Python：實現平均值意味着95％置信區間？

this solution如何使用pandas/python來實現？這個問題涉及使用此stats.stackexchange solution圍繞平均值找到95％CI的實現。Python：實現平均值意味着95％置信區間？

import pandas as pd 
from IPython.display import display 
import scipy 
import scipy.stats as st 
import scikits.bootstrap as bootstraps 

data = pd.DataFrame({ 
    "exp1":[34, 41, 39] 
    ,"exp2":[45, 51, 52] 
    ,"exp3":[29, 31, 35] 
}).T 

data.loc[:,"row_mean"] = data.mean(axis=1) 
data.loc[:,"row_std"] = data.std(axis=1) 
display(data)

<table border="1" class="dataframe"> <thead> <tr style="text-align: right;">  <th></th>  <th>0</th>  <th>1</th>  <th>2</th>  <th>row_mean</th>  <th>row_std</th> </tr> </thead> <tbody> <tr>  <th>exp1</th>  <td>34</td>  <td>41</td>  <td>39</td>  <td>38.000000</td>  <td>2.943920</td> </tr> <tr>  <th>exp2</th>  <td>45</td>  <td>51</td>  <td>52</td>  <td>49.333333</td>  <td>3.091206</td> </tr> <tr>  <th>exp3</th>  <td>29</td>  <td>31</td>  <td>35</td>  <td>31.666667</td>  <td>2.494438</td> </tr> 
 
</tbody> </table>

mean_of_means = data.row_mean.mean() 
std_of_means = data.row_mean.std() 
confidence = 0.95 
print("mean(means): {}\nstd(means):{}".format(mean_of_means,std_of_means))

平均值（裝置）：39.66666666666667
STD（裝置）：8.950481054731702

第一不正確嘗試（zscore）：

zscore = st.norm.ppf(1-(1-confidence)/2) 
lower_bound = mean_of_means - (zscore*std_of_means) 
upper_bound = mean_of_means + (zscore*std_of_means) 
print("95% CI = [{},{}]".format(lower_bound,upper_bound))

95％CI = [22.1,57.2]（不正確溶液）

第二不正確嘗試（tscore）：

tscore = st.t.ppf(1-0.05, data.shape[0]) 
lower_bound = mean_of_means - (tscore*std_of_means) 
upper_bound = mean_of_means + (tscore*std_of_means) 
print("95% CI = [{},{}]".format(lower_bound,upper_bound))

95％CI = [18.60,60.73]（不正確溶液）

第三不正確嘗試（自舉）：

CIs = bootstraps.ci(data=data.row_mean, statfunction=scipy.mean,alpha=0.05)

95％CI = [31.67，49.33（不正確解決方案）

this solution如何使用pan das/python在下面得到正確的解決方案？

95％CI = [17.4 61.9]（正確溶液）

來源

2017-08-14 blehman

也許'scikits-bootstrap'你想要做什麼？ – xaav

@xaav，剛剛添加了一個使用這個建議的例子，很遺憾，我沒有提供正確的解決方案，儘管我可能會錯誤地使用它。我不確定alpha是否應該設置爲0.05或0.025，但無論如何，這是不正確的。 – blehman

謝謝喬恩貝茨。

import pandas as pd 
import scipy 
import scipy.stats as st 

data = pd.DataFrame({ 
    "exp1":[34, 41, 39] 
    ,"exp2":[45, 51, 52] 
    ,"exp3":[29, 31, 35] 
}).T 

data.loc[:,"row_mean"] = data.mean(axis=1) 
data.loc[:,"row_std"] = data.std(axis=1) 

tscore = st.t.ppf(1-0.025, data.shape[0]-1) 

print("mean(means): {}\nstd(means): {}\ntscore: {}".format(mean_of_means,std_of_means,tscore)) 

lower_bound = mean_of_means - (tscore*std_of_means/(data.shape[0]**0.5)) 
upper_bound = mean_of_means + (tscore*std_of_means/(data.shape[0]**0.5)) 

print("95% CI = [{},{}]".format(lower_bound,upper_bound))

平均值（手段）：39.66666666666667
STD（手段）：8.950481054731702
tscore：4.302652729911275
95％CI = [17.432439139464606,61.90089419386874]

來源

2017-08-16 03:51:19 blehman

Python：實現平均值意味着95％置信區間？

回答

相關問題