一家乡村商店的经理对顾客的购物习惯感兴趣。根据以往记录，每位顾客所购物品数量的中位数为 $8$。经理认为，每位顾客所购物品数量的中位数已经增加。他随机抽取 $15$ 位顾客，记录每位顾客在店内购买的物品数量。结果如茎叶图所示。

(a) 提出一个理由，说明在这种情况下，使用 Wilcoxon 符号秩检验来考察经理的看法可能不合适。

(b) 在 $5\%$ 的显著性水平下，使用符号检验考察经理的看法。

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。

查看茎叶图的尾部。符号秩检验要求总体关于其中位数对称。

The observations have a long upper tail, with large values such as $56$ and $74$. This suggests a right-skewed population, so the symmetry assumption of the Wilcoxon signed-rank test may be inappropriate.

从茎叶图数出高于 $8$ 的人数。用单侧精确二项概率检验是否有超过一半的顾客购买多于 $8$ 件。

Let $m$ be the population median. Test $H_0\colon m=8$ against $H_1\colon m>8$. From the diagram, $12$ observations exceed $8$ and $3$ are below it, with no observations equal to $8$. Under the null hypothesis, the probability of exceeding $8$ is at most $1/2$. The upper-tail probability is largest at the boundary $1/2$, so we use $S\sim\operatorname{Binomial}(15,1/2)$ for the test. The tail probability is

Since this is below $0.05$, we reject $H_0$. There is sufficient evidence at the $5\%$ level that the population median number of items bought per customer has increased.

把高于基准的个数近似为正态变量，作连续性修正后进行单侧检验。

Test $H_0\colon m=8$ against $H_1\colon m>8$, where $m$ is the population median. There are $12$ observations above $8$ out of $15$, with no ties. At the null boundary, $S\sim\operatorname{Binomial}(15,1/2)$, so its normal approximation has mean $7.5$ and variance $3.75$.

For the upper tail $S\ge12$, the continuity-corrected statistic is

Since $z>1.645$, we reject $H_0$ at the $5\%$ level. There is sufficient evidence that the population median number of items bought per customer has increased.