我正在使用R中的power.prop.test
函数。
我正在进行A / B测试,我确定每组最小印象数的升力,以便A / B测试显着。
当我运行如下函数时,我得到第二个比例(p2)为0.0001870215,其中n为每组2,571,429:
original_conversion_rate<-0.00009
power.prop.test(n=2571429,
p1=original_conversion_rate,
power=0.8,
sig.level=0.05)
我的回答是:
Two-sample comparison of proportions power calculation
n = 2571429
p1 = 9e-05
p2 = 0.0001870215
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number in *each* group
当我用p2(0.0001870215)的答案重新运行来解决n时,出现了不同的n(230,952.6):
original_conversion_rate<-0.00009
power.prop.test(
p1=original_conversion_rate,
p2=0.0001870215,
power=0.8,
sig.level=0.05)
我的更改为:
Two-sample comparison of proportions power calculation
n = 230952.6
p1 = 9e-05
p2 = 0.0001870215
sig.level = 0.05
power = 0.8
alternative = two.sided
为什么在这种情况下会改变?
答案 0 :(得分:1)
您的频率非常小,微小的变化会产生巨大的影响 - 尤其是power.prop.test
中使用的默认容差级别。
如果再多花一次,你就得到了
power.prop.test(n=230952.5,
p1=original_conversion_rate,
power=0.8,
sig.level=0.05)
Two-sample comparison of proportions power calculation
n = 230952.5
p1 = 9e-05
p2 = 0.0001870215
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number in *each* group
与p2
的值相同。现在,如果我们回到你的第一次计算并降低我们得到的公差
power.prop.test(n=2571429,
p1=original_conversion_rate,
power=0.8,
sig.level=0.05, tol=.Machine$double.eps^.8)
Two-sample comparison of proportions power calculation
n = 2571429
p1 = 9e-05
p2 = 0.0001150142
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number in *each* group
您可以看到p2
已更改。如果我们插入p2
的更新值并求解n
,我们就会得到
power.prop.test(
p1=original_conversion_rate,
p2=0.0001150142,
power=0.8,
sig.level=0.05)
Two-sample comparison of proportions power calculation
n = 2571429
p1 = 9e-05
p2 = 0.0001150142
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number in *each* group
你得到的是n
。