Seva 正在研究澳大利亚两个地区 X 和 Y 中成年袋鼠尾巴的长度。他从地区 X 随机抽取 50 只成年袋鼠，并记录它们尾巴的长度 $x$ cm。他还从地区 Y 随机抽取 40 只成年袋鼠，并记录它们尾巴的长度 $y$ cm。结果概括如下。

不能假设两个分布的总体方差相同。

(a) 求地区 X 与地区 Y 成年袋鼠尾巴总体平均长度之差的 90% 置信区间。

地区 X 和地区 Y 成年袋鼠尾巴的总体平均长度分别为 $\mu_X$ cm 和 $\mu_Y$ cm。

(b) 在 10% 显著性水平下，检验原假设 $\mu_Y - \mu_X = 1.1$，备择假设为 $\mu_Y - \mu_X > 1.1$。结合题目背景说明你的结论。

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。

用两个独立大样本构造均值差区间。总体方差不要求相同，因此分别估计方差。

The sample means are $1080/50=21.6$ and $940/40=23.5$. The unbiased variance estimates are

The estimated standard error of the difference is $\sqrt{s_X^2/50+s_Y^2/40}=0.381279\ldots$. For $\mu_Y-\mu_X$, the $90\%$ interval is

Thus it is $(1.27,2.53)$ cm.

用 (a) 的标准误。零假设中的均值差是 $1.1$，因此先从样本均值差中减去 $1.1$。

For the stated hypotheses, the large-sample test statistic is

The upper $10\%$ critical value is $1.282$. Since $z>1.282$, we reject $H_0$. There is sufficient evidence at the $10\%$ level that adult wallabies in region Y have a population mean tail length more than $1.1$ cm greater than that in region X.

(a) 的 $90\%$ 双侧区间下限也是 $95\%$ 单侧置信下限。若它已超过 $1.1$，便可在较严格的 $5\%$ 水平拒绝，因此也能在 $10\%$ 水平拒绝。

From (a), the lower $95\%$ one-sided confidence bound for $\mu_Y-\mu_X$ is $1.272795\ldots$ cm. Since $1.1$ cm lies below this bound, the one-sided test rejects $H_0\colon\mu_Y-\mu_X=1.1$ in favour of $\mu_Y-\mu_X>1.1$ at the $5\%$ significance level. Rejection at $5\%$ also implies rejection at the less stringent $10\%$ level.

There is sufficient evidence at the $10\%$ level that the population mean tail length in region Y exceeds that in region X by more than $1.1$ cm.