📚 Case Study Mastery: Analysing Study Habits and Passing Rates | 案例分析实战：学习习惯与通过率分析
In this case study, we will apply key statistical tools from the Year 12 Edexcel specification to a realistic scenario. An online learning platform wants to know whether weekly study hours predict test performance, and whether its new course meets a 70% pass-rate claim. You will see how data collection, descriptive statistics, correlation, regression and binomial hypothesis testing combine to answer real business questions.
本案例分析中，我们将运用 Year 12 Edexcel 统计课程中的核心工具，处理一个真实场景。一家在线学习平台想了解每周学习时间是否能预测测验成绩，并且其新课程是否真的达到了声称的 70% 通过率。你将看到数据收集、描述统计、相关性、回归以及二项假设检验如何协同回答实际的商业问题。
1. The Case Scenario | 案例场景
A new revision course was launched by an EdTech company. The data team randomly selected 10 students and recorded their weekly self-study hours (x) and their end-of-module test scores (y, out of 100). Separately, a random sample of 40 students was taken after the course, and 24 of them passed the final assessment. The company’s claim is that ‘at least 70% of students will pass’. Your task is to analyse the bivariate data and evaluate the pass-rate claim using the AS-level statistical toolkit.
一家教育科技公司推出了新的复习课程。数据团队随机选取了 10 名学生，记录了他们每周自主学习时间 (x) 和单元末测试成绩 (y, 满分 100)。此外，在课程结束后另随机抽取了 40 名学生，其中 24 人通过最终评估。公司的声称是「至少有 70% 的学生会通过」。你的任务是运用 AS 阶段的统计工具包，分析这组双变量数据并评估通过率声称。
2. Data Collection and Presentation | 数据收集与展示
The study-hour and test-score data were obtained by a simple random sample from the course participants. The raw pairs are listed below. For the hypothesis test, the sample of 40 students is also assumed to be random and independent.
学习时间与测试成绩数据通过简单随机抽样从课程参与者中获得。原始数据对如下。用于假设检验的 40 名学生样本同样假设为随机且独立。
| Student | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Hours (x) | 5 | 8 | 12 | 3 | 10 | 7 | 14 | 6 | 9 | 15 |
| Score (y) | 62 | 78 | 88 | 54 | 85 | 72 | 92 | 68 | 80 | 95 |
For the pass-rate investigation, we define X = number of students who pass out of 40. Observed X = 24.
对于通过率调查，我们定义 X 为 40 名学生中通过的人数。观测到 X = 24。
3. Descriptive Statistics: Study Hours | 描述统计：学习时间
We begin by summarising the study hours (x). The sorted data are: 3, 5, 6, 7, 8, 9, 10, 12, 14, 15. The five-number summary and measures of spread are computed as follows.
我们首先概括学习时间 (x)。排序后的数据为：3, 5, 6, 7, 8, 9, 10, 12, 14, 15。五数概括和离散程度计算如下。
Mean: x̄ = (5+8+12+3+10+7+14+6+9+15) / 10 = 89/10 = 8.9 hours. Median: for n=10, median = (8+9)/2 = 8.5 hours. Lower quartile Q₁ = (5+6)/2 = 5.5, upper quartile Q₃ = (10+12)/2 = 11. Hence IQR = 11 – 5.5 = 5.5 hours.
均值：x̄ = 8.9 小时。中位数：8.5 小时。下四分位数 Q₁ = 5.5，上四分位数 Q₃ = 11，四分位距 IQR = 5.5 小时。
The standard deviation is found using s = √[Sxx/(n-1)]. Here Σx² = 929, Sxx = 929 – 89²/10 = 136.9. Therefore s = √(136.9/9) ≈ √15.211 ≈ 3.90 hours. The distribution is slightly positively skewed, as the median (8.5) is slightly below the mean (8.9).
标准差使用 s = √[Sxx/(n-1)]。Σx² = 929，Sxx = 136.9，故 s ≈ 3.90 小时。分布轻微右偏，因为中位数 (8.5) 略低于均值 (8.9)。
4. Descriptive Statistics: Test Scores | 描述统计：测试成绩
The test scores (y) have the following summary: Σy = 774, so mean ȳ = 77.4. Sorted scores: 54, 62, 68, 72, 78, 80, 85, 88, 92, 95. Median = (78+80)/2 = 79. Q₁ = (68+72)/2 = 70, Q₃ = (88+92)/2 = 90, giving an IQR of 20. The range is 95 – 54 = 41.
测试成绩 (y) 的概括：Σy = 774，均值 ȳ = 77.4。中位数 = 79，Q₁ = 70，Q₃ = 90，IQR = 20。极差为 41。
Calculating Syy: Σy² = 61510, Syy = 61510 – 774²/10 = 1602.4. The standard deviation s_y = √(1602.4/9) ≈ √178.044 ≈ 13.34. A box plot would show the central 50% of scores lying between 70 and 90, with no outliers on the standard 1.5×IQR rule.
计算 Syy = 1602.4，标准差 s_y ≈ 13.34。箱线图将显示中间 50% 的成绩落在 70 到 90 之间，根据 1.5×IQR 规则没有异常值。
5. Correlation Analysis | 相关分析
To measure the strength and direction of the linear relationship between study hours and test scores, we compute the product-moment correlation coefficient (PMCC). The needed statistics are already at hand.
为了衡量学习时间与测试成绩之间线性关系的强度和方向，我们计算积矩相关系数 (PMCC)。所需统计量已在手边。
Sxy = Σxy – (Σx)(Σy)/n = 7347 – (89×774)/10 = 7347 – 6888.6 = 458.4
r = Sxy / √(Sxx × Syy) = 458.4 / √(136.9 × 1602.4) = 458.4 / √219372.56 ≈ 458.4 / 468.37 ≈ 0.979
An r value of 0.979 indicates a very strong positive linear correlation. As study hours increase, the test score tends to increase almost perfectly linearly. The correlation is significant at the 1% level for n=10 (critical value ≈ 0.765), confirming a genuine association in the sample.
r 值为 0.979 表明极强的正线性相关。随着学习时间增加，测试成绩几乎完美地线性上升。对于 n=10，该相关系数在 1% 显著性水平下是显著的（临界值约 0.765），证实样本中存在真实关联。
6. Linear Regression Analysis | 线性回归分析
Since the correlation is strong, we can fit a least-squares regression line to predict test score from study hours. The line has equation y = a + bx.
鉴于相关性很强，我们可以拟合一条最小二乘回归线，用学习时间预测测试成绩。直线方程为 y = a + bx。
b = Sxy / Sxx = 458.4 / 136.9 ≈ 3.348 (3 d.p.)
a = ȳ – b x̄ = 77.4 – 3.348 × 8.9 ≈ 77.4 – 29.80 = 47.60 (2 d.p.)
The regression equation is: predicted score = 47.6 + 3.35 × (hours). For a student studying 11 hours per week, the predicted score is 47.6 + 3.35×11 = 84.45. Interpolation is reliable here because 11 is within the data range; extrapolation beyond 15 hours would be risky.
回归方程为：预测成绩 = 47.6 + 3.35 × (学习小时数)。对于每周学习 11 小时的学生，预测成绩为 84.45。因为 11 在数据范围内，内插法是可靠的；外推到超过 15 小时则有风险。
7. Binomial Setting for the Pass-Rate Claim | 通过率声称的二项分布设定
We now switch to the hypothesis test. The company states that at least 70% of students pass the new course. Let p be the true population pass proportion. The test is based on a random sample of n = 40 independent students, with observed number of passes X = 24. Under the null hypothesis, each student’s outcome is a Bernoulli trial with success probability 0.7, so X ~ B(40, 0.7).
我们现在转向假设检验。公司声称至少 70% 的学生通过了新课程。设 p 为真实的总体通过比例。检验基于 n=40 名独立学生的随机样本，观测通过人数 X=24。在原假设下，每名学生的结果是一次成功概率为 0.7 的伯努利试验，因此 X ~ B(40, 0.7)。
We check binomial conditions: fixed number of trials (40), two outcomes (pass/fail), constant probability (assumed 0.7 under H₀), and independence (random sample smaller than 10% of population). All are satisfied.
检查二项条件：试验次数固定 (40)，两种结果（通过/不通过），概率恒定（H₀ 下假定为 0.7），以及独立性（随机样本小于总体的 10%）。所有条件均满足。
8. Hypothesis Test Set-Up | 假设检验设置
The observed sample proportion is 24/40 = 0.6, which is below 0.7. We test whether this provides sufficient evidence to reject the company’s claim. Because the claim contains ‘at least’, a one-tailed test is appropriate.
观测样本比例为 24/40 = 0.6，低于 0.7。我们检验这是否提供了拒绝公司声称的充分证据。因为声称包含「至少」，适合采用单尾检验。
H₀: p = 0.7
H₁: p < 0.7
Significance level α = 0.05
The test statistic is the number of passes, X. Under H₀, X ~ B(40, 0.7). The p-value will be P(X ≤ 24), because values smaller than or equal to the observed count are less likely under the null and point toward the alternative.
检验统计量为通过人数 X。在 H₀ 下，X ~ B(40, 0.7)。p 值将是 P(X ≤ 24)，因为小于或等于观测计数的值在原假设下更不可能出现，并指向备择假设。
9. Computing the p-Value and Conclusion | 计算 p 值与结论
Using binomial tables or technology, we find P(X ≤ 24) for B(40, 0.7). The cumulative probability is approximately 0.0962. (If calculating manually, the normal approximation gives z ≈ -1.38 with a one-tail p-value of about 0.084, but the exact binomial value is more accurate.)
使用二项分布表或技术工具，我们找到 B(40, 0.7) 的 P(X ≤ 24) ≈ 0.0962。（若手动计算，正态近似给出 z ≈ -1.38，单尾 p 值约 0.084，但精确二项值更准确。）
p-value = P(X ≤ 24 | p=0.7) ≈ 0.0962
Since 0.0962 > 0.05, we do not reject H₀. There is insufficient evidence to conclude that the pass rate is less than 70%. The observed result, though lower, is not statistically significant at the 5% level. The company’s claim stands.
由于 0.0962 > 0.05，我们不拒绝 H₀。没有足够证据得出通过率低于 70% 的结论。观测结果虽然较低，但在 5% 显著性水平下并不显著。公司声称成立。
10. Sampling, Limitations and Further Considerations | 抽样、局限与进一步思考
All analyses depend on the assumption that both samples were truly random and representative. The small sample size of 10 for correlation limits the generalisability of the regression model; a larger sample would yield more reliable estimates. For the hypothesis test, the p-value is close to the boundary – a slightly larger sample might have led to rejection. Non-sampling errors, such as self-reported study hours, could introduce bias.
所有分析都依赖于两个样本确实是随机且具有代表性这一假设。用于相关性分析的样本容量仅为 10，限制了回归模型的推广性；更大的样本会得出更可靠的估计。在假设检验中，p 值接近边界——稍微增大样本量可能导致拒绝。非抽样误差，例如自报的学习时间，可能引入偏差。
In a full investigation you would also check residuals and linear model assumptions for the regression, and consider whether a continuity correction or a two-tailed test might be more appropriate in other contexts. The case study demonstrates the power of integrating data description, correlation, regression and hypothesis testing – all essential skills in the Year 12 Edexcel statistics toolkit.
在一个完整的调查中，你还会
Published by TutorHao | Year 12 统计 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