Why divide by the root n when constructing the u statistic

Question

[octpus616]

For a normal population, the following equation can be used to convert them to a standard normal distribution

if, X ~ N(μ, σ^2).

When we apply hypothesis testing between a normal sample population and a normal population with a konwn σ^2, we use following equation to compute U or z:

The question here is how to explain the difference in the denominator and why the latter is appropriate here. I noted some website (such as statisticshowto) explain this as following:

You always divide by sqrt(n). However, occasionally the square root of n sometimes equals 1 (making it just σ in the denominator. for example, if you are choosing one person and trying to figure out the probability their weight is under x pounds, then n=1. In other words, if you are calculating a z-score, you can always use sqrt(n). (https://www.statisticshowto.com/sigma-sqrt-n-used/)

But this looks like just a few handy tips to remember, right?

I would like to know the correct understanding of this issue, it would be better if it is more general and also rigorous.

Thank your for your attentions.

zhang

Answer 1

[Jean-Karim Heriche]

This is because you're computing the standard error of the mean, i.e. the standard deviation of the estimate of the mean. See the corresponding wikipedia article. In the first case, the standard deviation corresponds to the spread of the sample distribution, e.g. throw a die 1000 times and take the mean of all throws as the estimate of the population mean, whereas in the second case, this is the spread of the estimates of the mean, e.g. throw a die 10 times and take the average, repeat 100 times and estimate the population mean by taking the average of the 100 experiments.

[octpus616]

Hi @Jean-Karim Heriche，

Thanks for your reply, my English is not well, so thank you for understanding what I mean also.

The standard error (SE) of a statistic (usually an estimate of a parameter) is the standard deviation of its sampling distribution. so in fact, the standard deviation of X-bar is standard error of X(i) or samples. Is it right?In the latter expression, we are actually transforming the mean of the sample, so we need the standard error of the sample (σ/sqrt(n)) as the standard deviation of the mean (X-bar), in fact it is σ(X-bar) = σ/sqrt(n).

[Jean-Karim Heriche]

Yes I think you got it.

[octpus616]

Thank you for your reply.