📚 Case Study in Statistics: Practical Analysis Workout | 统计案例分析：实战演练
This article presents a hands-on case study to help Year 9 students practise key statistical skills, including data collection, organisation, representation, and interpretation. We will work through a real-life dataset, applying measures of central tendency, spread, and graphical displays. By the end, you will be confident in tackling statistical investigations.
本文通过一个动手案例，帮助 9 年级学生练习核心统计技能，包括数据收集、整理、表示和解释。我们将处理一个贴近生活的数据集，应用集中趋势、离散程度和图形展示。读完本文，你将能够自信地应对统计调查。
1. Understanding the Research Question | 理解研究问题
Our school wants to investigate the screen time habits of Year 9 students to promote digital well-being. The research question is: 'How many hours per week do Year 9 students spend on screens for leisure?' We need to collect data, summarise it and draw evidence-based conclusions.
学校想调查 9 年级学生的屏幕时间习惯，以促进数字健康。研究问题是：“9 年级学生每周用于休闲的屏幕时间是多少小时？”我们需要收集数据、汇总信息并得出基于证据的结论。
2. Data Collection Method | 数据收集方法
We decide to conduct a survey using a random sample of 30 students from the year group. To reduce bias and ensure fair representation, we use stratified sampling by gender, selecting 15 boys and 15 girls. Each student reports their leisure screen time for the previous week, rounded to the nearest whole hour.
我们决定从年级组中随机抽取 30 名学生进行调查。为减少偏差并保证代表性，我们采用按性别分层抽样，选出 15 名男生和 15 名女生。每位学生报告他们上一周的休闲屏幕时间，四舍五入到最接近的整数小时。
3. Raw Data and Organisation | 原始数据与整理
The collected data (in hours) are shown below in the order they were recorded: 23, 15, 28, 12, 34, 18, 27, 9, 31, 14, 22, 26, 19, 35, 8, 17, 29, 11, 33, 20, 16, 25, 30, 13, 38, 21, 24, 10, 36, 7. To work efficiently, we first arrange the values in ascending order.
收集到的数据（单位：小时）按记录顺序排列如下：23, 15, 28, 12, 34, 18, 27, 9, 31, 14, 22, 26, 19, 35, 8, 17, 29, 11, 33, 20, 16, 25, 30, 13, 38, 21, 24, 10, 36, 7。为了高效处理，我们首先将这些数值按升序排列。
Sorted data: 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 38.
排序后的数据：7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 38。
4. Frequency Distribution Table | 频数分布表
To see the overall pattern, we group the data into class intervals of width 10 hours: 0–9, 10–19, 20–29, 30–39. A frequency table helps us calculate the midpoint of each class and later estimate the mean.
为了观察整体模式，我们将数据按 10 小时为组距进行分组：0–9、10–19、20–29、30–39。频数表能帮助我们计算每组组中值，并在之后估算平均数。
| Class Interval (hours) | Frequency (f) | Midpoint (x) | f × x |
|---|---|---|---|
| 0 – 9 | 3 | 4.5 | 13.5 |
| 10 – 19 | 10 | 14.5 | 145 |
| 20 – 29 | 10 | 24.5 | 245 |
| 30 – 39 | 7 | 34.5 | 241.5 |
| Total | 30 | 645 |
the grouped data, we estimate the mean as 645 ÷ 30 = 21.5 hours. The modal classes are 10–19 and 20–29, both with frequency 10, suggesting a bimodal shape.
从分组数据估算的平均数为 645 ÷ 30 = 21.5 小时。众数区间为 10–19 和 20–29，频数均为 10，表明分布呈现双峰形状。
5. Measures of Central Tendency | 集中趋势度量
by 30. The total is 7+8+9+...+38 = 651 hours. Therefore, the mean screen time = 651 ÷ 30 = 21.7 hours (to one decimal place).
使用原始数据，精确平均数通过求和并除以 30 得出。总和为 7+8+9+...+38 = 651 小时。因此，平均屏幕时间 = 651 ÷ 30 = 21.7 小时（保留一位小数）。
ian lies between the 15th and 16th values: both are 21 and 22. So median = (21 + 22) ÷ 2 = 21.5 hours. There is no single mode in this dataset because every value appears exactly once, making it mode-less for individual values.
为求中位数，我们找出中间位置的数值。对于 30 个有序数，中位数位于第 15 和第 16 个数值之间：分别是 21 和 22。因此中位数 = (21 + 22) ÷ 2 = 21.5 小时。此数据集没有单一的众数，因为每个数值都只出现一次，故个体值无众数。
6. Measures of Spread | 离散程度度量
The range is the difference between the maximum and minimum: 38 − 7 = 31 hours. This tells us the overall spread but is affected by extreme values.
极差是最大值与最小值之差：38 − 7 = 31 小时。它反映了整体散布情况，但易受极端值影响。
give a more robust picture. Q₁ is at position ¼(30+1)=7.75, so Q₁ = 13 + 0.75×(14 − 13) = 13.75. Q₂ (median) = 21.5. Q₃ is at position ¾(30+1)=23.25, so Q₃ = 27 + 0.25×(28 − 27) = 27.25. The interquartile range IQR = Q₃ − Q₁ = 27.25 − 13.75 = 13.5 hours.
四分位数提供了更稳健的描述。Q₁ 位于 ¼(30+1)=7.75 位置，因此 Q₁ = 13 + 0.75×(14 − 13) = 13.75。Q₂（中位数）= 21.5。Q₃ 位于 ¾(30+1)=23.25 位置，因此 Q₃ = 27 + 0.25×(28 − 27) = 27.25。四分位距 IQR = Q₃ − Q₁ = 27.25 − 13.75 = 13.5 小时。
7. Box-and-Whisker Plot | 箱线图
he box spans from Q₁ to Q₃, with a line at the median. Whiskers extend to the min and max.
箱线图概括了五数综合：最小值 = 7，Q₁ = 13.75，中位数 = 21.5，Q₃ = 27.25，最大值 = 38。箱体从 Q₁ 延伸至 Q₃，并在中位数处画线。须线伸出到最小值和最大值。
se fences, there are no outliers. The box plot would show a right-skewed distribution, as the upper whisker is longer than the lower one.
我们使用 1.5×IQR 法则检查异常值。下界 = 13.75 − 1.5×13.5 = −6.5，上界 = 27.25 + 1.5×13.5 = 47.5。由于所有数据都在界内，不存在异常值。箱线图将呈现右偏分布，因为上须线比下须线更长。
8. Comparative Analysis by Gender | 按性别比较分析
Suppose the 30 students were also classified by gender, giving two independent samples. Here are the results for boys' and girls' leisure screen times:
假设这 30 名学生也按性别分类，得到两个独立样本。以下是男生和女生休闲屏幕时间的数据：
Boys (15): 18, 22, 25, 19, 28, 30, 21, 26, 23, 27, 33, 20, 24, 29, 31
女生 (15): 12, 14, 16, 11, 15, 9, 13, 17, 8, 18, 14, 10, 19, 20, 7
27,28,29,30,31,33. Mean ≈ 25.07 hours, median = 25, Q₁ = 21, Q₃ = 29, IQR = 8, range = 15.
对于男生，排序后为 18,19,20,21,22,23,24,25,26,27,28,29,30,31,33。平均数 ≈ 25.07 小时，中位数 = 25，Q₁ = 21，Q₃ = 29，IQR = 8，极差 = 15。
15,16,17,18,19,20. Mean ≈ 13.53 hours, median = 14, Q₁ = 10, Q₃ = 17, IQR = 7, range = 13.
对于女生，排序后为 7,8,9,10,11,12,13,14,14,15,16,17,18,19,20。平均数 ≈ 13.53 小时，中位数 = 14，Q₁ = 10，Q₃ = 17，IQR = 7，极差 = 13。
Comparing the two groups, boys have a higher average screen time and greater spread. The median score for boys (25) lies above the entire girls' box from Q₁ to Q₃ (10–17), confirming a clear difference in leisure habits between genders in this sample.
对比两组，男生的平均屏幕时间更高，分布也更分散。男生的中位数（25）完全处于女生箱体（Q₁ 至 Q₃ 为 10–17）之上，证实了此样本中两性休闲习惯存在明显差异。
9. Probability from a Random Selection | 随机选取的概率
If we select one student at random from the 30, what is the probability that the student's screen time exceeds 20 hours? values (21,22,23,24,25,26,27,28,29,30,31,33,34,35,36,38). So P(screen time > 20) = 16/30 = 8/15 ≈ 0.533.
如果从 30 人中随机选取一名学生，其屏幕时间超过 20 小时的概率是多少？在排序列表中数出大于 20 的数值：共有 16 个。因此 P(屏幕时间 > 20) = 16/30 = 8/15 ≈ 0.533。
We could also ask: 'What is the probability of picking a student with screen time below the median?' xactly half the values lie below the median, that probability is 15/30 = 0.5, as expected.
我们还可以问：“抽到屏幕时间低于中位数的学生的概率是多少？”由于恰好一半的数值低于中位数，该概率为 15/30 = 0.5，与预期一致。
10. Drawing Conclusions and Evaluation | 得出结论与评估
The analysis shows that Year 9 students in our sample spend an average of about 21.7 hours per week on leisure screens, with a median of 21.5 hours. The distribution is slightly right-skewed, meaning a few students have very high usage, pulling the mean above the median. Gender comparison suggests boys tend to have higher screen time than girls.
分析显示，样本中的 9 年级学生平均每周休闲屏幕时间约为 21.7 小时，中位数为 21.5 小时。分布略呈右偏，意味着少数学生使用时间非常高，将平均数拉升至中位数之上。性别比较表明，男生往往比女生屏幕时间更长。
However, there are limitations: self-reported data may be inaccurate, the sample size is small (only 30), and results may not represent all Year 9 students. In a real
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