估计样品平均值的标准错误
从大量人群中选择了143名女孩和127名男孩的随机样本。测量每个儿童的血红蛋白水平（以g/dl的测量），并以下结果进行测量。
女孩n = 143平均= 11.35 SD = 1.41 男孩n = 127平均11.01 SD = 1.32 估计样品平均值之间差异的标准误差
Random sample of 143 girl and 127 boys were selected from a large population.A measurement was taken of the haemoglobin level(measured in g/dl) of each child with the following result.
girl n=143 mean = 11.35 sd = 1.41
boys n=127 mean 11.01 sd =1.32
estimate the standard error of the difference between the sample means
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
从本质上讲，我们通过添加标准错误来汇总标准错误。这意味着我们回答了一个问题：考虑到两个样本 vairation是什么？
sd = sqrt（（（sd₁** 2 /n₁） +（sd₂** 2 /n₂）\
sd = sqrt（（（1.41 ** 2 /143） +（1.32 ** 2/127）
≈0.1662平方只是每个样本的差异，在我们的情况下，值很小。
鉴于我们测量的变化和是样本的大小。
观察到的差异是我们预期的平均差异的2.046倍， T-临界值可以通过使用T值图来轻松计算：一个人需要了解Alpha（通常为0.05，除非另有说明），需要知道原始的替代假设的行之间有差异，然后我们将应用两个尾巴分布 - 如果沿着性别x的线的某些东西具有大/小于性别x 然后，我们将使用单个尾巴分布）。
如果 t-value＆gt;然后，我们会声称均值之间的差异在统计上是显着的，从而有足够的明显拒绝零假设。或者，如果 t-value＆lt; t-Cricition ，我们不会有统计上有明显的证据来反对零假设，因此我们不会拒绝零假设。
In essence, we'd pool the standard errors by adding them. This implies that we´re answering the question: what is the vairation of the sampling distribution considering both samples?
SD = sqrt( (sd₁**2 / n₁) + (sd₂**2 / n₂) \
SD = sqrt( (1.41**2 / 143) + (1.32**2 / 127) ≈ 0.1662
Notice that the standrad deviation squared is simply the variance of each sample. As you can see, in our case the value is quite small, which indicates that the difference between sampled means doesn´t need to be that large for there to be a larger than expected difference between obervations.
We´d calculate the difference between means as 0.34 (or -0.34 depending on the nature of the question) and divide this difference by the standrad error to get a t-value. In our case 2.046 (or -2.046) indicates that the observed difference is 2.046 times larger than the average difference we would expect given the variation the variation that we measured AND the size of our sample.
However, we need to verify whether this observation is statistically significant by determining the t-critical value. This t-critical can be easily calculated by using a t-value chart: one needs to know the alpha (typically 0.05 unless otherwise stated), one needs to know the original alternative hypothesis (if it was something along the lines of there is a difference between genders then we would apply a two tailed distribution - if it was something along the lines of gender X has a hameglobin level larger/smaller than gender X then we would use a single tailed distribution).
If the t-value > t-critical then we would claim that the difference between means is statistically significant, thereby having sufficient evident to reject the null hypothesis. Alternatively, if t-value < t-critical, we would not have statistically significant evidence against the null hypothesis, thus we would fail to reject the null hypothesis.