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I'm working with the sampling distribution of the sample mean and sample variance. For my data, I have the following sample mean = 0.418, standard deviation = 0.306, and degrees of freedom = 44. I've determined the 95% confidence interval (CI) is [0.3266,0.5099]

Below is the plot the Student t-distribution for these parameters. My questions are as follows:

  1. How can I color the area under this distribution to show the 95% CI?
  2. How can I use NIntegrate to verify the area between the 95% CI is 0.95?
Plot[PDF[StudentTDistribution[0.418, 0.306, 44], x], {x, -1, 2}, 
 PlotRange -> All, Filling -> Axis]
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  • $\begingroup$ To find confidence intevals, look at 36827. $\endgroup$
    – Syed
    Nov 19, 2022 at 4:34

1 Answer 1

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Try

pl1 = Plot[
   PDF[StudentTDistribution[mean = 0.418, stDev = 0.306, degFr = 44], 
    x], {x, -1, 2}, PlotRange -> All];
pl2 = Plot[
   PDF[StudentTDistribution[mean, stDev, degFr], x], {x, 0.3266, 
    0.5099}, PlotRange -> All, Filling -> Axis, 
   AxesOrigin -> {0, 0}];
Show[{pl1, pl2}]

But your confidence interval does not seem right, because the above gives: enter image description here

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  • $\begingroup$ ya; I will have to look into that ... thank u Nicholas! $\endgroup$
    – PRG
    Nov 18, 2022 at 23:42
  • $\begingroup$ @PRG And the corresponding NIntegrate is NIntegrate[ PDF[StudentTDistribution[mean, stDev, degFr], x], {x, 0.3266, 0.5099}]. It is considered good form to accept and like answers. $\endgroup$
    – Nicholas G
    Nov 18, 2022 at 23:55
  • 1
    $\begingroup$ yes, Nicholas ... I hadn't finished the correspondence. $\endgroup$
    – PRG
    Nov 19, 2022 at 0:33

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