📚 A-Level Further Mathematics: Hypothesis Testing for the Geometric Parameter p | 进阶数学：几何分布参数p的假设检验
In A-Level Further Mathematics (Edexcel), hypothesis testing for the geometric distribution parameter p is a core topic in statistical inference. Unlike binomial or Poisson tests, the geometric distribution models the number of trials needed to achieve the first success, and its parameter p represents the probability of success on each trial. This article provides a comprehensive, exam-focused guide to conducting one-tailed and two-tailed hypothesis tests for p, including test statistics, critical regions, and decision rules using both calculator and manual methods.
在Edexcel进阶数学（A-Level Further Mathematics）中，几何分布参数p的假设检验是统计推断部分的核心内容。与二项分布或泊松分布检验不同，几何分布用于描述首次成功所需的试验次数，参数p表示每次试验成功的概率。本文将系统讲解如何对p进行单尾和双尾假设检验，包括检验统计量、临界区域、决策规则，以及使用计算器与手算两种方法，帮助考生从容应对考试。
1. The Geometric Distribution and Its Parameter p | 几何分布及其参数p
A discrete random variable X follows a geometric distribution with parameter p if X counts the number of trials required to obtain the first success. The probability mass function is given by P(X = x) = p(1 − p)ˣ⁻¹ for x = 1, 2, 3, … . The parameter p lies strictly between 0 and 1, and the mean of the distribution is E(X) = 1/p, while the variance is Var(X) = (1 − p)/p².
若离散随机变量X服从参数为p的几何分布，则X表示获得首次成功所需的试验次数。其概率质量函数为P(X = x) = p(1 − p)ˣ⁻¹，其中x = 1, 2, 3, …。参数p严格位于0与1之间，分布的期望为E(X) = 1/p，方差为Var(X) = (1 − p)/p²。
The geometric distribution is memoryless: P(X > a + b | X > a) = P(X > b). This property implies that the probability of waiting additional trials does not depend on how many trials have already occurred. In hypothesis testing, we use the tail probabilities P(X ≥ c) or P(X ≤ c) to determine whether an observed value is extreme under the null hypothesis.
几何分布具有无记忆性：P(X > a + b | X > a) = P(X > b)。这意味着在已经等待了a次之后，还需要等待b次的概率与之前已进行的试验次数无关。在假设检验中，我们使用尾部概率P(X ≥ c)或P(X ≤ c)来判断在零假设下观测值是否极端。
2. Formulating Hypotheses for p | 建立关于p的假设
In hypothesis testing for the geometric parameter p, the null hypothesis is always a specific numerical claim, typically written as H₀: p = p₀. The alternative hypothesis depends on the context of the problem and may be one-tailed or two-tailed. A one-tailed test uses H₁: p < p₀ or H₁: p > p₀, while a two-tailed test uses H₁: p ≠ p₀.
在几何分布参数p的假设检验中，零假设通常是对参数取值的具体主张，一般写为H₀: p = p₀。备择假设取决于问题的实际背景，可以是单尾或双尾检验。单尾检验使用H₁: p < p₀ 或 H₁: p > p₀，而双尾检验使用H₁: p ≠ p₀。
To choose the correct alternative hypothesis, examine the wording of the question. If a researcher suspects that the probability of success has increased, use H₁: p > p₀. If the suspicion is that the probability has decreased, use H₁: p < p₀. If the claim merely states that the probability has changed, without direction, use H₁: p ≠ p₀.
选择正确的备择假设时，需仔细审题。若研究者怀疑成功概率增大，则使用H₁: p > p₀；若怀疑成功概率减小，则使用H₁: p < p₀；若只是声称概率发生了改变，而未指明方向，则使用H₁: p ≠ p₀。
For geometric distributions, a smaller p means a larger expected waiting time for the first success, and therefore observed values of X tend to be larger. Conversely, a larger p means success is more likely on each trial, so observed values of X tend to be smaller. This relationship is essential for determining the direction of the critical region.
对于几何分布而言，p越小意味着首次成功所需的期望等待时间越长，因此观测到的X值往往偏大；反之，p越大意味着每次试验成功的可能性更高，因此观测到的X值往往偏小。理解这一关系对于确定临界区域的方向至关重要。
3. Test Statistic and Distributional Assumptions | 检验统计量与分布假设
When testing H₀: p = p₀, the test statistic is the observed value of X itself, where X ~ Geo(p₀) under the null hypothesis. We do not standardise X into a z-score or t-statistic; instead, we work directly with the geometric probabilities. This is a fundamental difference from tests involving the normal distribution.
在检验H₀: p = p₀时，检验统计量就是X的观测值本身，且在零假设下X ~ Geo(p₀)。我们不需要将X标准化为z分数或t统计量，而是直接使用几何分布的概率。这是与正态分布检验的根本区别。
The cumulative distribution function of the geometric distribution has a convenient closed form. For any positive integer x, P(X ≤ x) = 1 − (1 − p)ˣ, and consequently P(X > x) = (1 − p)ˣ. These formulas are invaluable for computing tail probabilities quickly without summing infinite series.
几何分布的累积分布函数具有简洁的闭式表达式。对于任意正整数x，P(X ≤ x) = 1 − (1 − p)ˣ，因此P(X > x) = (1 − p)ˣ。利用这些公式可以快速计算尾部概率，而无需对无穷级数求和。
P(X ≤ x) = 1 − (1 − p)ˣ , P(X > x) = (1 − p)ˣ
These identities follow from the fact that X > x means the first x trials all resulted in failure. Since each trial is independent and has failure probability 1 − p, the probability of x consecutive failures is exactly (1 − p)ˣ. This insight is the key to solving most geometric hypothesis testing problems.
上述恒等式源于X > x意味着前x次试验全部失败。由于每次试验独立，且失败概率为1 − p，因此连续x次失败的概率恰好为(1 − p)ˣ。这一认识是解决大多数几何分布假设检验问题关键。
4. One-Tailed Lower-Tail Test: H₁: p < p₀ | 单尾左尾检验：H₁: p < p₀
Consider the test H₀: p = p₀ against H₁: p < p₀. Since a smaller p makes large values of X more likely, the critical region lies in the upper tail of the distribution of X. That is, we reject H₀ if X ≥ c, where c is the smallest integer such that P(X ≥ c) ≤ α under H₀.
考虑检验H₀: p = p₀ 对 H₁: p < p₀。由于p越小，X取较大值的可能性越高，因此临界区域位于X分布的右尾。也就是说，当X ≥ c时拒绝H₀，其中c是满足在H₀下P(X ≥ c) ≤ α的最小整数。
Because P(X ≥ c) = (1 − p₀)^(c−1), the critical value c can be found by solving (1 − p₀)^(c−1) ≤ α. Taking natural logarithms of both sides gives (c − 1)ln(1 − p₀) ≤ ln α. Since ln(1 − p₀) is negative, the inequality reverses when dividing, yielding c ≥ 1 + ln α / ln(1 − p₀). The critical value is therefore c = ⌈1 + ln α / ln(1 − p₀)⌉.
由于P(X ≥ c) = (1 − p₀)^(c−1)，临界值c可通过求解(1 − p₀)^(c−1) ≤ α获得。两边取自然对数得(c − 1)ln(1 − p₀) ≤ ln α。由于ln(1 − p₀)为负数，除以该数时不等式方向反转，得到c ≥ 1 + ln α / ln(1 − p₀)。因此临界值为c = ⌈1 + ln α / ln(1 − p₀)⌉。
Critical region: X ≥ c , where c = ⌈1 + ln α / ln(1 − p₀)⌉
Alternatively, and more reliably in an exam, use the cumulative probability tables or a calculator’s inverse geometric distribution function. The Edexcel formula booklet provides geometric tail probability tables for selected values of p, and your calculator may directly compute the smallest x such that P(X ≥ x) ≤ α.
另一种更可靠的考试方法是使用累积概率表或计算器的逆几何分布函数。Edexcel公式手册针对若干常用p值提供了几何分布尾部概率表，而计算器可以直接计算满足P(X ≥ x) ≤ α的最小x值。
5. One-Tailed Upper-Tail Test: H₁: p > p₀ | 单尾右尾检验：H₁: p > p₀
Now consider H₀: p = p₀ against H₁: p > p₀. A larger p makes small values of X more likely, so the critical region lies in the lower tail of the distribution of X. We reject H₀ if X ≤ c, where c is the largest integer such that P(X ≤ c) ≤ α under H₀.
现在考虑H₀: p = p₀ 对 H₁: p > p₀。p越大，X取较小值的可能性越高，因此临界区域位于X分布的左尾。当X ≤ c时拒绝H₀，其中c是满足在H₀下P(X ≤ c) ≤ α的最大整数。
Using the cumulative formula, P(X ≤ c) = 1 − (1 − p₀)ᶜ. Thus we require 1 − (1 − p₀)ᶜ ≤ α, which rearranges to (1 − p₀)ᶜ ≥ 1 − α. Taking logarithms, c ln(1 − p₀) ≥ ln(1 − α). Because ln(1 − p₀) is negative, the inequality reverses, giving c ≤ ln(1 − α) / ln(1 − p₀). Therefore the critical value is c = ⌊ln(1 − α) / ln(1 − p₀)⌋.
利用累积分布公式，P(X ≤ c) = 1 − (1 − p₀)ᶜ。因此要求1 − (1 − p₀)ᶜ ≤ α，整理得(1 − p₀)ᶜ ≥ 1 − α。两边取对数得c ln(1 − p₀) ≥ ln(1 − α)。由于ln(1 − p₀)为负，不等式反转，得到c ≤ ln(1 − α) / ln(1 − p₀)。因此临界值为c = ⌊ln(1 − α) / ln(1 − p₀)⌋。
Critical region: X ≤ c , where c = ⌊ln(1 − α) / ln(1 − p₀)⌋
It is crucial to remember the direction: when testing p > p₀, small values of X provide evidence against the null hypothesis. A common mistake is placing the critical region on the wrong side. Always reason from the relationship between p and the expected number of trials before deciding the tail.
务必牢记方向：当检验p > p₀时，X取较小值为拒绝零假设提供证据。常见错误是将临界区域放在错误的一侧。在确定尾部之前，务必从p与期望试验次数之间的关系出发进行推理。
6. Two-Tailed Test: H₁: p ≠ p₀ | 双尾检验：H₁: p ≠ p₀
For a two-tailed test of H₀: p = p₀ against H₁: p ≠ p₀ at significance level α, we split the significance level equally between the two tails, each tail having probability α/2. The critical region consists of two disjoint sets: X ≤ c₁ for small values and X ≥ c₂ for large values.
对于H₀: p = p₀ 对 H₁: p ≠ p₀的双尾检验，在显著性水平α下，将显著性水平均分为两部分，每个尾部各占α/2。临界区域由两个不相交的集合构成：X ≤ c₁针对较小值，X ≥ c₂针对较大值。
The lower critical value c₁ is the largest integer such that P(X ≤ c₁) ≤ α/2. Using the cumulative formula, 1 − (1 − p₀)^(c₁) ≤ α/2, so (1 − p₀)^(c₁) ≥ 1 − α/2, giving c₁ = ⌊ln(1 − α/2) / ln(1 − p₀)⌋. The upper critical value c₂ is the smallest integer such that P(X ≥ c₂) ≤ α/2, namely (1 − p₀)^(c₂−1) ≤ α/2, yielding c₂ = ⌈1 + ln(α/2) / ln(1 − p₀)⌉.
下临界值c₁是满足P(X ≤ c₁) ≤ α/2的最大整数。由累积公式1 − (1 − p₀)^(c₁) ≤ α/2，得(1 − p₀)^(c₁) ≥ 1 − α/2，因此c₁ = ⌊ln(1 − α/2) / ln(1 − p₀)⌋。上临界值c₂是满足P(X ≥ c₂) ≤ α/2的最小整数，即(1 − p₀)^(c₂−1) ≤ α/2，因此c₂ = ⌈1 + ln(α/2) / ln(1 − p₀)⌉。
Reject H₀ if X ≤ c₁ or X ≥ c₂
In practice, two-tailed tests for the geometric distribution appear less frequently in Edexcel exams than one-tailed tests, but they are entirely within the specification. When using a calculator, be aware that some calculators only provide one-tailed inverse probabilities; you must manually adjust by using α/2 in each tail.
实际上，在Edexcel考试中，几何分布的双尾检验出现频率低于单尾检验，但它完全在大纲范围内。使用计算器时需注意，部分计算器只提供单尾逆概率功能，此时必须手动将α/2分别应用于两个尾部。
7. Worked Example 1: p < p₀ | 例题1：p < p₀
A manufacturer claims that the probability of a defective item is p = 0.2. An inspector believes the defect probability has decreased. She tests items until finding the first defective item and observes that it takes 10 trials. Test at the 5% significance level whether the defect probability has decreased.
某制造商声称产品缺陷概率为p = 0.2。检验员认为缺陷概率已经下降。她逐一测试产品直到发现第一个缺陷品，结果第10次试验才出现缺陷。在5%显著性水平下检验缺陷概率是否确实下降。
Step 1 — Hypotheses: H₀: p = 0.2, H₁: p < 0.2. Since p is suspected to be smaller, larger values of X provide evidence against H₀. This is a one-tailed upper-tail test with α = 0.05.
第一步 —— 建立假设： H₀: p = 0.2，H₁: p < 0.2。由于怀疑p变小，X取较大值为拒绝H₀提供证据。这是一个显著性水平α = 0.05的单尾右尾检验。
Step 2 — Find the critical region: We need the smallest c such that P(X ≥ c) ≤ 0.05 under X ~ Geo(0.2). Compute P(X ≥ c) = (0.8)^(c−1). For c = 14: (0.8)¹³ = 0.05498. For c = 15: (0.8)¹⁴ = 0.04398. Thus the critical region is X ≥ 15.
第二步 —— 求临界区域： 需要找到满足P(X ≥ c) ≤ 0.05的最小c，其中X ~ Geo(0.2)。计算P(X ≥ c) = (0.8)^(c−1)。当c = 14时，(0.8)¹³ = 0.05498；当c = 15时，(0.8)¹⁴ = 0.04398。因此临界区域为X ≥ 15。
Step 3 — Compare and conclude: The observed value is X = 10, which does not lie in the critical region. Therefore we do not reject H₀. There is insufficient evidence at the 5% level to conclude that the defect probability has decreased.
第三步 —— 比较并作出结论： 观测值为X = 10，不在临界区域内。因此不能拒绝H₀。在5%显著性水平下没有充分证据表明缺陷概率已经下降。
Step 4 — Interpret in context: The observed waiting time of 10 trials is not unusually long under the assumption p = 0.2. The probability of waiting at least 10 trials, given p = 0.2, is (0.8)⁹ = 0.134, which is well above 0.05, so the result is not statistically significant.
第四步 —— 结合背景解释： 在p = 0.2的假设下，等待10次试验并不异常。给定p = 0.2时，至少等待10次的概率为(0.8)⁹ = 0.134，远高于0.05，因此结果不具有统计显著性。
8. Worked Example 2: p > p₀ | 例题2：p > p₀
In a lottery game, the organiser claims that the probability of winning on each ticket is p = 0.1. A player suspects that the probability of winning is higher than claimed. He buys tickets one by one until he wins, and his first win occurs on the 2nd ticket. Test at the 1% significance level whether the win probability is greater than 0.1.
在某彩票游戏中，组织者声称每张彩票的中奖概率为p = 0.1。一位玩家怀疑中奖概率高于所声称的值。他一张一张购票直到中奖，首次中奖出现在第2张票。在1%显著性水平下检验中奖概率是否大于0.1。
Step 1 — Hypotheses: H₀: p = 0.1, H₁: p > 0.1. Since p is suspected to be larger, smaller values of X provide evidence against H₀. This is a one-tailed lower-tail test with α = 0.01.
第一步 —— 建立假设： H₀: p = 0.1，H₁: p > 0.1。由于怀疑p更大，X取较小值为拒绝H₀提供证据。这是一个显著性水平α = 0.01的单尾左尾检验。
Step 2 — Find the critical region: We need the largest c such that P(X ≤ c) ≤ 0.01 under X ~ Geo(0.1). Compute P(X ≤ c) = 1 − (0.9)ᶜ. For c = 1: 1 − 0.9 = 0.10. This already exceeds 0.01. Hence the critical region is empty at the 1% level.
第二步 —— 求临界区域： 需要找到满足P(X ≤ c) ≤ 0.01的最大c，其中X ~ Geo(0.1)。计算P(X ≤ c) = 1 − (0.9)ᶜ。当c = 1时，1 − 0.9 = 0.10，已经超过0.01。因此在1%水平下临界区域为空集。
Step 3 — Compare and conclude: Since the critical region is empty, it is impossible to reject H₀ at the 1% significance level, regardless of the observed value. The observed X = 2 is certainly not in the critical region. We do not reject H₀. There is no evidence that the win probability exceeds 0.1.
第三步 —— 比较并作出结论： 由于临界区域为空，在1%显著性水平下无论观测值是多少都不可能拒绝H₀。观测值X = 2当然不在临界区域内。不能拒绝H₀，没有证据表明中奖概率超过0.1。
Step 4 — Interpretation: Even the smallest possible value X = 1 has a probability of 0.10 under H₀, which is greater than the significance level 0.01. This means that no observation can provide sufficiently strong evidence against H₀ at the 1% level. This illustrates why very small significance levels may lead to non-rejection for geometric tests with small p₀.
第四步 —— 解释： 即使在H₀下可能的最小值X = 1的概率也有0.10，大于显著性水平0.01。这意味着任何观测值都无法在1%水平下提供足够强的拒绝H₀的证据。这说明了当p₀较小时，很小的显著性水平可能导致几何检验无法拒绝零假设。
9. Worked Example 3: Two-Tailed Test | 例题3：双尾检验
A biologist claims that the probability of a certain bird species nesting in a particular tree is p = 0.4. A researcher suspects that this probability has changed. She observes the number of trees examined until finding the first nesting pair and obtains X = 6. Test at the 5% significance level whether p differs from 0.4.
一位生物学家声称某种鸟类在特定树上筑巢的概率为p = 0.4。一位研究员怀疑该概率已发生变化。她观察直到发现第一对筑巢鸟为止所检查的树数，得到X = 6。在5%显著性水平下检验p是否不等于0.4。
Step 1 — Hypotheses: H₀: p = 0.4, H₁: p ≠ 0.4. Significance level α = 0.05, so each tail has α/2 = 0.025.
第一步 —— 建立假设： H₀: p = 0.4，H₁: p ≠ 0.4。显著性水平α = 0.05，因此每个尾部各占α/2 = 0.025。
Step 2 — Lower critical value: Find the largest c₁ such that P(X ≤ c₁) ≤ 0.025. Since P(X ≤ c₁) = 1 − (0.6)^(c₁), compute: for c₁ = 1, 1 − 0.6 = 0.400; for c₁ = 2, 1 − 0.36 = 0.640. Both exceed 0.025. Thus no positive integer c₁ satisfies the condition; the lower tail is empty.
第二步 —— 下临界值： 找到满足P(X ≤ c₁) ≤ 0.025的最大c₁。由于P(X ≤ c₁) = 1 − (0.6)^(c₁)，计算得：c₁ = 1时，1 − 0.6 = 0.400；c₁ = 2时，1 − 0.36 = 0.640。两者均超过0.025。因此不存在满足条件的正整数c₁，左尾为空。
Step 3 — Upper critical value: Find the smallest c₂ such that P(X ≥ c₂) ≤ 0.025. Since P(X ≥ c₂) = (0.6)^(c₂−1), compute: for c₂ = 10, (0.6)⁹ = 0.01008; for c₂ = 9, (0.6)⁸ = 0.01680; for c₂ = 8, (0.6)⁷ = 0.02799. Thus the upper critical value is c₂ = 9, giving the critical region X ≥ 9.
第三步 —— 上临界值： 找到满足P(X ≥ c₂) ≤ 0.025的最小c₂。由于P(X ≥ c₂) = (0.6)^(c₂−1)，计算得：c₂ = 10时，(0.6)⁹ = 0.01008；c₂ = 9时，(0.6)⁸ = 0.01680；c₂ = 8时，(0.6)⁷ = 0.02799。因此上临界值为c₂ = 9，临界区域为X ≥ 9。
Step 4 — Conclusion: The observed value X = 6 does not fall in the critical region. We do not reject H₀. There is insufficient evidence at the 5% level to conclude that the nesting probability has changed from 0.4.
第四步 —— 结论： 观测值X = 6不在临界区域内，不能拒绝H₀。在5%显著性水平下没有充分证据表明筑巢概率已从0.4发生变化。
10. Using p-values in Geometric Tests | 几何检验中的p值方法
Instead of finding a critical region, many exam questions require you to calculate a p-value, which is the probability of observing a test statistic at least as extreme as the one obtained, assuming H₀ is true. For a one-tailed test H₁: p < p₀ with observed value x, the p-value is P(X ≥ x) = (1 − p₀)^(x−1). For H₁: p > p₀, the p-value is P(X ≤ x) = 1 − (1 − p₀)ˣ.
除了求临界区域，许多考题还要求计算p值。p值是在H₀为真的前提下，观测到至少与当前检验统计量一样极端的结果的概率。对于单尾检验H₁: p < p₀，观测值为x时，p值等于P(X ≥ x) = (1 − p₀)^(x−1)。对于H₁: p > p₀，p值等于P(X ≤ x) = 1 − (1 − p₀)ˣ。
For a two-tailed test H₁: p ≠ p₀, the p-value is computed by doubling the smaller of the two one-tailed p-values. Specifically, define pₗₒₐᵢ = P(X ≥ x) = (1 − p₀)^(x−1) and pᵤₚₚₑᵣ = P(X ≤ x) = 1 − (1 − p₀)ˣ. The two-tailed p-value is 2 × min(pₗₒₐᵢ, pᵤₚₚₑᵣ).
对于双尾检验H₁: p ≠ p₀，p值等于将两个单尾p值中较小者乘以2。具体地，令pₗₒₐᵢ = P(X ≥ x) = (1 − p₀)^(x−1)，pᵤₚₚₑᵣ = P(X ≤ x) = 1 − (1 − p₀)ˣ。双尾p值为2 × min(pₗₒₐᵢ, pᵤₚₚₑᵣ)。
The decision rule is straightforward: if the p-value is less than or equal to the significance level α, reject H₀; otherwise, do not reject H₀. The p-value method is particularly useful when the observed value lies close to the boundary of the critical region, as it gives a more precise measure of the strength of evidence.
决策规则非常简单：若p值小于或等于显著性水平α，则拒绝H₀；否则不能拒绝H₀。当观测值接近临界区域边界时，p值方法尤其有用，因为它给出了证据强度的更精确度量。
Reject H₀ if p-value ≤ α ; otherwise do not reject H₀
Always present your p-value with at least three significant figures, and compare it explicitly to α in your written conclusion. Examiners award method marks for the correct p-value formula and accuracy marks for the final numerical comparison.
书写p值时至少保留三位有效数字，并在结论中明确将其与α进行比较。考官会对正确的p值公式给出方法分，对最终数值比较给出准确性分。
11. Common Pitfalls and Exam Tips | 常见错误与考试技巧
One of the most frequent errors is confusing the direction of the critical region. Remember: p < p₀ leads to a right-tail test (reject for large X), while p > p₀ leads to a left-tail test (reject for small X). Another common mistake is using P(X ≥ x) instead of P(X > x) or P(X ≤ x) instead of P(X < x); always verify whether the boundary value is included.
最常见的错误之一是混淆临界区域的方向。记住：p < p₀对应右尾检验（X大时拒绝），而p > p₀对应左尾检验（X小时拒绝）。另一个常见错误是混淆P(X ≥ x)与P(X > x)、P(X ≤ x)与P(X < x)；务必确认边界值是否包含在内。
When using the formula P(X ≥ c) = (1 − p)^(c−1), note that the exponent is c − 1, not c. This is because X ≥ c means the first c − 1 trials are all failures, and the c-th trial may be either a success or a failure. In contrast, P(X ≤ c) = 1 − (1 − p)ᶜ uses exponent c directly.
使用公式P(X ≥ c) = (1 − p)^(c−1)时，注意指数是c − 1而不是c。这是因为X ≥ c意味着前c − 1次试验全部失败，而第c次试验可能成功也可能失败。相比之下，P(X ≤ c) = 1 − (1 − p)ᶜ直接使用指数c。
Another pitfall is assuming the geometric distribution counts failures before the first success rather than trials until the first success. The Edexcel specification uses the latter convention: X ~ Geo(p) means X is the number of trials up to and including the first success. Always state your convention clearly in your working.
另一个陷阱是混淆几何分布是“首次成功前的失败次数”还是“直到首次成功的试验次数”。Edexcel大纲采用后一种约定：X ~ Geo(p)表示X是直到并包括首次成功为止的试验次数。在解题中务必明确你的约定。
Finally, be careful with calculator settings. Some calculators require you to specify whether you want P(X ≤ x) or P(X ≥ x). If the calculator returns the lower tail cumulative probability by default, you need to compute 1 − P(X ≤ x − 1) to obtain P(X ≥ x) for the geometric distribution.
最后，注意计算器的设置。部分计算器默认返回P(X ≤ x)的累积概率，此时需要计算1 − P(X ≤ x − 1)才能得到几何分布的P(X ≥ x)。
12. Summary and Revision Checklist | 总结与复习清单
Hypothesis testing for the geometric parameter p is a systematic process: formulate hypotheses, determine the direction of the test, compute the critical value or p-value using geometric tail probabilities, compare with the observed value, and draw a conclusion in context. The key formulas are P(X ≤ x) = 1 − (1 − p)ˣ and P(X ≥ x) = (1 − p)^(x−1).
几何分布参数p的假设检验是一个系统性过程：建立假设、确定检验方向、利用几何尾部概率计算临界值或p值、与观测值比较，并结合背景作出结论。核心公式为P(X ≤ x) = 1 − (1 − p)ˣ和P(X ≥ x) = (1 − p)^(x−1)。
| Test | 检验类型 | Alternative hypothesis | 备择假设 | Critical region | 临界区域 | p-value | p值 |
|---|---|---|---|
| One-tailed lower | 单尾左尾 | H₁: p < p₀ | X ≥ c (right tail) | (1 − p₀)^(x−1) |
| One-tailed upper | 单尾右尾 | H₁: p > p₀ | X ≤ c (left tail) | 1 − (1 − p₀)ˣ |
| Two-tailed | 双尾 | H₁: p ≠ p₀ | X ≤ c₁ or X ≥ c₂ | Published by TutorHao | A-Level 进阶数学 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信） CommentsMore posts |
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