Normal distribution confidence intervals and sample standard deviation problem
$begingroup$
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$$d(x_1,mu)leqmu+2sigma-(mu-2sigma)=4sigmaapprox4frac{sigma_1}{sqrt{n_1-1}}$
$x_2in B(mu,2sigma):sigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow d(x_2,mu)leq4frac{sigma_2}{sqrt{n_2-1}}$
$d(x_1,x_2)leq d(x_1,mu)+d(x_2,mu)leq 4frac{sigma_1}{sqrt{n_1-1}}+4frac{sigma_2}{sqrt{n_2-1}}$
But, $frac{sigma_1}{sqrt{n_1-1}}approxsigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow frac{sigma_1}{sqrt{n_1-1}}approxfrac{sigma_2}{sqrt{n_2-1}}$
This implies
$d(x_1,x_2)leq 4(frac{sigma_1}{sqrt{n_1-1}}+frac{sigma_1}{sqrt{n_1-1}})= 8frac{sigma_1}{sqrt{n_1-1}}$
This implies that for any $n_2$, if $n_1$ is very large, $d(x_1,x_2)leq0$
This result does not make sense in reality.
real-analysis probability-theory
asked Jan 4 at 7:37
FrankFrank
16210
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add a comment |
$begingroup$
I am confused by this relationship that in a normal distribution of sample means $x_i : x_i$~$N(mu,sigma)$ the sample mean $x_i$ is within $2$ standard deviations of the true mean $mu$ with $95$% confidence.
I am confused because of this second relationship that $sigma approx frac{sigma_i}{sqrt{n_i-1}}$
$n_i$ is the sample size corresponding to sample mean $x_i$
$x_1in B(mu,2sigma):sigmaapproxfrac{sigma_1}{sqrt{n_1-1}}rightarrow$$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
asked Jan 4 at 7:37
FrankFrank
16210
$endgroup$
add a comment |
$begingroup$
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$$d(x_1,mu)leqmu+2sigma-(mu-2sigma)=4sigmaapprox4frac{sigma_1}{sqrt{n_1-1}}$
$x_2in B(mu,2sigma):sigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow d(x_2,mu)leq4frac{sigma_2}{sqrt{n_2-1}}$
$d(x_1,x_2)leq d(x_1,mu)+d(x_2,mu)leq 4frac{sigma_1}{sqrt{n_1-1}}+4frac{sigma_2}{sqrt{n_2-1}}$
But, $frac{sigma_1}{sqrt{n_1-1}}approxsigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow frac{sigma_1}{sqrt{n_1-1}}approxfrac{sigma_2}{sqrt{n_2-1}}$
This implies
$d(x_1,x_2)leq 4(frac{sigma_1}{sqrt{n_1-1}}+frac{sigma_1}{sqrt{n_1-1}})= 8frac{sigma_1}{sqrt{n_1-1}}$
This implies that for any $n_2$, if $n_1$ is very large, $d(x_1,x_2)leq0$
This result does not make sense in reality.
real-analysis probability-theory
asked Jan 4 at 7:37
FrankFrank
16210
$endgroup$
I am confused by this relationship that in a normal distribution of sample means $x_i : x_i$~$N(mu,sigma)$ the sample mean $x_i$ is within $2$ standard deviations of the true mean $mu$ with $95$% confidence.
I am confused because of this second relationship that $sigma approx frac{sigma_i}{sqrt{n_i-1}}$
$n_i$ is the sample size corresponding to sample mean $x_i$
$x_1in B(mu,2sigma):sigmaapproxfrac{sigma_1}{sqrt{n_1-1}}rightarrow$$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
asked Jan 4 at 7:37
FrankFrank
16210
asked Jan 4 at 7:37
FrankFrank
16210
asked Jan 4 at 7:37
FrankFrank
16210
asked Jan 4 at 7:37
FrankFrank
16210
asked Jan 4 at 7:37
FrankFrank
16210
16210
add a comment |
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