Wolfram Mathematica的置信区间

Question

我有兴趣计算数据的置信区间。例如，使用R：

data = c(1,0,2,1,2,0,1,0)
confidence_interv = (sd(data)/sqrt(8))*qnorm(.95)

其中8是数据数组的长度。我也想对Mathematica做同样的事情，也许用一种更快的方法。那么，是否有可能在单个数据周围绘制数据和置信区间呢？谢谢

Answer 1

data = {1, 0, 2, 1, 2, 0, 1, 0};
lm = LinearModelFit[data, {1, x}, x];
lm["SinglePredictionConfidenceIntervalTable", ConfidenceLevel -> 0.95]

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Show[ListPlot[data], 
 Plot[{lm[x], 
   lm["SinglePredictionBands", ConfidenceLevel -> 0.95]}, {x, 0, 9}, 
  Filling -> {2 -> {1}}], PlotRange -> {Automatic, {-4, 6}}, 
 Frame -> True, ImageSize -> 300]

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另见Showing the correlation of two variables using a plot

Answer 2

对这个问题的解释有点不同。

 Needs["HypothesisTesting`"]
 data = Table[RandomVariate[NormalDistribution[1, 1]], {2000}];
 ci = NormalCI[ Mean[data] , StandardDeviation[data], 
          ConfidenceLevel -> .99]

{-1.55307，3.58071}

 Histogram[data,
   Epilog -> {Text[Style["99% confidence range", FontSize -> 20], {1, 100}],
              Red,Line[{{#, 0}, {#, 100}}] & /@ ci ,
              Arrowheads[{-.05, .05}], Arrow[{{ci[[1]], 70}, {ci[[2]], 70}}]}] 

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顺便说一句，您也可以得到这样的间隔：(w/o加载HypothesisTesting包)

  InverseCDF[
        NormalDistribution[Mean[data], StandardDeviation[data]],
            (1 + # {-1, 1})/2 &@.99 ]

{-1.55307，3.58071}