Case Study in Action: Applying Statistics to Cat Data | 案例分析实战演练：猫数据统计分析

📚 Case Study in Action: Applying Statistics to Cat Data | 案例分析实战演练：猫数据统计分析

In this case study, we will follow the journey of a pet food company as it analyses data on cats to make evidence-based decisions. Working through a real dataset helps to connect statistical techniques with meaningful conclusions, a vital skill for your IGCSE CIE Statistics exam. Each step mirrors what you might be asked to do in a practical or written paper, from collecting data to interpreting results.

在这个案例中，我们将跟随一家宠物食品公司，分析有关猫咪的数据，以做出基于证据的决策。通过一个真实数据集，将统计方法与有意义的结论联系起来，这是 IGCSE CIE 统计考试中的一项关键技能。每一步都对应着你在实践或笔试中可能被要求完成的任务，从收集数据到解读结果。

1. Introduction to the Case Study | 案例研究介绍

PurrfectPet, a specialist cat food producer, wants to understand the relationship between a cat’s body mass and its daily dry food intake. The company collected data from 15 adult domestic shorthair cats, recording weight in kilograms and food consumption in grams over one week. The goal is to describe the data, investigate any association, and use findings to guide portion recommendations.

PurrfectPet 是一家专业猫粮生产商，希望了解猫咪的体重与每日干粮摄入量之间的关系。该公司从 15 只成年家养短毛猫收集了数据，记录了一周内的体重（千克）和食物消耗量（克）。其目标是描述这些数据，研究是否存在关联，并利用发现来指导喂食建议。

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2. Data Collection and Sampling | 数据收集与抽样方法

The 15 cats were selected randomly from a local veterinary clinic’s database of healthy adult cats. This is a simple random sample, which helps avoid bias and makes the sample representative of the wider adult cat population. The variables measured are continuous quantitative data: weight to the nearest 0.1 kg and food consumption to the nearest gram.

这 15 只猫是从当地兽医诊所健康成年猫数据库中随机选取的。这是一个简单随机样本，有助于避免偏倚，使样本对更广泛的成年猫群体具有代表性。测量变量是连续型定量数据：体重精确到 0.1 kg，食物消耗量精确到克。

Cat	Weight (kg)	Food (g)
1	3.2	45
2	4.5	60
3	3.8	52
4	5.0	70
5	2.9	38
6	4.1	55
7	5.5	75
8	3.5	48
9	4.8	65
10	5.2	72
11	3.0	40
12	4.3	58
13	5.8	80
14	3.7	50
15	4.0	54

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3. Organising Data: Frequency Tables | 数据整理：频数分布表

When dealing with continuous data, a grouped frequency table makes patterns clearer. We group the cat weights into equal-width intervals: 2.5 – 3.4 kg, 3.5 – 4.4 kg, 4.5 – 5.4 kg, and 5.5 – 6.4 kg. The class width is 1.0 kg. Tallying the data gives us the frequency distribution shown below.

处理连续型数据时，分组频数表能让模式更加清晰。我们将猫的体重分成等宽区间：2.5–3.4 kg、3.5–4.4 kg、4.5–5.4 kg 和 5.5–6.4 kg。组距为 1.0 kg。清点数据后得到下面的频数分布。

Weight (kg)	Frequency, f
2.5 – 3.4	3
3.5 – 4.4	6
4.5 – 5.4	4
5.5 – 6.4	2

Remember, the lower class boundary is 2.45 for the first group and the upper class boundary 3.45, ensuring no gaps between bars in a histogram.

请记住，第一组的组下限是 2.45，组上限是 3.45，这能确保直方图中各矩形之间没有空隙。

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4. Visualising Data: Histograms and Frequency Polygons | 数据可视化：直方图与频数多边形

To draw a histogram, the horizontal axis shows weight boundaries and the vertical axis shows frequency density if class widths are unequal — here they are equal, so we can plot frequency directly. The bars touch each other to indicate continuous data. A frequency polygon can be added by joining the midpoints of each bar top.

绘制直方图时，横轴显示体重边界，如果组距不相等，纵轴需显示频数密度——但这里组距相等，因此可以直接绘制频数。矩形之间相互接触，表示连续数据。通过连接每个矩形顶部的组中点，可以添加频数多边形。

The distribution is slightly skewed to the right, with a peak in the 3.5–4.4 kg class. This tells us that most cats in this sample are of moderate weight.

分布略呈右偏，峰值出现在 3.5–4.4 kg 组。这说明样本中大多数猫体重适中。

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5. Cumulative Frequency and Box Plots | 累积频率与箱线图

Cumulative frequency is obtained by adding each frequency to the sum of previous frequencies. We plot the upper class boundary against cumulative frequency and draw a smooth curve. From this curve we can estimate the median, lower quartile (Q₁) and upper quartile (Q₃).

累积频数是将每个频数加到之前所有频数之和上得到的。以组上限为横坐标、累积频数为纵坐标作图，并绘制平滑曲线。从这条曲线上我们可以估计中位数、下四分位数 (Q₁) 和上四分位数 (Q₃)。

Upper class boundary	Cumulative frequency
3.45	3
4.45	9
5.45	13
6.45	15

From the cumulative frequency curve, median ≈ 4.1 kg, Q₁ ≈ 3.4 kg, Q₃ ≈ 5.1 kg. The interquartile range (IQR) is Q₃ – Q₁ = 1.7 kg. A box-and-whisker plot shows the minimum (2.9 kg), Q₁, median, Q₃ and maximum (5.8 kg), giving a clear picture of spread and skewness.

根据累积频率曲线，中位数 ≈ 4.1 kg，Q₁ ≈ 3.4 kg，Q₃ ≈ 5.1 kg。四分位距 (IQR) 为 Q₃ – Q₁ = 1.7 kg。箱线图展示了最小值 (2.9 kg)、Q₁、中位数、Q₃ 和最大值 (5.8 kg)，清晰地反映了数据的分散程度和偏态。

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6. Averages and Measures of Spread | 中心趋势和离散度：平均值、中位数、标准差

We calculate the mean weight: sum of all weights = 63.5 kg, so mean x̄ = 63.5 ÷ 15 ≈ 4.23 kg. The median is the 8th ordered value = 4.1 kg, close to the mean. The mode lies in the modal class 3.5–4.4 kg.

我们计算体重平均值：所有体重之和为 63.5 kg，因此均值 x̄ = 63.5 ÷ 15 ≈ 4.23 kg。中位数是排序后的第 8 个值 = 4.1 kg，与均值接近。众数落在众数组 3.5–4.4 kg 内。

For spread, the range = 5.8 – 2.9 = 2.9 kg. The IQR = 1.7 kg. To measure consistency, we find the standard deviation (using the formula s = √[Σ(x – x̄)²/(n – 1)]). The computed standard deviation is approximately 0.96 kg. This tells us typical deviation from the mean.

对于离散程度，极差 = 5.8 – 2.9 = 2.9 kg，IQR = 1.7 kg。为了衡量一致性，我们计算标准差（使用公式 s = √[Σ(x – x̄)²/(n – 1)]）。经计算，标准差约为 0.96 kg。这表明了相对于均值的典型偏离程度。

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7. Scatter Diagrams and Correlation | 散点图与相关性

Now we examine the relationship between weight (x) and food consumption (y). Plotting the 15 paired values reveals a clear upward trend: heavier cats tend to eat more. This suggests a positive correlation. We can describe it as strong, positive and approximately linear.

现在我们研究体重 (x) 与食物消耗量 (y) 之间的关系。绘制 15 个成对数据点后，可看到明显的上升趋势：较重的猫往往吃得更多。这表明存在正相关。我们可以将其描述为强正向近似线性相关。

An initial visual assessment can be supported by calculating Spearman’s rank correlation coefficient, or simply Pearson’s correlation coefficient if we assume linearity. For IGCSE, you may be asked to comment on the scatter, draw a line of best fit and interpret the strength without calculating r.

除了初步的直观评估，还可以计算斯皮尔曼等级相关系数，或如果假设线性关系，可计算皮尔逊相关系数。在 IGCSE 中，你可能只需要对散点图进行评论，画一条最佳拟合线，并在不计算 r 的情况下解释相关强度。

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8. Linear Regression and Line of Best Fit | 线性回归与最佳拟合线

Using the method of least squares (or estimating by eye and passing through the mean point), we find the equation of the line of best fit. For our data, the approximate regression line is y = 10.5x + 5.8, where y is food in grams and x is weight in kg. This line can be used for prediction: for a 4.5 kg cat, predicted consumption ≈ 10.5×4.5 + 5.8 ≈ 55 g per day.

使用最小二乘法（或通过目测并穿过均值点），我们得到最佳拟合线的方程。对于我们的数据，回归线约为 y = 10.5x + 5.8，其中 y 为食物克数，x 为体重公斤数。这条线可用于预测：对于一只 4.5 kg 的猫，预测消耗量 ≈ 10.5×4.5 + 5.8 ≈ 每天 55 g。

Remember, the line of best fit must pass through (x̄, ȳ). For our data, x̄ ≈ 4.23 kg, ȳ ≈ 57.5 g, and this point lies on the line. When drawing, always label axes, use a sensible scale and plot points accurately.

请记住，最佳拟合线必须通过 (x̄, ȳ)。对于我们的数据，x̄ ≈ 4.23 kg, ȳ ≈ 57.5 g，该点确实在线上。绘制时务必标记坐标轴、使用合理的刻度并准确描点。

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9. Probability from Relative Frequency | 相对频率与概率

We can use the data to estimate probabilities. For instance, what is the probability that a randomly chosen cat from this sample weighs more than 5.0 kg? There are 4 out of 15 cats meeting this condition (cats 4,7,10,13). So estimated P(weight > 5.0 kg) = 4/15 ≈ 0.267.

我们可以利用数据来估计概率。例如，从该样本中随机选出一只猫，体重大于 5.0 kg 的概率是多少？满足条件的猫有 4 只（第 4、7、10、13 号）。因此估计的 P(体重 > 5.0 kg) = 4/15 ≈ 0.267。

Similarly, for food consumption above 60 g, 5 cats qualify, giving estimated probability 5/15 = 1/3. These are experimental probabilities based on observed relative frequency.

类似地，食物消耗量超过 60 g 的猫有 5 只，估计概率为 5/15 = 1/3。这些都是基于观测相对频率的实验概率。

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10. Conditional Probability & Tree Diagrams | 条件概率与树状图

Conditional probability questions often appear in IGCSE. For example, given that a cat eats more than 60 g, what is the probability that it also weighs more than 5.0 kg? From our data, the 5 cats with high consumption are: 4, 7, 9, 10, 13. Among these, the ones weighing >5.0 kg are cats 4, 7, 10, 13 — that’s 4 out of 5. So P(weight >5.0 | food >60) = 4/5 = 0.8.

条件概率问题在 IGCSE 中经常出现。例如，已知一只猫食量超过 60 g，它体重也超过 5.0 kg 的概率是多少？从数据中可知，高食量的 5 只猫为：4、7、9、10、13 号。其中体重 >5.0 kg 的是 4、7、10、13 号，共 4 只。因此 P(体重 >5.0 | 食量 >60) = 4/5 = 0.8。

You can also extend this to a tree diagram, showing branches for ‘high weight’ and ‘not high weight’, with probabilities updating at the second branch based on the first outcome. Tree diagrams help solve multi-stage probability problems.

你还可以将此扩展到树状图上，显示“高体重”和“非高体重”的分支，并根据第一次的结果在第二级分支更新概率。树状图有助于解决多阶段概率问题。

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11. Interpretation and Conclusions | 解读与结论

The analysis confirms a strong positive association between cat weight and food consumption. Heavier cats require more calories, which is biologically plausible. The regression model can guide feeding recommendations, but we must be cautious: the sample is small (n=15) and includes only one breed. Extrapolating beyond the data range (e.g., for kitten or obese cats) may be unreliable.

分析证实了猫的体重与食物消耗量之间存在强正相关。较重的猫需要更多热量，这在生物学上是合理的。回归模型可以为喂食建议提供指导，但我们必须谨慎：样本量较小（n=15），且仅包含一个品种。超出数据范围的外推（例如对幼猫或极度肥胖的猫）可能不可靠。

In your exam answer, always link statistical findings back to the original context. State whether the evidence supports the hypothesis, point out limitations and suggest improvements such as a larger or more varied sample.

在考试作答中，务必将统计发现与原始情境联系起来。说明证据是否支持假设，指出局限性，并提出改进建议，如增大样本量或纳入更多样的样本。

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12. Tips for Tackling Case Study Questions | 实战演练技巧

When given a case study in IGCSE Statistics, read the scenario carefully and identify the type of data. Plan your steps: organise, represent, calculate, analyse, conclude. Use correct terminology such as ‘positive correlation’, ‘outlier’, ‘interquartile range’. Show working clearly, label diagrams and state conclusions in context.

遇到 IGCSE 统计中的案例题时，仔细阅读情境并识别数据类型。规划好步骤：整理、展示、计算、分析、结论。使用正确术语，如“正相关”、“异常值”、“四分位距”。清晰展示计算过程，给图表添加标签，并在情境中陈述结论。

Practise with real datasets, even small ones like this example. The more you link numbers to meaning, the more confident you will be. Always ask: what story does the data tell?

用真实数据集进行练习，即使是像这样的小样本。你将数字与含义联系得越多，就会越自信。始终要问：数据在讲述什么故事？

Published by TutorHao | Statistics Revision Series | aleveler.com

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