Year 12 OCR Statistics: Case Study in Action | OCR Year 12 统计：案例分析实战演练

📚 Year 12 OCR Statistics: Case Study in Action | OCR Year 12 统计：案例分析实战演练

Welcome to a full walkthrough of a statistical investigation designed for Year 12 OCR Statistics students. In this case study, we examine a basketball player’s claim about his free-throw shooting accuracy. We will collect data, summarise it with charts, model the number of successes with a binomial distribution, and perform a hypothesis test to see whether the data cast doubt on the player’s stated success rate. Every step mirrors the content you meet in the OCR AS Statistics syllabus, helping you see how classroom theory turns into real-world practice.

欢迎来到专为 OCR Year 12 统计学生设计的完整统计分析演练。本案例中，我们将考察一名篮球运动员对其罚球命中率的声称。我们将收集数据、用图表加以概括、用二项分布对成功次数建模，并进行假设检验，以判断数据是否对球员声称的成功率产生质疑。每一步都对应 OCR AS 统计课程的内容，帮助你直观理解课堂理论如何转为实际应用。

1. Case Introduction: A Claim about Free-Throw Accuracy | 案例介绍：有关罚球命中率的声称

A professional basketball player states that his free-throw success rate is 80%. In other words, he claims that on any given free-throw attempt, the probability of scoring is p = 0.8. We want to investigate whether there is statistical evidence that his true success rate is actually lower than 80%. To do this, we randomly select 20 of his free-throw attempts from recent matches and count how many are successful.

一位职业篮球运动员称其罚球命中率为 80%。也就是说，他声称在任意一次罚球中，得分的概率为 p = 0.8。我们希望调查是否有统计证据表明其真实命中率实际上低于 80%。为此，我们从其近期比赛中随机选取 20 次罚球，统计其中成功的次数。

2. Sampling and Data Collection: Ensuring a Fair Test | 抽样与数据收集：确保检验公正

We decided to use a simple random sample of 20 free-throw attempts. The team video analyst assigned a number to every free-throw taken by the player in the last three months and used a random number generator to pick 20 distinct attempts. This method avoids selection bias; attempts from high-pressure moments and routine situations are equally likely to be chosen. It also approximates the independence condition needed for the binomial model, because the outcomes of widely separated attempts are less likely to influence each other.

我们决定采用简单随机抽样，从该球员过去三个月内的所有罚球中抽取 20 次。球队视频分析师为每一次罚球编号，利用随机数生成器选出 20 次互不相同的尝试。这一方法避免了选择偏倚；高压时刻的罚球与一般情况下的罚球被选中的概率相同。同时，它也为二项模型所需的独立性条件提供了近似保证，因为间隔较远的罚球结果不太可能相互影响。

3. Representing the Data: Frequency Table and Bar Chart | 数据表示：频数表与柱状图

After watching the 20 selected attempts, we recorded the outcome of each as either ‘success’ or ‘failure’. The raw data are summarised in a frequency table and a bar chart.

观看所选 20 次罚球后，我们将每次结果记录为“成功”或“失败”。原始数据通过频数表和柱状图进行总结。

A bar chart (not shown here) would have two bars of height 13 and 7, instantly showing that the sample contains more successes than failures, but with a success proportion of 13/20 = 0.65. This is noticeably below the claimed 0.80.

一幅柱状图（此处未显示）将有高度为 13 和 7 的两根柱，瞬间展现出样本中成功多于失败，但成功比例为 13/20 = 0.65，明显低于声称的 0.80。

4. Summary Statistics: Sample Proportion and Variability | 摘要统计量：样本比例与变异性

Let p̂ = 13/20 = 0.65 denote the sample proportion of successes. This is a point estimate of the player’s true success probability. Although a single number gives a snapshot, we also want a sense of how much it might vary from sample to sample. Under the assumed model with p = 0.8, the standard deviation of the number of successes, X, would be √(np(1-p)) = √(20 × 0.8 × 0.2) = √3.2 ≈ 1.7889, so the standard deviation of the sample proportion is 1.7889/20 = 0.0894.

令 p̂ = 13/20 = 0.65 表示样本成功比例。它是球员真实成功概率的点估计。尽管单个样本比例提供了一个快照，我们也希望了解不同样本之间可能的波动程度。在假定 p = 0.8 的模型下，成功次数 X 的标准差为 √(np(1-p)) = √(20 × 0.8 × 0.2) = √3.2 ≈ 1.7889，因此样本比例的标准差为 1.7889/20 = 0.0894。

5. The Binomial Distribution as a Model | 二项分布模型

The number of successful free-throws in a fixed number of independent attempts with the same probability of success can be modelled by a binomial distribution. Let X be the random variable ‘number of successes in 20 attempts’. If the player’s claim is true, then X ~ B(20, 0.8). This means the probability that X equals a specific value x is given by:

在相同成功概率的独立尝试中，固定次数的成功次数可以用二项分布建模。设随机变量 X 为“20 次尝试中的成功次数”。如果球员的声称属实，则 X ~ B(20, 0.8)。这意味着 X 取某个特定值 x 的概率由下式给出：

P(X = x) = C(20, x) × 0.8ˣ × 0.2²⁰⁻ˣ

Here C(20, x

Published by TutorHao | Year 12 统计 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）