一名学生记录了若干堂讲座的时长 $t$（以分钟为单位）。随机抽取的 $80$ 堂讲座的结果汇总如下。

(a) 计算讲座总体平均时长的一个 $96\%$ 置信区间。

(b) 由于样本量较大，(a) 中采用的方法是适用的。说明如果样本量较小，为什么这种方法将不适用。

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。

总体分布未说明，先估计均值与无偏方差，再用大样本正态近似构造 $96\%$ 区间。

The unbiased estimates are

and

Since the random sample is large, the Central Limit Theorem justifies a normal approximation for the sample mean. For a $96\%$ confidence interval, the two tails each have probability $0.02$, so $z=\Phi^{-1}(0.98)\approx2.054$.

The interval is

Thus an approximate $96\%$ confidence interval is

小样本且总体未给正态，不能保证样本均值有可用的正态近似。

For a small sample, the Central Limit Theorem would not justify approximating the sample-mean distribution by a normal distribution. The population distribution is not stated to be normal, so normality of the sample mean could not otherwise be assumed.