in some exam past papers I have been doing I have come across the statistics equation
z=(sample mean - mean)/standard deviation
as well as the equation
z = (sample mean - mean)/(standard deviation/sqrt n).
could anyone please explain when to use which equation and what the difference is? thanks so much!
On
The $z$-score calculation is designed to answer the question: "How far from typical is this result?"
When dealing with a single datum, the single-value formula $z=\frac{x-\mu}{\sigma}$ gives us the answer. But when dealing with a whole pile of individual measurements, we expect the Law of Large Numbers, and its more formal cousin the Central Limit Theorem, to take over: more attempts means the mean of our results is going to look more like the population mean. To model this, we divide the usual standard deviation by $\sqrt{n}$: $z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}$.
On
First, Dilip's suggestion should be heeded - without context there is no clear answer to your question. However, in general, the standard deviation in the denominator should match the standard deviation of the variable being normalized. otherwise a Z-score will not be approximated by a standard normal.
I think it is possible, that the two are identical, but with different notation:
$Z=\frac{\bar X - \mu}{\sigma_{\bar X}}=\sqrt{n}\frac{\bar X - \mu}{\sigma_{X}}$.
Note that the first standard dev. is of the estimator while the second is of the data.
On
Suppose $X \sim \mathrm{Normal}(\mu,\sigma)$ (mean $\mu$ and standard deviation $\sigma$). Then $Z = \dfrac{X-\mu}{\sigma} \sim \mathrm{Normal}(0,1)$. This much should be clear. Now suppose we observe $n$ independent and identically distributed values drawn from $X$; i.e., we consider these $x_1, x_2, \ldots, x_n \sim \mathrm{Normal}(\mu,\sigma)$. The distribution of their sum is $\mathrm{Normal}(n \mu, \sigma \sqrt{n})$, since $$\mathrm{E}[X_1 + X_2 + \cdots + X_n] = \mathrm{E}[X_1] + \mathrm{E}[X_2] + \cdots + \mathrm{E}[X_n]$$ by the linearity of expectation, and $$\mathrm{Var}[X_1 + X_2 + \cdots + X_n] \overset{\mathrm{ind}}{=} \mathrm{Var}[X_1] + \mathrm{Var}[X_2] + \cdots + \mathrm{Var}[X_n]$$ when $X_1, \ldots, X_n$ are independent. So the distribution of the sample mean $$\bar x = \frac{1}{n}(x_1 + x_2 + \cdots + x_n) \sim \mathrm{Normal}(\mu, \sigma/\sqrt{n})$$ because $$\mathrm{E}[cX] = c \mathrm{E}[X]$$ and $$\mathrm{Var}[cX] = c^2 \mathrm{Var}[X]$$ for a scalar constant $c$. So if we standardize the sample mean, $$Z = \frac{\bar x - \mu}{\sigma/\sqrt{n}} \sim \mathrm{Normal}(0,1)$$ is the required transformation. Note that if $n = 1$ in the above formula, we recover the standardization for a single observation, described at the beginning of this answer.
On
Simply said by @S2000magician in this thread
If you're dealing with a single observation (i.e. next month's return), then you use σ.
If you're dealing with a mean of observations (i.e. hypothesis test of a sample mean), so σ/√n
When you don't know what to use, ask yourself
“Is this a confidence interval for a single observation, or for the mean of a bunch of observations?”
It is the difference between the $z$ score for a datum from an entire population and a sampling.
The $z$ score for a datum $x$ is $z = (x - \mu)/\sigma$ where $\mu$ is the population mean and $\sigma$ is the population standard deviation.
If the datum $x$ is not from the entire population but rather from a sampling from that population then the standard deviation is divided by the square root of the sample size $n$.