📚 Case Study in Statistics: Analysing Screen Time and Sleep Patterns | 统计案例分析实战演练：探究屏幕时间与睡眠
In this case study, we will carry out a complete statistical investigation to explore the relationship between daily screen time (in hours) and nightly sleep duration (in hours) among Year 11 students. This realistic scenario will guide you through each stage of the statistical enquiry cycle: design, data collection, analysis, and interpretation. By working through this example, you will see how the techniques from your Cambridge Statistics course come together in a practical, exam-style context.
在本案例分析中，我们将完成一项完整的统计调查，探究11年级学生每日屏幕时间（小时）与夜间睡眠时间（小时）之间的关系。这一真实情境将带你经历统计探究周期的每一个阶段：设计、数据收集、分析与解读。通过这个实例，你将看到剑桥统计课程中的各种技巧如何在一个实用的、贴近考试的背景中协同运用。
1. Designing the Questionnaire | 问卷设计
We began by designing a simple questionnaire to record two numerical variables: “On a typical school day, how many hours do you spend looking at screens (phone, tablet, computer, TV)?” and “On a typical school night, how many hours of sleep do you get?”. The questions were kept closed and quantitative to ensure accurate, comparable data. Ethical considerations included anonymity and informed consent; students were told the purpose of the study and that participation was voluntary.
我们首先设计了一份简单的问卷，用于记录两个数值型变量：“在一个普通上学日，你花费多少小时看屏幕（手机、平板、电脑、电视）？”以及“在一个普通上学日的夜晚，你通常睡多少小时？”问题保持封闭和定量，以确保数据准确、可比。伦理方面的考量包括匿名和知情同意；学生被告知研究目的，且参与是完全自愿的。
2. Data Collection and Sampling | 数据收集与抽样
We used simple random sampling to select 30 students from the Year 11 cohort of 200. Each student was assigned a number from 1 to 200, and a random number generator was used to pick 30 unique numbers. This method gave every student an equal chance of being chosen, reducing selection bias. The final sample consisted of 30 paired observations: (screen time, sleep time).
我们从200名11年级学生中采用简单随机抽样选取了30人。每位学生被赋予1至200的编号，然后使用随机数生成器选出30个不重复的号码。这种方法让每个学生被选中的机会均等，减少了选择偏差。最终样本包含30组成对观测值：（屏幕时间，睡眠时间）。
3. Organising Data: Frequency Table | 数据整理：频数分布表
To summarise the screen time data, we grouped the values into class intervals of width 1 hour. The frequency table below shows the distribution of screen time among the 30 students. Such a table helps us see the shape of the data at a glance.
为了汇总屏幕时间数据，我们将数值分组，组距为1小时。下面的频数表展示了30名学生屏幕时间的分布情况。这样的表格能帮助我们一眼看出数据的形态。
| Screen time (hours) | Frequency | Cumulative frequency |
|---|---|---|
| 0 ≤ t < 1 | 2 | 2 |
| 1 ≤ t < 2 | 5 | 7 |
| 2 ≤ t < 3 | 8 | 15 |
| 3 ≤ t < 4 | 7 | 22 |
| 4 ≤ t < 5 | 6 | 28 |
| 5 ≤ t < 6 | 2 | 30 |
The grouped data show a rough bell shape, with most students reporting between 2 and 4 hours of screen time. Only a few students had very low or very high usage.
分组数据大致呈钟形，大多数学生报告的屏幕时间在2至4小时之间。只有少数学生的使用量偏低或偏高。
4. Visualising Data: Histograms and Scatter Plots | 数据可视化：直方图与散点图
We constructed a histogram of screen time using the frequency table above. The histogram confirmed a slightly right-skewed distribution, with the modal class being 2–3 hours. A scatter plot of sleep time against screen time was then drawn, treating screen time as the explanatory variable (x-axis) and sleep time as the response variable (y-axis). The scatter plot showed a clear downward trend: as screen time increased, sleep time tended to decrease. Several students who reported more than 5 hours of screen time slept fewer than 6 hours.
我们利用上述频数表绘制了屏幕时间的直方图。直方图确认了分布略呈右偏，众数所在组为2–3小时。接着绘制了睡眠时间对屏幕时间的散点图，将屏幕时间作为解释变量（x轴），睡眠时间作为响应变量（y轴）。散点图呈现出明显的下降趋势：屏幕时间越长，睡眠时间往往越短。有好几位报告屏幕时间超过5小时的学生，睡眠时间不足6小时。
5. Measures of Central Tendency | 集中趋势的度量
For the screen time variable, we calculated the mean, median, and mode. The raw data had a total sum of 86.3 hours for 30 students, giving a mean screen time of x̄ = 86.3 / 30 ≈ 2.88 hours. The median was found by ordering the 30 values; the 15th and 16th ordered values were 2.8 and 2.9, so the median was (2.8 + 2.9)/2 = 2.85 hours. The mode, taken from the grouped table, fell in the 2–3 hour interval. The closeness of the mean and median suggests a roughly symmetric distribution with a slight positive skew.
对于屏幕时间变量，我们计算了均值、中位数和众数。原始数据30名学生的屏幕时间总和为86.3小时，因此均值 x̄ = 86.3 / 30 ≈ 2.88 小时。将30个数值排序后，第15和第16个有序值分别是2.8和2.9，因此中位数为 (2.8 + 2.9)/2 = 2.85 小时。从分组表看，众数落在2–3小时区间。均值与中位数接近，表明分布大致对称且带有轻微正偏。
For sleep time, the mean was 7.35 hours and the median was 7.4 hours. Both were very close, and the typical student got between 7 and 8 hours of sleep per night.
对于睡眠时间，均值为7.35小时，中位数为7.4小时。两者非常接近，典型学生每晚睡眠时间在7至8小时之间。
6. Measures of Dispersion: Range, Quartiles, Standard Deviation | 离散程度的度量：极差、四分位数、标准差
We computed the range of screen time as 5.1 − 0.8 = 4.3 hours. The lower quartile (Q₁) was 1.9 hours and the upper quartile (Q₃) was 3.7 hours, giving an interquartile range of 1.8 hours. To measure the average spread around the mean, we calculated the sample standard deviation using:
我们计算屏幕时间的极差为 5.1 − 0.8 = 4.3 小时。下四分位数（Q₁）为1.9小时，上四分位数（Q₃）为3.7小时，四分位距为1.8小时。为了度量围绕均值的平均离散程度，我们用下面公式计算了样本标准差：
s = √[ Σ(x – x̄)2 / (n – 1) ]
After substituting the values, we obtained s ≈ 1.12 hours for screen time. This tells us that a typical student’s screen time deviates from the average by about 1.12 hours. The sleep time standard deviation was 0.95 hours, showing slightly less relative scatter.
代入数值后，我们求得屏幕时间的 s ≈ 1.12 小时。这告诉我们，一名典型学生的屏幕时间与平均值大约相差1.12小时。睡眠时间的标准差为0.95小时，显示相对离散程度略小。
7. Box Plots and Outliers | 箱线图与异常值
We drew box plots for both variables. For screen time, the five‑number summary was: min = 0.8, Q₁ = 1.9, median = 2.85, Q₃ = 3.7, max = 5.1. Fences were calculated as Q₁ − 1.5×IQR = −0.8 and Q₃ + 1.5×IQR = 6.4. No data point fell outside these fences, so there were no outliers. The box plot confirmed a slight skew to the right. The sleep time box plot was fairly symmetric, with no outliers either. Comparing the two box plots side by side highlighted the inverse relationship: higher screen times corresponded to lower median sleep times when we split the sample into low and high screen time groups.
我们为两个变量绘制了箱线图。对于屏幕时间，五数概括为：最小值=0.8，Q₁=1.9，中位数=2.85，Q₃=3.7，最大值=5.1。计算得出的内限为 Q₁ − 1.5×IQR = −0.8 和 Q₃ + 1.5×IQR = 6.4。没有数据点落在这些界限之外，因此没有异常值。箱线图确认了分布略右偏。睡眠时间的箱线图相当对称，同样没有异常值。当我们把样本按屏幕时间高低分成两组时，将两个箱线图并排比较，明显看出逆相关关系：屏幕时间较高的组，睡眠时间中位数较低。
8. Correlation Analysis: Pearson’s r | 相关性分析：皮尔逊相关系数
To quantify the linear relationship, we calculated Pearson’s product‑moment correlation coefficient. The formula is:
为了量化线性关系，我们计算了皮尔逊积矩相关系数。公式如下：
r = Σ(x – x̄)(y – ȳ) / √[ Σ(x – x̄)2 Σ(y – ȳ)2 ]
Using our data, we found Σ(x – x̄)(y – ȳ) = −18.24, Σ(x – x̄)2 = 36.30, and Σ(y – ȳ)2 = 26.18. Thus:
根据我们的数据，算得 Σ(x – x̄)(y – ȳ) = −18.24，Σ(x – x̄)2 = 36.30，Σ(y – ȳ)2 = 26.18。因此：
r = −18.24 / √(36.30 × 26.18) ≈ −0.592
An r of approximately −0.59 indicates a moderate negative linear correlation: as screen time goes up, sleep time tends to go down. The value is sufficiently far from 0 to be of practical interest, but we need a formal test to determine statistical significance.
r 约为 −0.59，表明存在中等程度的负线性相关：屏幕时间增加时，睡眠时间往往减少。这个值离0足够远，值得实际关注，但需要正式检验来确定其统计显著性。
9. Regression Analysis: Line of Best Fit | 回归分析：最佳拟合线
We fitted a least-squares regression line of sleep time (y) on screen time (x). The equation of the line is y = a + bx, where:
我们用最小二乘法拟合了睡眠时间（y）对屏幕时间（x）的回归直线。直线方程为 y = a + bx，其中：
b = Σ(x – x̄)(y – ȳ) / Σ(x – x̄)2 = −18.24 / 36.30 ≈ −0.502
a = ȳ – b x̄ = 7.35 − (−0.502 × 2.88) ≈ 8.80
So the regression equation is: Sleep = 8.80 − 0.502 × Screen time. This model predicts that for each additional hour of screen time, sleep time decreases by about 0.5 hours, on average. The intercept of 8.80 hours represents the estimated sleep time for a student with zero screen time – a value that may not be realistic but is mathematically useful for the line.
于是回归方程为：睡眠时间 = 8.80 − 0.502 × 屏幕时间。该模型预测，屏幕时间每增加1小时，睡眠时间平均减少约0.5小时。截距8.80小时表示屏幕时间为零的学生的估计睡眠时间——这个值可能不现实，但在数学上对直线有用。
We used the equation to predict sleep time for a student reporting 3.5 hours of screen time: 8.80 − 0.502 × 3.5 ≈ 7.04 hours. The observed values for students near this screen time were indeed around 7 hours, confirming the model’s fit.
我们用该方程预测了一名报告3.5小时屏幕时间的学生的睡眠：8.80 − 0.502 × 3.5 ≈ 7.04小时。该屏幕时间附近学生的观测值确实在7小时左右，验证了模型的拟合效果。
10. Hypothesis Testing for Correlation | 相关系数的假设检验
We performed a hypothesis test to check whether the sample correlation is significantly different from zero in the population. Let ρ denote the population correlation coefficient.
我们进行假设检验，以判断样本相关系数是否与零存在显著差异。令ρ表示总体相关系数。
- H₀: ρ = 0 (no linear correlation)
- H₁: ρ ≠ 0 (two‑tailed test)
- H₀：ρ = 0（无线性相关）
- H₁：ρ ≠ 0（双侧检验）
Significance level α = 0.05. With n = 30, degrees of freedom = n − 2 = 28. From critical value tables for Pearson’s r, the critical value at α = 0.05 (two‑tailed) is approximately 0.361. Our calculated |r| = 0.592 exceeds 0.361, so we reject the null hypothesis. There is sufficient evidence at the 5% level to conclude that a significant linear correlation exists between screen time and sleep time in the population of Year 11 students.
显著性水平取 α = 0.05。由于 n = 30，自由度 = n − 2 = 28。查阅皮尔逊 r 的临界值表，在 α = 0.05（双侧）下，临界值约为0.361。我们计算所得的 |r| = 0.592 大于 0.361，因此拒绝零假设。有足够的证据在5%显著水平下推断，在11年级学生总体中，屏幕时间与睡眠时间之间存在显著的线性相关。
We also computed the p‑value using a t‑transform: t = r √(n−2) / √(1−r²) ≈ 3.89, with corresponding p ≈ 0.0006, which is far less than 0.05, reinforcing the conclusion.
我们还用 t 变换计算了 p 值：t = r √(n−2) / √(1−r²) ≈ 3.89，相应的 p ≈ 0.0006，远小于0.05，强化了这一结论。
11. Interpretation and Limitations | 解释与局限性
Although we found a significant negative correlation, we must be careful not to claim that more screen time causes less sleep. Correlation does not imply causation. There may be confounding variables, such as homework load, extracurricular commitments, or students’ chronotypes (morningness‑eveningness), that affect both screen use and sleep duration. Moreover, our sample size of 30, while adequate for this school‑based investigation, may not be representative of all Year 11 students in the country. The data were self‑reported, so recall bias could affect accuracy. The regression model explains only r² ≈ 0.35 (35%) of the variation in sleep time, leaving 65% unexplained. These limitations should be acknowledged in any real‑world statistical report.
虽然我们发现了显著的负相关，但必须谨慎，不能宣称更多的屏幕时间导致睡眠减少。相关不意味着因果。可能存在混杂变量，比如作业量、课外活动或学生的昼夜节律类型（清晨型‑夜晚型），这些因素既影响屏幕使用也影响睡眠时长。此外，我们的样本量30虽然对这次校内研究足够，但可能无法代表全国所有11年级学生。数据是自报的，回忆偏差可能影响准确性。回归模型仅解释了睡眠时间变异的大约35%（r² ≈ 0.35），还有65%无法解释。在任何现实世界的统计报告中，都应承认这些局限性。
12. Conclusion and Reflection | 结论与反思
This case study has guided you through a full statistical investigation: from careful design and ethical data collection, through descriptive statistics and graphical displays, to inferential techniques including correlation, regression, and hypothesis testing. The analysis revealed a moderate, statistically significant negative relationship between screen time and sleep among the Year 11 students sampled. Skills such as selecting appropriate measures, interpreting computer output or hand calculations, and communicating findings with honesty about limitations are precisely what examiners look for. Revisiting real datasets like this will build your confidence for both coursework and written examinations.
本案例分析带你经历了一次完整的统计调查：从精心设计、合乎伦理的数据收集，到描述性统计与图形展示，再到推断技术，包括相关、回归和假设检验。分析表明，在所抽样的11年级学生中，屏幕时间与睡眠之间存在中等程度且统计上显著的负相关关系。选择适当的度量标准、解释计算机输出或手算结果、并诚实地结合局限性传达发现，正是考官看重的技能。重温类似的真实数据集，将为你的课程作业和笔试带来自信。
Published by TutorHao | Statistics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply