Normal distribution confidence intervals and sample standard deviation problem












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I am confused by this relationship that in a normal distribution of sample means $x_i : x_i$~$N(mu,sigma)$ the sample mean $x_i$ is within $2$ standard deviations of the true mean $mu$ with $95$% confidence.



I am confused because of this second relationship that $sigma approx frac{sigma_i}{sqrt{n_i-1}}$



$n_i$ is the sample size corresponding to sample mean $x_i$



$x_1in B(mu,2sigma):sigmaapproxfrac{sigma_1}{sqrt{n_1-1}}rightarrow$$d(x_1,mu)leqmu+2sigma-(mu-2sigma)=4sigmaapprox4frac{sigma_1}{sqrt{n_1-1}}$



$x_2in B(mu,2sigma):sigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow d(x_2,mu)leq4frac{sigma_2}{sqrt{n_2-1}}$



$d(x_1,x_2)leq d(x_1,mu)+d(x_2,mu)leq 4frac{sigma_1}{sqrt{n_1-1}}+4frac{sigma_2}{sqrt{n_2-1}}$



But, $frac{sigma_1}{sqrt{n_1-1}}approxsigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow frac{sigma_1}{sqrt{n_1-1}}approxfrac{sigma_2}{sqrt{n_2-1}}$



This implies



$d(x_1,x_2)leq 4(frac{sigma_1}{sqrt{n_1-1}}+frac{sigma_1}{sqrt{n_1-1}})= 8frac{sigma_1}{sqrt{n_1-1}}$



This implies that for any $n_2$, if $n_1$ is very large, $d(x_1,x_2)leq0$



This result does not make sense in reality.






real-analysis probability-theory







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    0














    I am confused by this relationship that in a normal distribution of sample means $x_i : x_i$~$N(mu,sigma)$ the sample mean $x_i$ is within $2$ standard deviations of the true mean $mu$ with $95$% confidence.



    I am confused because of this second relationship that $sigma approx frac{sigma_i}{sqrt{n_i-1}}$



    $n_i$ is the sample size corresponding to sample mean $x_i$



    $x_1in B(mu,2sigma):sigmaapproxfrac{sigma_1}{sqrt{n_1-1}}rightarrow$$d(x_1,mu)leqmu+2sigma-(mu-2sigma)=4sigmaapprox4frac{sigma_1}{sqrt{n_1-1}}$



    $x_2in B(mu,2sigma):sigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow d(x_2,mu)leq4frac{sigma_2}{sqrt{n_2-1}}$



    $d(x_1,x_2)leq d(x_1,mu)+d(x_2,mu)leq 4frac{sigma_1}{sqrt{n_1-1}}+4frac{sigma_2}{sqrt{n_2-1}}$



    But, $frac{sigma_1}{sqrt{n_1-1}}approxsigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow frac{sigma_1}{sqrt{n_1-1}}approxfrac{sigma_2}{sqrt{n_2-1}}$



    This implies



    $d(x_1,x_2)leq 4(frac{sigma_1}{sqrt{n_1-1}}+frac{sigma_1}{sqrt{n_1-1}})= 8frac{sigma_1}{sqrt{n_1-1}}$



    This implies that for any $n_2$, if $n_1$ is very large, $d(x_1,x_2)leq0$



    This result does not make sense in reality.






    real-analysis probability-theory







    share|cite|improve this question

























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      I am confused by this relationship that in a normal distribution of sample means $x_i : x_i$~$N(mu,sigma)$ the sample mean $x_i$ is within $2$ standard deviations of the true mean $mu$ with $95$% confidence.



      I am confused because of this second relationship that $sigma approx frac{sigma_i}{sqrt{n_i-1}}$



      $n_i$ is the sample size corresponding to sample mean $x_i$



      $x_1in B(mu,2sigma):sigmaapproxfrac{sigma_1}{sqrt{n_1-1}}rightarrow$$d(x_1,mu)leqmu+2sigma-(mu-2sigma)=4sigmaapprox4frac{sigma_1}{sqrt{n_1-1}}$



      $x_2in B(mu,2sigma):sigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow d(x_2,mu)leq4frac{sigma_2}{sqrt{n_2-1}}$



      $d(x_1,x_2)leq d(x_1,mu)+d(x_2,mu)leq 4frac{sigma_1}{sqrt{n_1-1}}+4frac{sigma_2}{sqrt{n_2-1}}$



      But, $frac{sigma_1}{sqrt{n_1-1}}approxsigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow frac{sigma_1}{sqrt{n_1-1}}approxfrac{sigma_2}{sqrt{n_2-1}}$



      This implies



      $d(x_1,x_2)leq 4(frac{sigma_1}{sqrt{n_1-1}}+frac{sigma_1}{sqrt{n_1-1}})= 8frac{sigma_1}{sqrt{n_1-1}}$



      This implies that for any $n_2$, if $n_1$ is very large, $d(x_1,x_2)leq0$



      This result does not make sense in reality.






      real-analysis probability-theory







      share|cite|improve this question













      I am confused by this relationship that in a normal distribution of sample means $x_i : x_i$~$N(mu,sigma)$ the sample mean $x_i$ is within $2$ standard deviations of the true mean $mu$ with $95$% confidence.



      I am confused because of this second relationship that $sigma approx frac{sigma_i}{sqrt{n_i-1}}$



      $n_i$ is the sample size corresponding to sample mean $x_i$



      $x_1in B(mu,2sigma):sigmaapproxfrac{sigma_1}{sqrt{n_1-1}}rightarrow$$d(x_1,mu)leqmu+2sigma-(mu-2sigma)=4sigmaapprox4frac{sigma_1}{sqrt{n_1-1}}$



      $x_2in B(mu,2sigma):sigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow d(x_2,mu)leq4frac{sigma_2}{sqrt{n_2-1}}$



      $d(x_1,x_2)leq d(x_1,mu)+d(x_2,mu)leq 4frac{sigma_1}{sqrt{n_1-1}}+4frac{sigma_2}{sqrt{n_2-1}}$



      But, $frac{sigma_1}{sqrt{n_1-1}}approxsigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow frac{sigma_1}{sqrt{n_1-1}}approxfrac{sigma_2}{sqrt{n_2-1}}$



      This implies



      $d(x_1,x_2)leq 4(frac{sigma_1}{sqrt{n_1-1}}+frac{sigma_1}{sqrt{n_1-1}})= 8frac{sigma_1}{sqrt{n_1-1}}$



      This implies that for any $n_2$, if $n_1$ is very large, $d(x_1,x_2)leq0$



      This result does not make sense in reality.






      real-analysis probability-theory




      real-analysis probability-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 4 at 7:37









      FrankFrank

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