z****k
发帖数: 1057 | 1 假设我想test是不是两个选项中左面的那个总是相对来说有更大的机会被选中
于是我找了两组人
第一组人我让他们 A 和 B 之间选一个
第二组人我让他们 B 和 A 之间选一个
以下两种统计检验方法，哪种是对的？区别在哪里？
1) 把左面那个选项的比例（即第一组人选A的比例和第二组人选B的比例）平均起来，
检验其是否大于50%
2) 检验是否第一组人选A的比例大于第二组人选A的比例（用2x2 chi-square或者
fisher's exact test)
谢谢指教~ | s******h
发帖数: 539 | 2 I assume that your subjects are randomly assigned to the two designs (i.e.,
[A B] and [B A]). If the sample size of the two groups are the same, say n,
then 1) and 2) are essentially equivalent, imo.
To see this, assume that X1, ..., Xn are the Bernoulli responses of choosing
A in design [A B], and Y1, ..., Yn are the response of choosing A in design
[B A]. Assume that E(X1) = P1 and E(Y1) = P2. You want to test if P1 = P2.
So I think 2) makes more sense.
Method 1) is then {(X1 + ... + Xn) + (n - Y1 - ... - Yn)}/(2n) > 0.5, which
is equivalent to (X1 + ... + Xn)/n > (Y1 + ... + Yn)/n. Thus, it'll be
equivalent to your Method 2) (in asymptotic sense, since you may use fisher'
s exact test for Method 2).
However, when two groups do not have equal sample size, you would expect 1)
to be different from 2). Because, e.g. n2 = 2n1 = 2n, if you use simple
average as in 1), you are testing something like P1 + 0.5 > 2P2. |
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