📚 Cross-Disciplinary Integrated Question Training for A-Level Edexcel Statistics | A-Level Edexcel 统计：跨学科综合题型训练
Mastering Statistics at A-Level goes beyond textbook exercises. The Edexcel specification increasingly embeds statistical methods within real-world contexts drawn from biology, economics, psychology, and environmental science. This article provides structured, exam-style practice with cross-disciplinary integrated questions, sharpening your ability to identify appropriate statistical tests, interpret outputs, and communicate findings effectively. Each section pairs an English explanation with a Chinese translation, followed by a worked example that bridges theory and application.
掌握A-Level统计绝不仅仅是完成课本练习。Edexcel考试大纲越来越多地将统计方法嵌入到生物学、经济学、心理学和环境科学等真实世界情境中。本文提供结构化的跨学科综合题型训练，帮助你提高识别合适统计检验、解读结果并有效传达发现的能力。每个小节均以英文说明搭配中文翻译，并通过一个跨越理论与实践的例题进行解析。
1. Statistical Modelling in Biology: Hardy–Weinberg Equilibrium | 生物学中的统计建模：哈代-温伯格平衡
The Hardy–Weinberg principle predicts allele frequencies in a non-evolving population. In an A-Level Statistics question, you may be given observed genotype counts and asked to test whether the population is in equilibrium using a chi-squared goodness-of-fit test. The expected frequencies are derived from p², 2pq, and q² under the assumption of random mating. A significant result suggests evolutionary forces at work.
哈代-温伯格原理可预测非进化群体中的等位基因频率。在A-Level统计题中，你可能会得到观察到的基因型计数，并被要求用卡方拟合优度检验来判断群体是否处于平衡状态。期望频率是根据随机交配假设下的p²、2pq和q²公式推导出来的。显著的结果表明进化力量在起作用。
Example: In a sample of 500 individuals, genotype counts are: AA = 245, Aa = 210, aa = 45. Test at the 5% significance level whether the population is in Hardy–Weinberg equilibrium.
例题：在500个个体样本中，基因型计数为：AA = 245，Aa = 210，aa = 45。在5%显著性水平下检验该群体是否符合哈代-温伯格平衡。
Step 1: Estimate allele frequencies. Let p = frequency of A. p = (2×245 + 210) / (2×500) = 700/1000 = 0.7, so q = 0.3. Expected: AA = 500×0.49 = 245, Aa = 500×0.42 = 210, aa = 500×0.09 = 45. Here observed exactly equals expected, so chi-squared = 0. With 2 df, critical value at 5% is 5.991. Since 0 < 5.991, we do not reject H₀. The population appears to be in equilibrium.
步骤1：估计等位基因频率。设p = A的频率。p = (2×245 + 210) / (2×500) = 700/1000 = 0.7，则q = 0.3。期望值：AA = 500×0.49 = 245，Aa = 500×0.42 = 210，aa = 500×0.09 = 45。此处观察值恰等于期望值，因此卡方值为0。自由度为2，5%临界值为5.991。因0 < 5.991，不拒绝零假设。该群体似乎处于平衡状态。
2. Correlation and Regression in Economics: Demand Elasticity | 经济学中的相关与回归：需求弹性
In economics, the relationship between price and quantity demanded is often modelled using linear regression. The product-moment correlation coefficient (PMCC) quantifies the strength and direction of the linear relationship. A hypothesis test for the correlation coefficient can determine whether the observed negative correlation is statistically significant. Furthermore, the regression equation can predict demand at a given price, and residuals can highlight outliers.
在经济学中，价格与需求量之间的关系常用线性回归建模。积矩相关系数(PMCC)量化了线性关系的强度和方向。对相关系数进行假设检验可以判断观察到的负相关是否具有统计显著性。此外，回归方程可以预测给定价格下的需求量，残差则能揭示异常值。
Example: Data from a bakery: price (£, x): 1.0, 1.5, 2.0, 2.5; demand (units, y): 200, 150, 120, 90. Test H₀: ρ = 0 vs H₁: ρ < 0 at α = 0.05. Calculate the regression line and predict demand when price = £1.80.
例题：一家面包店的数据：价格（英镑，x）：1.0, 1.5, 2.0, 2.5；需求量（个，y）：200, 150, 120, 90。检验H₀: ρ = 0 vs H₁: ρ < 0，显著性水平α = 0.05。计算回归线并预测价格为1.80英镑时的需求量。
Compute PMCC: n=4, Σx=7, Σy=560, Σx²=13.5, Σy²=85400, Σxy=875. r = [4×875 – 7×560] / √[(4×13.5 – 7²)(4×85400 – 560²)] = (3500 – 3920) / √[(54 – 49)(341600 – 313600)] = -420 / √[5×28000] = -420 / √140000 ≈ -420 / 374.17 ≈ -1.122 impossible due to rounding. Recalculate precisely: Σxy = 1×200 + 1.5×150 + 2×120 + 2.5×90 = 200+225+240+225 = 890. Correct r = (4×890 – 7×560) / √[(54-49)(4×85400 – 560²)] = (3560 – 3920) / √(5×28000) = -360 / √140000 = -360 / 374.1657 ≈ -0.962. Critical value for n=4, one-tailed 5% is 0.9000 (from table). Since |r| > 0.9, reject H₀. Significant negative correlation. Regression line: b = Sxy/Sxx = (-360/5) / (5/4?) Let’s compute: Sxx = Σx² – (Σx)²/n = 13.5 – 49/4 = 13.5 – 12.25 = 1.25; Sxy = Σxy – (Σx Σy)/n = 890 – (7×560)/4 = 890 – 980 = -90. So b = -90/1.25 = -72; a = ȳ – b x̄ = 140 – (-72)×1.75 = 140 + 126 = 266. Equation: y = 266 – 72x. At x=1.8, demand = 266 – 129.6 = 136.4 ≈ 136 units.
计算PMCC：n=4, Σx=7, Σy=560, Σx²=13.5, Σy²=85400, Σxy正确值为890（1×200=200, 1.5×150=225, 2×120=240, 2.5×90=225）。r = (4×890 – 7×560) / √[(4×13.5 – 49)(4×85400 – 313600)] = (3560 – 3920) / √[(54-49)×28000] = -360 / √140000 ≈ -0.962。n=4时单尾5%临界值为0.9000，|r| > 0.9，拒绝H₀，存在显著负相关。回归线：Sxx=1.25, Sxy=-90, b=-72, a=266。方程y=266-72x。当价格=1.80时，需求量=266-129.6≈136个。
3. Normal Distribution in Psychology: IQ Scores and Percentiles | 心理学中的正态分布：智商分数与百分位数
IQ scores are typically modelled by a normal distribution with mean 100 and standard deviation 15. In psychological research, you may need to find the proportion of individuals scoring above a cutoff, or the score corresponding to a given percentile. The inverse normal function is often used to determine eligibility thresholds for gifted programmes.
智商分数通常用均值为100、标准差为15的正态分布建模。在心理学研究中，你可能需要找出得分高于某个临界值的人口比例，或者对应给定百分位数的分数。逆正态函数常被用来确定天才计划的入选门槛。
Example: A gifted programme requires an IQ in the top 2%. Find the minimum IQ required. Also, what proportion of the population has an IQ between 85 and 115?
例题：一个天才计划要求智商处于前2%。求所需的最低智商分数。另外，智商在85到115之间的人口比例是多少？
Let X ~ N(100, 15²). Top 2% means P(X > k) = 0.02. Using inverse normal: z-value for 0.98 is approximately 2.0537. k = 100 + 2.0537×15 ≈ 100 + 30.8055 = 130.8. So minimum IQ ≈ 131. For range: P(85 < X < 115) = P(-1 < Z < 1) ≈ 0.6826. About 68.3% of the population.
设X ~ N(100, 15²)。前2%意味着P(X > k) = 0.02。使用逆正态：对应0.98的z值约为2.0537。k = 100 + 2.0537×15 ≈ 130.81。最低智商约131。范围：P(85 < X < 115) = P(-1 < Z < 1) ≈ 0.6826，约68.3%的人口。
4. Discrete Random Variables in Environmental Science: Oil Spill Clean-up | 环境科学中的离散随机变量：溢油清理
Environmental incidents, such as the number of oil spills reaching a shoreline per year, can be modelled with discrete probability distributions. The Poisson distribution is appropriate for rare events occurring independently in time or space. A question may ask you to calculate probabilities of multiple incidents or test whether observed frequencies fit a Poisson model.
环境事件，如每年抵达海岸线的溢油次数，可以用离散概率分布建模。泊松分布适用于在时间或空间上独立发生的稀有事件。题目可能会要求你计算多次事件的概率，或者检验观察频率是否符合泊松模型。
Example: The number of oil spills per year at a port follows a Poisson distribution with mean 2.1. Find the probability that in a given year there are exactly two spills. Also, find the probability of more than three spills in a year.
例题：某港口每年溢油次数服从均值为2.1的泊松分布。求某一年恰好发生两次溢油的概率，以及一年内发生超过三次溢油的概率。
X ~ Po(2.1). P(X=2) = e⁻²·¹ × 2.1² / 2! = 0.122456 × 4.41 / 2 ≈ 0.270. P(X > 3) = 1 – P(X ≤ 3). P(X=0)=e⁻²·¹=0.1225; P(X=1)=0.1225×2.1=0.2572; P(X=2)=0.2700; P(X=3)=0.1225×2.1³/6 = use 0.1225×9.261/6=0.1891. Sum = 0.8388. P(X>3)=0.1612.
X ~ Po(2.1)。P(X=2) = e⁻²·¹ × 2.1² / 2! ≈ 0.270。P(X>3) = 1 – P(X≤3)。计算可得P(X=0)=0.1225，P(X=1)=0.2572，P(X=2)=0.2700，P(X=3)=0.1891，合计0.8388。故P(X>3)=0.1612。
5. Sampling and Estimation in Business: Consumer Preferences | 商业中的抽样与估计：消费者偏好
Businesses often use sample surveys to estimate the proportion of customers satisfied with a new product. Confidence intervals for a population proportion are constructed using the normal approximation to the binomial distribution. The width of the interval is influenced by sample size and confidence level. Questions may involve determining the minimum sample size for a desired margin of error.
企业常通过抽样调查来估计对新品满意的顾客比例。总体比例的置信区间利用二项分布的正态近似构建。区间的宽度受样本量和置信水平的影响。题目可能会涉及在给定误差边际条件下确定最小样本量。
Example: In a random sample of 200 customers, 142 are satisfied. Construct a 95% confidence interval for the population proportion. How large a sample is needed to estimate the proportion within ±3% with 95% confidence?
例题：一项随机调查了200名顾客，142人表示满意。构建总体比例的95%置信区间。要使得估计误差在±3%以内且置信水平为95%，需要多大的样本量？
p̂ = 142/200 = 0.71. Standard error = √[0.71×0.29 / 200] ≈ 0.0321. 95% CI: 0.71 ± 1.96×0.0321 = (0.647, 0.773). For sample size: n = (z² × p̂(1-p̂)) / E². Using p̂=0.71, z=1.96, E=0.03, n = (3.8416×0.2059)/0.0009 ≈ 0.791/0.0009 ≈ 879. So at least 879 customers.
p̂ = 142/200 = 0.71。标准误 ≈ 0.0321。95% CI: 0.71 ± 1.96×0.0321 = (0.647, 0.773)。样本量：n = (1.96² × 0.71×0.29) / 0.03² = (3.8416×0.2059)/0.0009 ≈ 879。至少需要879名顾客。
6. Contingency Tables in Medicine: Treatment Efficacy | 医学中的列联表：治疗效果
Medical trials often produce categorical data, such as improved/not improved across treatment and placebo groups. A chi-squared test for independence assesses whether the treatment and outcome are associated. Degrees of freedom are (rows-1)×(columns-1). You may also be asked to calculate expected frequencies and interpret the test statistic in context.
医学试验经常产生分类数据，比如治疗组与安慰剂组中改善/未改善的例数。卡方独立性检验评估治疗与结局是否相关。自由度为(行数-1)×(列数-1)。你可能还需要计算期望频率并结合背景解释检验统计量。
Example: Trial data: Treatment group 120 improved, 30 not improved; Placebo group 80 improved, 70 not improved. Test for association at 1% significance level.
例题：试验数据：治疗组120人改善，30人未改善；安慰剂组80人改善，70人未改善。在1%显著性水平下检验关联性。
Observed table: [120, 30; 80, 70]. Totals: row1=150, row2=150, col1=200, col2=100, grand=300. Expected: E₁₁=150×200/300=100, E₁₂=50; E₂₁=100, E₂₂=50. Chi-squared = (120-100)²/100 + (30-50)²/50 + (80-100)²/100 + (70-50)²/50 = 4 + 8 + 4 + 8 = 24. df=1, critical value at 1% is 6.635. Since 24 > 6.635, reject H₀. There is significant association between treatment and improvement.
观察表：[120, 30; 80, 70]。总计：行1=150，行2=150，列1=200，列2=100，总=300。期望值：E₁₁=100，E₁₂=50；E₂₁=100，E₂₂=50。卡方 = (20²/100)+( -20²/50)+( -20²/100)+(20²/50)=4+8+4+8=24。自由度1，1%临界值6.635。24 > 6.635，拒绝H₀。治疗与改善存在显著关联。
7. Paired and Unpaired t-tests in Sports Science | 运动科学中的配对与非配对t检验
Sports science experiments often compare performance before and after a training intervention (paired t-test) or between two independent groups (unpaired t-test). The choice depends on the study design. In paired tests, you work with differences, while unpaired tests involve pooled variance or Welch’s approximation when variances are unequal.
运动科学实验经常比较训练干预前后的表现（配对t检验）或两个独立组之间的表现（非配对t检验）。选择哪种方法取决于研究设计。配对检验中，你处理的是差值；而非配对检验在方差不齐时可能使用合并方差或Welch近似。
Example: Eight athletes complete a 40 m sprint before and after a 6-week plyometric programme. Times (seconds) before: 5.1,5.3,5.0,5.2,5.4,5.1,5.3,5.2; after: 5.0,5.1,4.9,5.1,5.2,5.0,5.1,5.1. Use a 5% significance level to test if the programme improves sprint time.
例题：8名运动员在为期六周的增强式训练前后完成40米冲刺。训练前时间（秒）：5.1,5.3,5.0,5.2,5.4,5.1,5.3,5.2；训练后：5.0,5.1,4.9,5.1,5.2,5.0,5.1,5.1。以5%显著性水平检验该训练是否提高了冲刺表现。
Paired t-test appropriate. Differences (before – after): 0.1,0.2,0.1,0.1,0.2,0.1,0.2,0.1. Mean d̄ = 0.1375, sₐ = 0.0518 (sample sd of differences). SE = 0.0518/√8 ≈ 0.0183. t = 0.1375/0.0183 ≈ 7.51. df=7, one-tailed critical t at 5% is 1.895. Since 7.51 > 1.895, reject H₀. Significant improvement.
适用配对t检验。差值（前 – 后）：0.1,0.2,0.1,0.1,0.2,0.1,0.2,0.1。均值d̄=0.1375，差值样本标准差sₐ≈0.0518。标准误=0.0183。t=7.51，自由度7，单尾5%临界值1.895。7.51 > 1.895，拒绝H₀。有显著改善。
8. Probability Trees and Bayes’ Theorem in Legal Studies | 法律研究中的概率树与贝叶斯定理
Forensic evidence, such as DNA matching, is evaluated using conditional probability. Bayes’ Theorem updates the probability of guilt given a positive test result. Questions might involve a tree diagram with prior probabilities, true positive rates, and false positive rates. You need to correctly compute the posterior probability and interpret it cautiously.
法医学证据，如DNA匹配，使用条件概率评估。贝叶斯定理可根据阳性检测结果更新有罪的概率。题目可能涉及包含先验概率、真阳性率和假阳性率的树状图。你需要正确计算后验概率并进行谨慎的解释。
Example: In a town, 1 in 10,000 people is guilty. A forensic test is 99.9% accurate on guilty individuals and gives a false positive rate of 0.1% on innocents. If a random person tests positive, what is the probability they are guilty?
例题：某城镇中，每万人中有1人有罪。一项法医测试对有罪者的准确性为99.9%，对无辜者的假阳性率为0.1%。若随机一人检测呈阳性，其确有罪的概率是多少？
Let G be guilty, + be positive. P(G)=0.0001, P(+|G)=0.999, P(+|G’)=0.001. P(+) = P(+|G)P(G) + P(+|G’)P(G’) = 0.999×0.0001 + 0.001×0.9999 ≈ 0.0000999 + 0.0009999 = 0.0010998. P(G|+) = (0.999×0.0001)/0.0010998 ≈ 0.0908. So only about 9.1% chance of guilt despite positive test.
设G为有罪，+为阳性。P(G)=0.0001，P(+|G)=0.999，P(+|G’)=0.001。P(+)=0.999×0.0001 + 0.001×0.9999 ≈ 0.0010998。P(G|+) = (0.999×0.0001)/0.0010998 ≈ 0.0908。因此即使检测呈阳性，有罪的概率也只有约9.1%。
9. Time Series Analysis in Finance: Stock Price Trends | 金融中的时间序列分析：股价趋势
Financial data often exhibit trends and seasonal effects. Moving averages smooth out short-term fluctuations to reveal longer-term patterns. In Edexcel Statistics, you may be asked to calculate centred moving averages, deseasonalise data, and forecast using trend lines and seasonal components. These skills directly apply to stock prices, sales figures, or economic indicators.
金融数据往往呈现趋势和季节效应。移动平均可平滑短期波动，揭示长期规律。在Edexcel统计中，你可能需要计算中心化移动平均值、对数据进行去季节化，并运用趋势线和季节成分进行预测。这些技能直接适用于股价、销售数据或经济指标。
Example: Quarterly sales (£’000) for two years: Q1 yr1=20, Q2=30, Q3=28, Q4=40; Q1 yr2=24, Q2=36, Q3=32, Q4=46. Calculate four-point centred moving averages and the average seasonal variation for Q3.
例题：两年间季度销售额（千英镑）：第1年Q1=20，Q2=30，Q3=28，Q4=40；第2年Q1=24，Q2=36，Q3=32，Q4=46。计算四点中心化移动平均值以及Q3的平均季节变动。
Four-point MA: (20+30+28+40)/4=29.5; next (30+28+40+24)/4=30.5; (28+40+24+36)/4=32; (40+24+36+32)/4=33; (24+36+32+46)/4=34.5. Centred: first centred MA corresponds to Q3 yr1? We average first two MAs: (29.5+30.5)/2=30, placed at Q3 yr1. Next (30.5+32)/2=31.25 at Q4 yr1; (32+33)/2=32.5 at Q1 yr2; (33+34.5)/2=33.75 at Q2 yr2. For Q3, we have only one seasonal variation: 28 – 30 = -2. So average seasonal variation for Q3 is -2. (Alternatively, could involve additional years, but with two years we get only one value.)
四点移动平均：(20+30+28+40)/4=29.5；下一组(30+28+40+24)/4=30.5；第三组(28+40+24+36)/4=32；第四组(40+24+36+32)/4=33；第五组(24+36+32+46)/4=34.5。中心化：前两个MA平均为30，对应第1年Q3；接着(30.5+32)/2=31.25对应第1年Q4；(32+33)/2=32.5对应第2年Q1；(33+34.5)/2=33.75对应第2年Q2。对于Q3，我们只有一个季节变动值：28 – 30 = -2。因此Q3的平均季节变动为-2。
10. Binomial Distribution in Quality Control | 质量控制中的二项分布
Manufacturing processes often accept or reject batches based on the number of defective items in a random sample. The binomial distribution models the number of defectives when sampling with replacement or from a large population. Critical regions and significance levels determine the decision rule for accepting or rejecting a batch. This is directly linked to hypothesis testing in an industrial context.
制造过程中，常根据随机样本中的缺陷品数量决定接受或拒收批次。当有放回抽样或总体很大时，二项分布可对缺陷品数量建模。拒绝域和显著性水平决定接受或拒收批次的决策规则。这直接关联到工业背景下的假设检验。
Example: A factory claims that at most 5% of its products are defective. A random sample of 20 items is taken, and if more than 2 are defective, the batch is rejected. Find the probability of a Type I error. Also, if the true defect rate is 15%, calculate the probability of a Type II error.
例题：一家工厂宣称其产品缺陷率至多为5%。随机抽取20件产品，若超过2件有缺陷，则拒收该批次。求犯第I类错误的概率。若真实缺陷率为15%，计算犯第II类错误的概率。
H₀: p=0.05, H₁: p>0.05. Test statistic X ~ B(20,0.05). Critical region X > 2. P(Type I) = P(X > 2 | p=0.05) = 1 – P(X ≤ 2). P(X=0)=0.3585, P(X=1)=0.3774, P(X=2)=0.1887, sum=0.9246. So P(Type I)=0.0754. If p=0.15, X ~ B(20,0.15). P(Type II) = P(X ≤ 2 | p=0.15) = P(X=0)=0.0388, P(X=1)=0.1368, P(X=2)=0.2293, sum=0.4049. So Type II error probability is about 0.405.
零假设p=0.05，备择假设p>0.05。检验统计量X ~ B(20,0.05)。拒绝域X > 2。第I类错误概率 = P(X > 2 | p=0.05) = 1 – 0.9246 = 0.0754。若p=0.15，X ~ B(20,0.15)。第II类错误概率 = P(X ≤ 2 | p=0.15) = 0.0388+0.1368+0.2293=0.4049。故第II类错误概率约为0.405。
11. Non-parametric Tests in Education Research | 教育研究中的非参数检验
When data are ordinal or assumptions of normality are violated, non-parametric tests like Spearman’s rank correlation or the Wilcoxon signed-rank test are used. In education, you might compare test scores before and after an intervention when the data are not normally distributed. Spearman’s rank assesses monotonic relationships between ranks, crucial for Likert-scale survey data.
当数据为序数或正态性假设不成立时，会使用诸如斯皮尔曼秩相关或威尔科克森符号秩检验等非参数方法。在教育领域，当数据不符合正态分布时，你可以用这些方法比较干预前后的考试成绩。斯皮尔曼秩相关评估秩次之间的单调关系，对于李克特量表调查数据至关重要。
Example: Ten students’ self-rated confidence (1-10 scale) before and after a revision workshop. Use the Wilcoxon signed-rank test at 5% significance to decide if there is a significant improvement. Data: before: 4,5,3,6,4,5,4,3,5,6; after: 6,7,5,7,5,8,6,4,6,7.
例题：10名学生在复习工作坊前后对自信心的自评（1-10分制）。使用5%显著性水平的威尔科克森符号秩检验，判断是否有显著提高。数据：前：4,5,3,6,4,5,4,3,5,6；后：6,7,5,7,5,8,6,4,6,7。
Differences (after – before): 2,2,2,1,1,3,2,1,1,1. Ignore zero differences, all non-zero. Ranks of absolute differences: difference of 3 gets rank 10. Differences of 2: ranks 5,6,7,8,9 (average rank 7.5 if needed? Since values identical, we assign average rank: for five 2’s, ranks 5 to 9 sum=35, avg=7. Each gets 7). Differences of 1: four occurrences, ranks 1,2,3,4 avg=2.5. Sum of positive ranks: all differences are positive, so W⁺ = sum of all ranks = 4×2.5 + 5×7 + 10 = 10 + 35 + 10 = 55. n=10. Critical value for two-tailed at 5% is 8 (from table, n=10, two-tailed 0.05, critical value 8). Since 55 > 8? Actually Wilcoxon signed-rank test: reject if T (smaller sum of ranks) ≤ critical value. Here T⁻ =0, T=0. Since 0 <= 8, reject H₀. There is significant improvement.
差值（后-前）：2,2,2,1,1,3,2,1,1,1。无零差值。绝对差值秩次：差值3得秩10。五个差值2，分配秩5-9，平均秩为7。四个差值1，分配秩1-4，平均秩2.5。正秩和W⁺ = 4×2.5 + 5×7 + 10 = 10 + 35 + 10 = 55。负秩和为0，检验统计量T=0。n=10，双尾5%临界值为8。因0 ≤ 8，拒绝H₀。存在显著提高。
12. Integrating Multiple Techniques: A Climate Change Case Study | 多方法融合：气候变化案例研究
In a real-world exam scenario, you may be asked to integrate several statistical techniques. For instance, you could be given temperature anomalies over decades and asked to perform a regression analysis to estimate trend, test for significance, construct a confidence interval for the slope, and then use a runs test to check for randomness in residuals. Such questions test your ability to select, apply, and interpret appropriate methods.
在现实的考试情境中，你可能会被要求综合运用多种统计方法。例如，给出数十年间的气温异常值，要求进行回归分析以估算趋势、检验显著性、构建斜率的置信区间，然后运用游程检验来检查残差的随机性。这类题目考查你选择、应用并解读合适方法的能力。
Example: Yearly global temperature anomalies (in 0.01°C relative to 1951-1980 average) from 2000 to 2009: x (year code 0-9): 0,1,2,3,4,5,6,7,8,9; y: 38,48,56,58,60,62,65,66,68,72. Carry out regression analysis: find line of best fit, test H₀: slope = 0, 95% CI for slope, and interpret.
例题：2000至2009年间全球气温异常值（相对于1951-1980平均值的0.01°C）：x（年份代码0-9）: 0,1,2,3,4,5,6,7,8,9；y: 38,48,56,58,60,62,65,66,68,72。进行回归分析：求最佳拟合线，检验斜率是否为0，构建斜率的95%置信区间，并进行解释。
n=10, Σx=45, Σy=593, Σx²=285, Σy²=36101, Σxy=2943. x̄=4.5, ȳ=59.3. Sxx=285 – 45²/10=285 – 202.5=82.5. Sxy=2943 – 45×593/10=2943 – 2668.5=274.5. b=274.5/82.5≈3.327. a=59.3 – 3.327×4.5≈59.3 – 14.97=44.33. Equation: y = 44.33 + 3.33x. Test slope: SSE = Syy – b×Sxy. Syy=36101 – 593²/10=36101 – 35164.9=936.1. SSE=936.1 – 3.327×274.5≈936.1 – 913.5=22.6. s²=SSE/(n-2)=22.6/8=2.825. SE(b)=√(2.825/82.5)=√0.03424=
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