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My question is: what is the correct value to use for "$n$" in formulas pertaining to the mean and variance of the sampling distribution of the sample mean.

Let's say we are taking $25$ samples from a population $64$ times. And we want to calculate $\mu_{\bar{X}}$

I would think we would estimate for each sample of $25$ elements $\bar{X}_k=\frac{\Sigma_{i=1}^{25}X_i}{25}$. And then $\mu_{\bar{X}}=\frac{\Sigma_{k=1}^{64}\bar{X}_k}{64}$

Similarly, for the variance, $\sigma_{\bar{X}}= \sqrt{\frac{\Sigma_{k=1}^{64}(\bar{X}_k- \mu_{\bar{X}})^2}{64}}$

What makes me skeptical that this is correct is that all the formulas for these parameter estimates have an "$n$" which seems to refer to the sample size. And In the context of what I'm asking, is that $25$ or $64$.

In particular in the relation: $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$ is $n=25$ or $64$, or maybe even $25\times 64= 1600$.

Thanks

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  • $\begingroup$ $$\mu_{\bar{X}}=\mathbb{E}(\bar{X})=\mu_{X}}$$ $\endgroup$
    – Karl
    Apr 10 '20 at 18:53
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    $\begingroup$ I'm finding the notation you are using confusing. Why have used $\bar{X}_i$? I can't understand what for example $\bar{X}_{2}$ would mean according to your definition. Id have written $\bar{X}_{25}$ to indicate sample size $25$ for your $\bar{X}_{I}$. It's a good question but I'm not sure your notation really makes sense. $\endgroup$
    – Karl
    Apr 10 '20 at 19:45
  • $\begingroup$ @Karl Each $\bar{X_i}$ is the mean of one of $64$ samplings of the population, each consisting of $25$ samples. Each element in that sample is an $X_i$. But I see your point, keeping track of the indices. I'll fix it. $\endgroup$
    – user663837
    Apr 10 '20 at 20:14
  • $\begingroup$ I think this is a great question. It forces you to think exactly about the definitions. +1 $\endgroup$
    – Karl
    Apr 10 '20 at 20:17
  • $\begingroup$ I'd write $Y=\bar{X_{25}$ a random variable for the mean of a sample size $25$ so you want $\mathbb{\bar{Y_60}}$ that is a sample of size 60 from the sample means of sample size 25. Hope this makes sense. $\endgroup$
    – Karl
    Apr 10 '20 at 20:25
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First of all, you are right with you answer. I think your confusion Côme from the fact you have two distinct situation.

First situation: you take sample of 25 induviduals from a population. Here $n=25$ and you used it to find $\bar X_k$.

Second situation: you have 64 of these d'amples. Here $n=64$. You used it to evalutate $\mu_{\bar X}$.

For your standard deviation $\sigma_{\bar X}$, you are dealing with values from the second situation, so $n=64$.

It is always a lot of fun to deal with those multileveled questions. It is important to label your variable to distinguish them.

EDIT OP added

In particular relating the actual variance of the population to the sample error of the mean: $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$ is $n=25$ or $64$?

Once again, it is a matter te define variable properly.

Level $0$: the population has a mean $\mu$ and a standard deviation of $\sigma$.

Level $1$: sample of size $25$. For each sample, you will have a mean of $\bar X_k$ and a standard deviation of $\sigma_k$ (the standard deviation of the sample).

The expected mean distribution has a mean of $E[\bar X_k]=\mu$ (same has population) and a standard deviation of $$\sqrt{\mathrm{Var}[\bar X_k]}=\frac{\sigma}{\sqrt{25}}$$

Level $2$: $64$ samples. You took $64$ means of samples evaluate the mean.

The expected mean of the value of the level $2$ mean is $$E[\bar X]=E[\bar X_k]=\mu$$ And the variance will be $$\sqrt{\mathrm{Var}[\bar X]}=\frac{\sqrt{\mathrm{Var}[\bar X_k]}}{\sqrt{64}}=\frac{\sigma}{\sqrt{25}\sqrt{64}}$$


TL;DR Now to answer about your $$\sigma_{\bar X}=\sqrt{\frac{\sum_{k=1}^{64}(\bar X_k -\mu_{\bar X})^2}{64}}$$ It is the standard deviation of your $64$ sample means, it refer to the expected distribution of the level $1$. So it should be around $\frac{\sigma}{\sqrt{25}}$.

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  • $\begingroup$ This is a nice answer but what is $\bar{X_i}$ exactly? As I understand only $$\sqrt{\mathrm{Var}[\bar X_25]}=\frac{\sigma}{\sqrt{25}}$$. $\endgroup$
    – Karl
    Apr 10 '20 at 20:09
  • $\begingroup$ I just notice OP uses $\bax X_k$. I'll update. As I understand it, it is the mean of each of first sample. $\endgroup$ Apr 10 '20 at 20:22
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Be careful. (This is not quite an answer, because it's not clear what your exact question is.)

The number $n$ represents sample size, but your question involves samples of different sizes from different distributions. Specifically, you talk about

  1. Samples of size $25$ from a distribution.
  2. A sample of size $64$ from a different distribution (the distribution of sample means for samples of size $25$ taken from your first distribution).

To ask “What is $n$?” when you are talking about two distributions and two sample sizes is impossible to answer.

Always be careful with your language. Since $\mu$ represents the population mean, you can never “calculate $\mu_{\bar X}$” from sample data, you can only estimate it. I think you know this, but you still shouldn't say “calculate $\mu_{\bar X}$

You also mention “the sample error of the mean.” I don't know what that is. It isn’t a clear description of a number. If you are extremely careful about language, things are sometimes clearer.

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  • $\begingroup$ Thank you, you are absolutely right in your suggestions and corrections. Estimate is not calculate. And it should be the standard error of the mean. With regards, $\endgroup$
    – user663837
    Apr 11 '20 at 1:01
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I think notation is key here so hopefully I've understood and not messed up myself.

Let $X$ be a random variable with $\mathbb{E}(X)=\mu$ and $Var(X)=\sigma^2$.

To deal with the sample of size 25:

Let $\bar{X}_{25}=\frac{X_1+...+X_{25}}{25}$ then $\mathbb{E}(\bar{X}_{25})=\mu$ and $Var(\bar{X}_{25})=\frac{\sigma^2}{25}$

Here each of the $X_i$ are identical and independently distributed random variables. For example $X_3$ is the 3rd observation in the sample of 25 etc.

Note that each $X_i$ is a random variable that is just $X$ in disguise. Hence $\bar{X}_{25}$ is a random variable.

The expectation and variance of $\bar{X}_{25}$ can be easily found using the algebra rules for expectation and variance.

Now for the second part. Repeating the sample 64 times.

Let $Y=\bar{X}_{25}$ then $Y$ is a random variable with $\mathbb{E}(Y)=\mu$ and $Var(Y)=\frac{\sigma^2}{25}$

Let $\bar{Y}_{64}=\frac{Y_1+..+Y_{64}}{64}$ and so $\bar{Y}_{64}$ is a random variable in a similar way to before.

We now have $\mathbb{E}(\bar{Y}_{64})=\mu$ and $Var(\bar{Y}_{64})=\frac{\sigma^2}{25\times64}$

This deals with repeating 64 times.

Hope this helps.

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  • $\begingroup$ The problem, and what motivated my question is that $\mu$, the population mean, is not necessarily equal to the mean of just one sample. It's better approximated by the mean of the means of several (the more the better) samples. The rest follows. Maybe that's what you intend. So (I'm certainly no authority) I would reconsider your line beginning "Let." With regards, $\endgroup$
    – user663837
    Apr 10 '20 at 21:19
  • $\begingroup$ I don't understand. I defined a random variable which is the mean of a sample size 25. I then found its expected value. $\endgroup$
    – Karl
    Apr 10 '20 at 21:24
  • $\begingroup$ Suppose I defined $\bar{Y}_n$ as $\frac{Y_1+..+Y_n}{n} then $Var(\bar{Y}_n=0$ as $n\to \infty$ then the variance tends to zero. $\endgroup$
    – Karl
    Apr 10 '20 at 21:42

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