购物习惯：双样本 t 检验的显著性水平

Shopping Habits

专题: Probability / 概率
难度: L6
来源: QuantQuestion

题目详情

研究者分别独立随机抽取 20 位周末购物者与 20 位工作日购物者。

周末：均值 78，标准差 22。
工作日：均值 67，标准差 20。

假设正态、方差齐性。检验两群体均值是否有显著差异（双侧）。求达到的显著性水平（p 值），保留 3 位有效数字。

A researcher is investigating the temporal effect of shopping habits at the mall. She independently and randomly selects 20 weekend and 20 weekday shoppers and finds that weekend shoppers spend 78 dollar on average with a standard deviation of 22dollar, while weekday shoppers spend $67 on average with a standard deviation of$ 20. The researcher would like to test if there is sufficient evidence to claim that there is a significant difference in the average amount spent by weekend and weekday shoppers. What is the attained significance level? Round to 3 significant figures. Assume simple random sampling, variance homogeneity, and that spending is approximately normally distributed.

解析

合并标准差

S_p=\sqrt{\frac{19\cdot 22^2+19\cdot 20^2}{38}}\approx 21.024.

t 统计量（自由度 38）：

t=\frac{78-67}{S_p\sqrt{1/20+1/20}}\approx 1.655.

双侧 p 值约为 $0.106$ （3 位有效数字）。

Original Explanation

We are testing the null hypothesis $H_0: \mu_1 = \mu_2$ against the alternative hypothesis $H_a: \mu_1 \neq \mu_2$ . Because neither sample size is sufficiently large, we must use a $t$ test with 38 degrees of freedom:

t = \frac{\bar{x_1} - \bar{x_2}}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \text{ where } S_p = \sqrt{\frac{(n_1-1)S_1 ^2 + (n_2-1)S_2 ^2}{n_1+n_2-2}}

Solving for $S_p$ :

S_p = \sqrt{\frac{(20-1) \times 22^2 + (20-1) \times 20^2}{20 + 20 - 2}} \approx 21.024

Solving for $t$ :

t = \frac{78 - 67}{21.024 \sqrt{\frac{1}{20} + \frac{1}{20}}} \approx 1.655

Because this is a two-tailed test, the attained significance level is double the p-value of a one-tailed test. Thus,

p\text{-value} = 2 \times P(t \geq 1.655) = 2 \times 0.05308 \approx 0.106