一家公司有两台机器 A 和 B，它们独立地向小瓶中灌装液体。机器 A 和 B 灌装的每瓶液体体积（以适当单位计）分别记为 $x$ 和 $y$。公司的一名科学家随机抽取 40 个由机器 A 灌装的小瓶，以及 50 个由机器 B 灌装的小瓶。结果汇总如下。

机器 A 和 B 灌装的小瓶中液体体积的总体均值分别记为 $\mu_A$ 和 $\mu_B$。

(a) 在 2% 显著性水平下，检验 $\mu_A$ 和 $\mu_B$ 之间是否存在差异。

(b) 求 $\alpha$ 的取值集合，使得在 $\alpha\%$ 显著性水平下，有证据表明 $\mu_A - \mu_B$ 大于 0.25。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

用两个大样本分别估计总体方差，再用均值差的正态近似。检验是否存在差异要取双侧。

Test $H_0\colon\mu_A-\mu_B=0$ against $H_1\colon\mu_A-\mu_B\ne0$. The sample means are $1120/40=28$ and $1370/50=27.4$. The unbiased variance estimates are

The estimated variance of the difference of sample means is

The estimated standard error is $\sqrt{s_A^2/40+s_B^2/50}=0.225714\ldots$, giving $z=0.6/0.225714\ldots=2.65822\ldots$. The two-sided $2\%$ critical values are $\pm2.326$. Since $|z|>2.326$, we reject $H_0$. There is sufficient evidence at the $2\%$ level that the machines have different population mean filling volumes.

零假设的差值改为 $0.25$，备择方向为较大。用 (a) 的标准误求单尾 $p$ 值，再换成百分数。

For $H_0\colon\mu_A-\mu_B=0.25$ against $H_1\colon\mu_A-\mu_B>0.25$, the statistic is

The upper-tail probability is $1-\Phi(1.550629\ldots)=0.0604952\ldots$. Using rejection when the $p$-value is no greater than the significance level, the required values satisfy $\alpha\ge100(1-\Phi(1.550629\ldots))$, whose boundary is about $6.05$.