Amala 认为，与不经常锻炼的男性相比，经常锻炼的男性静息心率更低。她测量了一组 50 名经常锻炼男性和一组 40 名不经常锻炼男性的静息心率 $h$。结果汇总如下表。

(a) 求 $\alpha$ 的值和 $\beta$ 的值。

(b) 在 5% 显著性水平下检验：是否有证据支持 Amala 的看法。清楚写出假设。

(c) 解释中心极限定理在 (b) 中检验的意义。

(d) 写出你在 (b) 中进行检验时作出的两个假设。

$\alpha$ 是样本均值，用 $\sum h/n$ 计算。$\beta$ 是总体方差的无偏估计，因此要用分母 $n-1$，并由 $\sum h^2-n\bar{h}^{\,2}$ 求离均差平方和。

For the men who exercise regularly,

The unbiased estimate of the variance is

设经常锻炼与不经常锻炼男性的总体平均静息心率分别为 $\mu_E$、$\mu_N$。Amala 的主张对应 $\mu_E<\mu_N$，所以进行左尾两样本均值检验。两个样本都较大，用样本方差估计总体方差并构造标准正态统计量。

Let $\mu_E$ and $\mu_N$ be the population mean resting heart rates of men who exercise regularly and men who do not exercise regularly, respectively.

Under $H_0$, the estimated standard error is

The test statistic is

For a one-tailed test at the $5\%$ significance level, the critical value is $-1.6449$.

Since

the test statistic lies in the critical region, so $H_0$ is rejected. There is sufficient evidence at the $5\%$ significance level to support Amala’s belief that the mean resting heart rate is lower in men who exercise regularly than in men who do not exercise regularly.

两个样本量分别为 $50$ 和 $40$，都足够大。中心极限定理说明，即使个体静息心率的总体分布不是正态分布，两个样本均值的抽样分布仍可近似看作正态分布，从而支持 (b) 中使用标准正态检验。

Both sample sizes are large. By the Central Limit Theorem, the sampling distribution of the mean resting heart rate is approximately normal for each group, even if the underlying resting heart rates are not normally distributed. This justifies the normal test used in part (b).

检验把两组视为彼此独立的随机样本，并用两个样本方差代替未知总体方差。因此需要明确写出独立性假设与“样本方差可视为相应总体方差”的假设。

The assumptions are:

1. The two samples are independent, and the observations within each sample are independent.
2. The sample variances can be treated as the corresponding population variances, so $\sigma_E^2=s_E^2$ and $\sigma_N^2=s_N^2$.