某网站平均每小时有 $\mu$ 次访问。过去 $\mu$ 的值为 $14.4$。对网站进行一些改进后，网站所有者想检验 $\mu$ 的值是否增加。他随机选取一段 $10$ 分钟的时间，发现这段时间内共有 $6$ 次访问。可以假设网站在任一给定时间段内收到的访问次数服从泊松分布。

(a) 在 $2.5\%$ 的显著性水平下进行检验。

(b) 说明在进行该检验时，是否可能犯第一类错误、第二类错误，或同时犯这两类错误。

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。

假设参数使用每小时平均点击数，实际观测却只有十分钟；在原假设下将 Poisson 参数除以六，再计算至少六次的右尾概率。

Test

where $\mu$ is the population mean number of website hits per hour. Under $H_0$, the number $N$ of hits in ten minutes has distribution

For the observation $N=6$,

Since $0.0357>0.025$, do not reject $H_0$. There is insufficient evidence at the $2.5\%$ significance level that the population mean hourly hit rate has increased.

本次没有拒绝原假设，所以不可能是第一类错误；若真实点击率已经增加，则可能是第二类错误。

Since $H_0$ was not rejected, a Type I error cannot have been made: that would require rejecting a true $H_0$.

A Type II error may have been made if the true mean hourly hit rate has increased but the test failed to detect this increase. Thus only a Type II error is possible for the decision reached in (a).