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我正在尝试确定两个接口之间是否存在显着差异。我有一个如下所示的文本文件：

group   conversion
A   0
A   0
A   1
A   0
A   0
A   1
A   1
A   0
A   0
A   1
A   1
A   1
A   1
A   1
A   1
A   0
A   0
A   0
A   0
A   0
A   1
A   0
A   1
A   0
A   1
A   1
A   0
A   1
A   0
A   1
A   1
A   0
A   0
A   0
A   0
A   0
A   1
A   1
A   0
A   0
A   1
A   1
A   0
A   1
A   1
A   0
A   0
A   0
A   1
A   1
A   0
A   0
A   0
A   0
A   1
A   1
A   0
A   1
A   1
A   1
A   1
A   1
A   1
A   1
A   0
A   0
A   0
A   1
A   1
A   0
A   1
A   1
A   0
A   0
A   1
A   0
A   0
A   0
A   1
A   0
A   1
A   1
A   1
A   0
A   0
A   0
A   0
A   0
A   0
A   0
A   1
A   1
A   1
A   1
A   1
A   1
A   0
A   0
A   1
A   1
B   0
B   0
B   1
B   0
B   0
B   0
B   1
B   0
B   0
B   0
B   0
B   1
B   0
B   1
B   0
B   1
B   0
B   1
B   0
B   0
B   1
B   1
B   1
B   1
B   1
B   1
B   1
B   1
B   1
B   0
B   0
B   1
B   0
B   0
B   1
B   0
B   0
B   0
B   0
B   0
B   1
B   1
B   0
B   0
B   0
B   0
B   1
B   1
B   0
B   0
B   1
B   0
B   1
B   0
B   0
B   0
B   1
B   1
B   1
B   1
B   0
B   1
B   0
B   0
B   1
B   1
B   0
B   0
B   0
B   0
B   0
B   0
B   0
B   1
B   0
B   0
B   1
B   0
B   0
B   0
B   0
B   0
B   0
B   0
B   0
B   1
B   1
B   1
B   0
B   0
B   0
B   0
B   1
B   0
B   1
B   1
B   1
B   1
B   1
B   1

现在我需要找出在执行此操作时应该使用哪种方法。到目前为止，我已经尝试了韦尔奇的两个样本 T 检验方法，我认为这是正确的。但这是确定是否存在意义的正确方法吗？顺便说一句，显着性水平为 5%。

这是我的代码：

# Load in the values from "test.txt"
dat = read.delim(“test.txt”)

# Calculate the amount of unique values
length(unique(dat$group))

# Calculate the p-value
t.test(dat$conversion ~ dat$group)

p 值的输出是：0.2586，大于0.05，这应该意味着没有意义，对吧？还是我做错了什么？我是R的初学者。

4

1 回答 1

1

I think that you are looking for the Fisher's T-test

using your data I created a data frame named x:

head(x)
  group conversion
1     A          0
2     A          0
3     A          1
4     A          0
5     A          0
6     A          1

then I made a frequency table: 

y<-table(x)  

# and previewed the count table:
y
     conversion
group  0  1
    A 50 50
    B 58 42

Then you run a Fisher's t-test:

fisher.test(y)  

    Fisher's Exact Test for Count Data

data:  y

p-value = 0.3207

alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.3989079 1.3135633

sample estimates:
odds ratio 
 0.7253254

And it even tells you that it is for comparing counts. It is a way of evaluating exactly the difference between two categorical identities.

于 2017-03-17T20:19:43.153 回答