee = (34/80) × 360° = 153°. Tea: (26/80) × 360° = 117°. Hot chocolate: (14/80) × 360° = 63°. Other: (6/80) × 360° = 27°. A pie chart visually emphasises how the whole market is divided into shares, making it immediately obvious that coffee and tea dominate. A bar chart, while accurate, compares categories in isolation and does not inherently show the part‑whole relationship, which is crucial for a market share presentation.

咖啡和茶合计的比例为 (34 + 26) ÷ 80 = 60 ÷ 80 = 0.75，即75%。饼图中，总角度为360°。咖啡扇形的角度 = (34/80) × 360° = 153°。茶：117°。热巧克力：63°。其他：27°。饼图能够直观地强调整个市场是如何分割成不同份额的，让人一眼就能看出咖啡和茶占据主导地位。条形图虽然准确，但孤立地比较各类别，本身并不体现部分与整体的关系，而这对于市场份额展示至关重要。

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4. Social Science: Opinion Polls and Sampling | 社会科学：民意测验与抽样

A school council wants to estimate the proportion of students who support a new uniform policy. The school has 1200 pupils across Years 7–11. They decide to use a stratified sample of 120 students. The numbers in each year group are: Year 7 240, Year 8 250, Year 9 230, Year 10 260, Year 11 220. Calculate the number of students to be sampled from each year. Discuss one advantage of stratified sampling over simple random sampling in this context. If, in the sample, 78 out of 120 students support the policy, estimate the total number of students in the school who are in favour.

学生会希望估计支持新校服政策的学生比例。学校共有1200名学生，分布在7到11年级。他们决定采用120名学生的分层抽样。各年级人数分别为：7年级240人，8年级250人，9年级230人，10年级260人，11年级220人。计算每个年级应抽取的学生人数。在此背景下，讨论分层抽样相对于简单随机抽样的一个优点。如果样本中120人里有78人支持该政策，估计全校支持该政策的学生总人数。

The sampling fraction is 120/1200 = 0.1. Stratum sizes: Year 7: 240 × 0.1 = 24; Year 8: 250 × 0.1 = 25; Year 9: 230 × 0.1 = 23; Year 10: 260 × 0.1 = 26; Year 11: 220 × 0.1 = 22. Stratified sampling ensures that each year group is represented proportionally, so the views of minority year groups cannot be completely missed by chance, as could happen with simple random sampling. The sample proportion in favour is 78/120 = 0.65. Estimated total supporters in the school: 0.65 × 1200 = 780 students.

抽样比例为120/1200 = 0.1。各分层人数：7年级：240 × 0.1 = 24；8年级：25；9年级：23；10年级：26；11年级：22。分层抽样能确保每个年级都按比例被代表，因此人数较少的年级的意见不会因偶然因素被完全遗漏，而简单随机抽样则可能出现这种情况。样本中的支持比例为78/120 = 0.65。全校支持该政策的估计总人数为：0.65 × 1200 = 780名学生。

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5. Physics: Reaction Time Experiment | 物理：反应时间实验

A physics class tested the reaction time (in milliseconds) of 10 students using a computer‑based ruler drop test under two conditions: with no distraction and while listening to music. The paired data are given. The teacher wants to know if music, on average, slows reaction time. Calculate the differences (music − no distraction) for each student. Find the mean difference and its 95% confidence interval if the standard deviation of the differences is 18 ms. Based on the interval, state whether there is evidence that music increases reaction time.

物理课上，10名学生在一个基于计算机的落尺测试中分别在无干扰和听音乐两种条件下进行了反应时间（单位：毫秒）的测试。配对数据如下。老师希望了解听音乐是否会平均减慢反应速度。计算每位学生的差值（音乐 − 无干扰），求出差值的平均数，并计算其95%置信区间（设差值的标准差为18 ms）。根据置信区间，判断是否有证据表明音乐会增加反应时间。

Student	No distraction (ms)	Music (ms)
1	215	228
2	198	210
3	240	255
4	205	218
5	222	236
6	210	221
7	230	244
8	195	207
9	248	260
10	212	225

Differences (music − no distraction) are: 13, 12, 15, 13, 14, 11, 14, 12, 12, 13. Mean difference d̄ = 12.9 ms. The 95% confidence interval for the mean difference is d̄ ± (t9,0.025 × s/√n). For 9 degrees of freedom, the t‑value is approximately 2.262. s = 18 ms (given), s/√n = 18/√10 ≈ 5.69. Margin of error = 2.262 × 5.69 ≈ 12.9. So the interval is roughly (0, 25.8) ms. Because the interval does not contain zero, we have evidence at the 5% level that the true mean difference is positive, i.e., music increases reaction time on average.

差值（音乐 − 无干扰）分别为：13, 12, 15, 13, 14, 11, 14, 12, 12, 13。平均差值 d̄ = 12.9 ms。平均差值的95%置信区间为 d̄ ± (t9,0.025 × s/√n)。自由度为9时 t 值约为2.262。s = 18 ms（给定），s/√n = 18/√10 ≈ 5.69。误差范围 = 2.262 × 5.69 ≈ 12.9。因此区间约为 (0, 25.8) ms。由于区间不包含零，我们有5%显著性水平的证据表明真正的平均差值为正，即听音乐平均会增加反应时间。

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6. Environmental Science: Pollution Levels | 环境科学：污染水平

An environmental group recorded the concentration of nitrogen dioxide (NO₂) in micrograms per cubic metre (µg/m³) at ten monitoring sites across a city. The readings were: 42, 38, 55, 61, 47, 44, 59, 63, 41, 48. The air quality standard is an annual mean of 40 µg/m³. The group wants to test if the city’s mean NO₂ level exceeds the standard. Perform a one‑sample t‑test at the 5% significance level, stating your hypotheses, the test statistic, the critical value and your conclusion. Assume the data are approximately normally distributed.

一个环保组织记录了某城市十个监测点的二氧化氮（NO₂）浓度（单位：微克每立方米，µg/m³）。读数分别为：42, 38, 55, 61, 47, 44, 59, 63, 41, 48。空气质量标准规定年平均值为40 µg/m³。该组织想检验该城市的平均NO₂水平是否超过该标准。在5%显著性水平下进行单样本t检验，写明假设、检验统计量、临界值以及你的结论。假设数据近似正态分布。

Hypotheses: H₀: µ = 40; H₁: µ > 40 (one‑tailed). Sample mean x̄ = (42+38+55+61+47+44+59+63+41+48)÷10 = 498÷10 = 49.8 µg/m³. Sample standard deviation s = √[Σ(x − x̄)²/(n−1)]. Deviations squared: (42−49.8)²=60.84, (38−49.8)²=139.24, (55−49.8)²=27.04, (61−49.8)²=125.44, (47−49.8)²=7.84, (44−49.8)²=33.64, (59−49.8)²=84.64, (63−49.8)²=174.24, (41−49.8)²=77.44, (48−49.8)²=3.24. Sum = 733.6. Variance = 733.6/9 ≈ 81.51. s ≈ 9.03. Test statistic t = (49.8 − 40) ÷ (9.03/√10) = 9.8 ÷ (9.03/3.162) ≈ 9.8 ÷ 2.856 ≈ 3.43. Degrees of freedom = 9. One‑tailed critical value at 5% from t‑table is 1.833. Since 3.43 > 1.833, we reject H₀ and conclude there is sufficient evidence that the city’s mean NO₂ level exceeds the standard.

假设：H₀: µ = 40；H₁: µ > 40（单侧检验）。样本平均数 x̄ = (42+38+55+61+47+44+59+63+41+48)÷10 = 498÷10 = 49.8 µg/m³。样本标准差 s = √[Σ(x − x̄)²/(n−1)]。离差平方：60.84, 139.24, 27.04, 125.44, 7.84, 33.64, 84.64, 174.24, 77.44, 3.24。总和 = 733.6。方差 = 733.6/9 ≈ 81.51，s ≈ 9.03。检验统计量 t = (49.8 − 40) ÷ (9.03/√10) = 9.8 ÷ (9.03/3.162) ≈ 9.8 ÷ 2.856 ≈ 3.43。自由度 = 9。单侧5%临界值查t表得1.833。因为3.43 > 1.833，拒绝H₀，有充分证据表明该城市的平均NO₂水平超过标准。

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7. Health & Medicine: Drug Trial Data | 健康与医学：药物试验数据

A randomised controlled trial compared a new migraine drug with a placebo. The reduction in the number of monthly migraine attacks was recorded for 20 patients in each group. The summary statistics are: Treatment group: mean reduction = 4.2 attacks, standard deviation = 1.8. Placebo group: mean reduction = 1.5 attacks, standard deviation = 2.1. Carry out a two‑sample t‑test assuming unequal variances to determine whether the drug is significantly more effective than the placebo. Use a 1% significance level. State the null and alternative hypotheses, compute the test statistic and the approximate degrees of freedom using Welch’s formula, and give your conclusion.

一项随机对照试验将一种新型偏头痛药物与安慰剂进行比较。记录了两组各20名患者每月偏头痛发作次数的减少量。统计量总结为：治疗组：平均减少4.2次，标准差1.8；安慰剂组：平均减少1.5次，标准差2.1。进行不等方差的两样本t检验，判断药物是否显著比安慰剂更有效。使用1%显著性水平。写明零假设与备择假设，计算检验统计量和使用韦尔奇公式计算的近似自由度，并给出结论。

H₀: µ₁ = µ₂; H₁: µ₁ > µ₂ (one‑tailed). Test statistic t = (x̄₁ − x̄₂) / √(s₁²/n₁ + s₂²/n₂) = (4.2 − 1.5) / √(1.8²/20 + 2.1²/20) = 2.7 / √(3.24/20 + 4.41/20) = 2.7 / √(0.162 + 0.2205) = 2.7 / √0.3825 = 2.7 / 0.6185 ≈ 4.36. Degrees of freedom using Welch’s formula: df = (s₁²/n₁ + s₂²/n₂)² / [ (s₁²/n₁)²/(n₁−1) + (s₂²/n₂)²/(n₂−1) ] = (0.3825)² / [ (0.162)²/19 + (0.2205)²/19 ] = 0.1463 / [ 0.02624/19 + 0.04862/19 ] = 0.1463 / [ (0.07486)/19 ] = 0.1463 / 0.00394 ≈ 37.1 → approximately 37 df. The one‑tailed critical value for t with 37 df at the 1% level is about 2.43. Since 4.36 > 2.43, we reject H₀ and conclude the drug is significantly more effective.

H₀: µ₁ = µ₂；H₁: µ₁ > µ₂（单侧）。检验统计量 t = (x̄₁ − x̄₂) / √(s₁²/n₁ + s₂²/n₂) = (4.2 − 1.5) / √(1.8²/20 + 2.1²/20) = 2.7 / √(3.24/20 + 4.41/20) = 2.7 / √(0.162 + 0.2205) = 2.7 / √0.3825 = 2.7 / 0.6185 ≈ 4.36。韦尔奇公式的自由度：df = (0.3825)² / [ (0.162)²/19 + (0.2205)²/19 ] = 0.1463 / [ 0.02624/19 + 0.04862/19 ] = 0.1463 / (0.07486/19) = 0.1463 / 0.00394 ≈ 37.1，取约37自由度。在1%显著性水平下，37自由度的单侧t临界值约为2.43。由于4.36 > 2.43，拒绝H₀，药物显著更有效。

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8. Integrated Problem‑Solving Strategies | 综合题解题策略

Across all the examples, a systematic approach is vital when you face a cross‑curricular statistics question. Start by identifying the context and the data type (categorical, discrete, continuous). Decide which representation is most informative—a bar chart for categories, a line graph for time series, a scatter graph for relationships. Always check for outliers and consider whether a resistant measure like the median is more suitable than the mean. When drawing inferences, state hypotheses clearly, choose the appropriate test, calculate the test statistic and compare it to the correct critical value. Finally, write your conclusion in plain English that refers back to the original problem; never just say ‘reject H₀’ without explaining what that means in context.

在以上所有例子中，面对跨学科统计问题时，系统的方法至关重要。首先确定题目背景和数据类型（类别型、离散型、连续型）。判断哪种图表最能提供信息——条形图适合分类数据，折线图适合时间序列，散点图适合探究关系。始终检查异常值，并考虑使用中位数这类抵抗性量数是否比平均数更合适。进行推断时，清晰陈述假设，选择合适的检验方法，计算检验统计量并与相应的临界值比较。最后，用简洁的英语写出结论，并回指到原始问题；决不要只说“拒绝H₀”而不解释这在具体情境中意味什么。

Time management is equally important. In multi‑part questions, spend a minute scanning the whole task before you start writing. Often a later part gives a clue about the approach needed earlier. Practise converting between different representations of data—for example, from a table to a cumulative frequency diagram, or from a scatter plot to a line of best fit. After you finish, sanity‑check your answers: does an average of 133 mm for plant heights sound plausible? Does a confidence interval that includes negative values make sense in a drug trial? These checks will catch careless mistakes and deepen your statistical intuition.

时间管理同样重要。在多部分题目中，动笔前先花一分钟浏览整个任务。后面的小题常常会给前面的解题方法提供线索。练习在不同数据表征之间转换——例如从表格到累积频率图，或从散点图到最佳拟合线。完成后，对自己的答案进行理性检查：植物高度的平均数133 mm听起来合理吗？药物试验的置信区间包含负值有意义吗？这些检查

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