Statistics 3
Pearson Edexcel · International A-Level · Further Mathematics Тема 9 5:08 Английское озвучивание · Английский + китайские субтитры (встроенные)
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Paper WST03: 1 hour 30 minutes, 75 marks, with S1 and S2 assumed.
WST03 试卷：1 小时 30 分钟，75 分，默认已学 S1 和 S2。
Memorise five things: how Normals combine, the sample mean's distribution, the unbiased variance, the chi-squared statistic, and Spearman's coefficient.
记住五件事：正态变量如何组合、样本均值的分布、无偏方差、卡方统计量，以及斯皮尔曼系数。
Two cases C1 and C2 each have standard deviation 0.03.
C1 和 C2 的标准差都是 0.03。
Their difference is Normal, centred on 0, with variance 0.0018: the variances add even though the variables subtract.
它们的差服从正态分布，以 0 为中心，方差 0.0018：变量相减，方差却相加。
By symmetry, both tails beyond 0.02 make twice one tail: 0.638.
由对称性，超出 0.02 的两个尾部是一个尾部的两倍：0.638。
Three bags sum to Normal 54, 0.0027, so P above 54.1 is 0.0274.
三袋之和服从 N(54, 0.0027)，所以超过 54.1 的概率是 0.0274。
For T below 30 P1 plus 190, take 30 P1 across: one P1 is already in T, leaving minus 29 P1.
对于 T < 30P₁ + 190，把 30P₁ 移过去：T 中已有一个 P₁，剩下 −29P₁。
The mean is 194 and the variance 33.6697, with 29 squared on P's variance, so the probability is 0.245.
均值 194，方差 33.6697，其中 P 的方差乘以 29²，所以概率是 0.245。
Simple random sampling uses random numbers on a sampling frame.
简单随机抽样在抽样框上使用随机数。
Systematic sampling takes every 18th from a random start: no selector bias, but neighbours never appear together.
系统抽样从随机起点开始每隔 18 个取一个：没有选择者偏差，但相邻成员永远不会同时入选。
Stratified sampling samples each group in proportion.
分层抽样按比例从每组抽取。
Quota sampling needs no frame but risks interviewer bias.
配额抽样不需要抽样框，但有访问者偏差的风险。
R is a half X1 plus three quarters X20, so its expectation is five quarters mu: biased, by mu over 4.
R = ½X₁ + ¾X₂₀，所以期望是 5μ/4：有偏，偏差为 μ/4。
Requiring S and T to be unbiased gives a minus b equals 2 and 14a plus 6b equals 20, so a is 1.6 and b is minus 0.4.
要求 S 和 T 无偏得 a − b = 2 和 14a + 6b = 20，所以 a = 1.6，b = −0.4。
The mean of one value, of four, of sixteen: every curve is centred on mu, but the spread halves each time n quadruples.
一个值、四个值、十六个值的平均：每条曲线都以 μ 为中心，但 n 每扩大四倍，分布宽度减半。
The standard error is sigma over root n.
标准误是 σ/√n。
The two probabilities give z-values minus 0.4 and 2.1.
两个概率给出 z 值 −0.4 和 2.1。
Subtracting the equations gives 2.5 standard errors equals 30, so the standard error is 12 and mu is 50.
两式相减得 2.5 个标准误等于 30，所以标准误为 12，μ = 50。
Then P of X bar above 59 is 0.2266.
于是 P(X̄ > 59) = 0.2266。
The true mean stays fixed.
真实均值固定不变。
Each sample gives its own interval, x bar plus or minus 1.96 standard errors; most cover mu, and now and then one misses it.
每个样本给出自己的区间：x̄ ± 1.96 个标准误；大多数覆盖 μ，偶尔有一个没有覆盖。
95 percent describes the procedure, not one interval.
95% 描述的是这个方法，而不是某一个区间。
Under the null hypothesis the test statistic is standard Normal.
在原假设下检验统计量服从标准正态分布。
The 5 percent one-tailed critical value 1.6449 cuts off the rejection region, and z is 2.46, inside it: the grey mean is more than 200 grams above the red.
5% 单尾临界值 1.6449 划出拒绝域，而 z = 2.46 落在其中：灰松鼠的均值比红松鼠高出 200 克以上。
The red sums give a mean of 290 and an unbiased variance of 169.
红松鼠的数据给出均值 290 和无偏方差 169。
The difference of means, 210, minus the hypothesised 200, over the root of the summed variances over n: 10 over 4.0698, which is 2.46.
均值差 210 减去假设的 200，除以各方差除以 n 之和的平方根：10/4.0698 = 2.46。
Males in 250 litters of 5, with p hat 0.488.
250 窝、每窝 5 只中雄性的数量，p̂ = 0.488。
The observed counts and the expected counts mostly agree, but the zero class has 2 against 8.80, contributing 5.25 on its own.
观察频数与期望频数大体一致，但 0 这一类是 2 对 8.80，单独贡献 5.25。
X squared is 10.95 on 4 degrees of freedom, above 9.488: the binomial is not a suitable model.
X² = 10.95，自由度 4，超过 9.488：二项分布不是合适的模型。
The diesel, month 1 to 6 cell expects 35 times 36 over 250, 5.04, and contributes 4.88.
柴油、第 1 至 6 个月这一格的期望频数是 35×36/250 = 5.04，贡献 4.88。
With 7.444 from the other cells, X squared is 12.325.
加上其他格的 7.444，X² = 12.325。
A 3 by 3 table has 4 degrees of freedom, and 12.325 exceeds 9.488: evidence of an association.
3×3 表有 4 个自由度，12.325 超过 9.488：有关联的证据。
Two judges rank 10 performances.
两位评委给 10 场表演排名。
The rank differences give sum d squared 66, so r s is 0.6.
等级差给出 Σd² = 66，所以 rₛ = 0.6。
At the 1 percent one-tailed level the critical value is 0.7455, higher: insufficient evidence of positive agreement between the judges.
在 1% 单尾水平下临界值是 0.7455，更高：没有足够证据表明两位评委的排名正相关。
Check yourself.
自我检测。
The variance of X minus Y is the sum of the variances.
X − Y 的方差是两方差之和。
The standard error is sigma over root n.
标准误是 σ/√n。
Combine cells whose expected value is below 5.
期望值低于 5 的格要合并。
A binomial fit with p estimated loses two degrees of freedom.
估计了 p 的二项拟合失去两个自由度。
And Spearman is 1 minus 6 sum d squared over n, n squared minus 1.
斯皮尔曼系数是 1 − 6Σd²/ 。