Normal distribution confidence intervals and sample standard deviation problem
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$$d(x_1,mu)leqmu+2sigma-(mu-2sigma)=4sigmaapprox4frac{sigma_1}{sqrt{n_1-1}}$
$x_2in B(mu,2sigma):sigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow d(x_2,mu)leq4frac{sigma_2}{sqrt{n_2-1}}$
$d(x_1,x_2)leq d(x_1,mu)+d(x_2,mu)leq 4frac{sigma_1}{sqrt{n_1-1}}+4frac{sigma_2}{sqrt{n_2-1}}$
But, $frac{sigma_1}{sqrt{n_1-1}}approxsigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow frac{sigma_1}{sqrt{n_1-1}}approxfrac{sigma_2}{sqrt{n_2-1}}$
This implies
$d(x_1,x_2)leq 4(frac{sigma_1}{sqrt{n_1-1}}+frac{sigma_1}{sqrt{n_1-1}})= 8frac{sigma_1}{sqrt{n_1-1}}$
This implies that for any $n_2$, if $n_1$ is very large, $d(x_1,x_2)leq0$
This result does not make sense in reality.
real-analysis probability-theory
asked 11 mins ago
Frank
15010
add a comment |
I am confused by this relationship that in a normal distribution of sample means $x_i : x_i$~$N(mu,sigma)$ the sample mean $x_i$ is within $2$ standard deviations of the true mean $mu$ with $95$% confidence.
I am confused because of this second relationship that $sigma approx frac{sigma_i}{sqrt{n_i-1}}$
$n_i$ is the sample size corresponding to sample mean $x_i$
$x_1in B(mu,2sigma):sigmaapproxfrac{sigma_1}{sqrt{n_1-1}}rightarrow$$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 11 mins ago
Frank
15010
add a comment |
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$$d(x_1,mu)leqmu+2sigma-(mu-2sigma)=4sigmaapprox4frac{sigma_1}{sqrt{n_1-1}}$
$x_2in B(mu,2sigma):sigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow d(x_2,mu)leq4frac{sigma_2}{sqrt{n_2-1}}$
$d(x_1,x_2)leq d(x_1,mu)+d(x_2,mu)leq 4frac{sigma_1}{sqrt{n_1-1}}+4frac{sigma_2}{sqrt{n_2-1}}$
But, $frac{sigma_1}{sqrt{n_1-1}}approxsigmaapproxfrac{sigma_2}{sqrt{n_2-1}}rightarrow frac{sigma_1}{sqrt{n_1-1}}approxfrac{sigma_2}{sqrt{n_2-1}}$
This implies
$d(x_1,x_2)leq 4(frac{sigma_1}{sqrt{n_1-1}}+frac{sigma_1}{sqrt{n_1-1}})= 8frac{sigma_1}{sqrt{n_1-1}}$
This implies that for any $n_2$, if $n_1$ is very large, $d(x_1,x_2)leq0$
This result does not make sense in reality.
real-analysis probability-theory
asked 11 mins ago
Frank
15010
I am confused by this relationship that in a normal distribution of sample means $x_i : x_i$~$N(mu,sigma)$ the sample mean $x_i$ is within $2$ standard deviations of the true mean $mu$ with $95$% confidence.
I am confused because of this second relationship that $sigma approx frac{sigma_i}{sqrt{n_i-1}}$
$n_i$ is the sample size corresponding to sample mean $x_i$
$x_1in B(mu,2sigma):sigmaapproxfrac{sigma_1}{sqrt{n_1-1}}rightarrow$$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 11 mins ago
Frank
15010
asked 11 mins ago
Frank
15010
asked 11 mins ago
Frank
15010
asked 11 mins ago
Frank
15010
asked 11 mins ago
Frank
15010
15010
add a comment |
add a comment |
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