Are $\sigma_{\bar x_{70\%}} = cte * \sigma /\sqrt{N}$, & $\sigma_{\bar x_{70\%}} = cte * \sigma /\sqrt{N}$ valid with non-normal distributions?

added mathjax; edited for greater readability

Source Link

edited Sep 3, 2012 at 19:56

gung - Reinstate Monica

147.5k
89
406
717

smean=spop /√n, but do the equations S70% Are $\sigma_{\bar x_{70\%}} = cte * spop\sigma /√n and Strimmed mean\sqrt{N}$, & $\sigma_{\bar x_{70\%}} = cte * spop\sigma /√n make also sense in\sqrt{N}$ valid with non-normal distributions?

Would you be so kind to help me further. I already looked on the internet, textbooks… But I can’t find the answer to my question.

My working population is non normally distributed and from time to time I need to take a sample of it. Sometimes my sample exists of 100 units, at other times of 150 units.

Of my sample I calculate the mean, $\mu$, and with bootstrapping I determine the sMeanSE, $\sigma_{\bar x}$.

Since sMean = spop /√n$\sigma_{\bar x} = \sigma /\sqrt{N}$, I calculate the spop$\sigma$. So far so good, by using control charts (mean, spop$\mu,\sigma$) I can now see if there is anything happening to worry about.

By use of deviation calculation rules youyou can prove quite easily that the formula sMean = spop /√n $\sigma_{\bar x} = \sigma /\sqrt{N}$ also can be used for non normal distributions.

I play with the idea to make control charts using the trimmed mean and the 70%-value. Now I am not so sure anymore… anymore. (Frankly, I have my doubts wetherwhether you can do this for normal distributions as well.) Can I use the same simple trick to remove the n out of$n$ from my equation?

Let me put it like this: Is what I am writing valid? S70% = cte * spop /√n with$$\sigma_{70\%} = cte * \sigma / \sqrt{N}$$ with a cte$cte$ independent of the value of nN? Strimmed mean = cte * spop /√n with$$\sigma_{\bar x_{70\%}} = cte * \sigma /\sqrt{N}$$ with a cte$cte$ independent of the value of n$N$?

To use my control charts I don’t need to know the value of my cte$cte$. For my s-chart I can just plot out cte* spop$cte* \sigma$ in function of my sample number. But I really need to get rid of the n!!$N$!

I hope you understand what I am getting at. I would already be happy with a yes or a no, you can not do that anymore. If you can direct me to a place where I can find more information about this problem I would be even more gratefullgrateful.

smean=spop /√n, but do the equations S70% = cte * spop /√n and Strimmed mean = cte * spop /√n make also sense in non-normal distributions?

Would you be so kind to help me further. I already looked on the internet, textbooks… But I can’t find the answer to my question.

My working population is non normally distributed and from time to time I need to take a sample of it. Sometimes my sample exists of 100 units, at other times of 150 units.

Of my sample I calculate the mean and with bootstrapping I determine the sMean.

Since sMean = spop /√n I calculate the spop. So far so good, by using control charts (mean, spop) I can now see if there is anything happening to worry about.

By use of deviation calculation rules you can prove quite easily that the formula sMean = spop /√n also can be used for non normal distributions.

I play with the idea to make control charts using the trimmed mean and the 70%-value. Now I am not so sure anymore… (Frankly, I have my doubts wether you can do this for normal distributions as well) Can I use the same simple trick to remove the n out of my equation?

Let me put it like this: Is what I am writing valid? S70% = cte * spop /√n with a cte independent of the value of n? Strimmed mean = cte * spop /√n with a cte independent of the value of n?

To use my control charts I don’t need to know the value of my cte. For my s-chart I can just plot out cte* spop in function of my sample number. But I really need to get rid of the n!!!

I hope you understand what I am getting at. I would already be happy with a yes or a no, you can not do that anymore. If you can direct me to a place where I can find more information about this problem I would be even more gratefull.

Are $\sigma_{\bar x_{70\%}} = cte * \sigma /\sqrt{N}$, & $\sigma_{\bar x_{70\%}} = cte * \sigma /\sqrt{N}$ valid with non-normal distributions?

My working population is non normally distributed and from time to time I need to take a sample of it. Sometimes my sample exists of 100 units, at other times of 150 units. Of my sample I calculate the mean, $\mu$, and with bootstrapping I determine the SE, $\sigma_{\bar x}$. Since $\sigma_{\bar x} = \sigma /\sqrt{N}$, I calculate $\sigma$. So far so good, by using control charts ($\mu,\sigma$) I can now see if there is anything happening to worry about.

By use of deviation calculation rules you can prove quite easily that the formula $\sigma_{\bar x} = \sigma /\sqrt{N}$ also can be used for non normal distributions.

I play with the idea to make control charts using the trimmed mean and the 70%-value. Now I am not so sure anymore. (Frankly, I have my doubts whether you can do this for normal distributions as well.) Can I use the same simple trick to remove the $n$ from my equation?

Let me put it like this: Is what I am writing valid? $$\sigma_{70\%} = cte * \sigma / \sqrt{N}$$ with a $cte$ independent of the value of N? $$\sigma_{\bar x_{70\%}} = cte * \sigma /\sqrt{N}$$ with a $cte$ independent of the value of $N$?

To use my control charts I don’t need to know the value of my $cte$. For my s-chart I can just plot out $cte* \sigma$ in function of my sample number. But I really need to get rid of the $N$!

I hope you understand what I am getting at. I would already be happy with a yes or a no, you can not do that anymore. If you can direct me to a place where I can find more information about this problem I would be even more grateful.

Source Link

asked Sep 3, 2012 at 19:05

siegfried

589
1
5
4