去年，Pete’s Pizza Pit 的比萨平均配送用时为 $32.4$ 分钟。今年，从 Pete’s Pizza Pit 的配送订单中随机抽取 $50$ 单，记录每单配送用时 $t$ 分钟。结果如下。

$n=50\qquad\sum t=1700\qquad\sum t^2=59\,050$

(a) 求总体均值和方差的无偏估计值。

(b) 在 $2\%$ 的显著性水平下检验平均配送用时与去年相比是否发生了变化。

(c) 在什么情况下，解答 (b) 时无需使用中心极限定理？

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。

使用样本均值及分母49的无偏方差公式。

The unbiased estimate of the population mean is

The unbiased estimate of the population variance is

是否改变对应双尾备择。中心极限定理近似样本均值分布，用未舍入方差求统计量，与每尾1%的临界值比较。

Let $\mu$ be this year’s population mean delivery time, in minutes.

Using the large-sample normal approximation,

For a two-tailed test at the 2% significance level, the critical values are $\pm2.326$. Since $|2.24|<2.326$, we do not reject $H_0$.

There is insufficient evidence at the 2% significance level that the population mean delivery time has changed since last year.

若原始总体已知正态，样本均值便精确正态，不需以中心极限定理近似。

The Central Limit Theorem would not be needed if the population distribution of delivery times were known to be normal.