3# lnlhckao123
在multiple comparison 下，貌似只能够估计各组标准差相等时候的N，当S=3时，各组N=335。当S=2时，各组N=192，对于标准差不一样的情况，有书上写可以用两两比较（调整显著性水平）的公式计算得到三个样本量，然后取最大值，具体操作还是看看HELP吧，上面有example，应该对你有帮助！

In a single factor ANOVA study, sample sizes of 335, 335, and 335 are obtained from the 3
groups whose means are to be compared. The total sample of 1005 subjects achieves 95% power to
detect a difference of at least 1.00 using the Tukey-Kramer (Pairwise) multiple comparison test
at a 0.05000 significance level. The common standard deviation within a group is assumed to be
3.00.

你好2楼谢谢你的回答！但是标准答案是总共样本量是570（每组190），power是90%，而且你也没教我怎样操作，你可否教我具体怎样操作！谢谢！

谢谢baoaibaobao！我摸索出来！

thanks for sharing