去年，某家商店的顾客在店内停留的时间以分钟为单位，其均值为 $10.5$，标准差为 $6.8$。今年，店主想检验顾客的平均停留时间是否发生变化。他决定选取 $20$ 名顾客作为样本。

(a) 一名店员建议选取某天早上最先进入商店的 $20$ 名顾客。

给出这种做法无法得到随机样本的两个理由。

店主决定在 $10\%$ 的显著性水平下进行检验。他随机抽取 $20$ 名顾客，发现这些顾客在店内停留的平均时间为 $8.4$ 分钟。应假设这些停留时间来自正态分布。

(b) 写出进行该检验所需的一个假设。

(c) 进行该检验。

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。

指出具体的抽样限制，并给出两条不同原因；只说“有偏”或“样本太小”不够。

Two reasons are:

1. Only customers arriving in the morning are included, so customers who visit at other times of day have no chance of selection.
2. Only one day’s customers are included, so customers who visit on other days have no chance of selection and different days may have different customer patterns.

正态分布已经由题目给出，额外需要的是沿用去年的已知标准差。

Assume that the population standard deviation of time spent in the shop has not changed and is still $\boxed{6.8\text{ minutes}}$.

“改变”对应双侧检验。显著性水平为 $10\%$，所以每侧为 $5\%$，临界值是 $\pm1.645$。

Let $\mu$ be the population mean time spent in the shop this year. Test

Using the normal-population assumption and the unchanged standard deviation, under $H_0$,

The test statistic is

For a two-tailed test at the $10\%$ level, the critical values are $\pm1.645$. Since $|-1.381|<1.645$, do not reject $H_0$.

There is insufficient evidence at the $10\%$ significance level that the population mean time spent in the shop has changed from last year.