In order to estimate population parameters, it is important to have measures of how good the estimates are likely to be. For example, how well the sample mean is likely to represent the true population mean.
A precise measure defined to measure the degree of spread (standard deviation) in a set of sample statistics (e.g. means) calculated from multiple samples is called standard error. But it can also be estimated in the frame of a single sample, so the corresponding counterpart in the one-sample statistics would be $\frac{\sigma}{n^{\frac{1}{2}}} ,\,\,$ where $\sigma$ is the sample standard deviation and $n$ the sample size.
$1.\,$ In accordance with the properties of normal distribution, for example $68.27\,\% $ of the repeated means fall between the true mean and $\pm$ one sample standard deviation. I am wondering, why instead of the sample standard deviation it is not taken into consideration the estimated standard error, $\frac{\sigma}{n^{\frac{1}{2}}} \,\,$ ?
$2.\,$ When one talks about the standard error of the mean, is it about the sample standard deviation, $\sigma, \,\,$the degree of spread in a set of samples, or the estimated population standard deviation out of a single sample, $\frac{\sigma}{n^{\frac{1}{2}}} \,\,$ ?
$3.\,$ In the formula $$P\,\{ \bar y- t_{0.05(n-1)}s_{\bar y}\leq \mu \leq\bar y +t_{0.05(n-1)} s_{\bar y}\}\,\,,$$ where $\mu $ is the unknown polpulation mean, $\bar y$ is the sample mean, what kind of error is $s_{\bar y}$, i.e. is it the sample $\sigma$, maybe $\frac{\sigma}{n^{\frac{1}{2}}},$ or something else ?
Many thanks.
In 1, it is about $68.27\%$ of the observations that are within one standard deviation of the mean. The standard deviation of your estimate of the mean is $\frac {\sigma}{n^{\frac 12}}$. Both are fine formulas, but they are referring to different things. I don't know how you estimate the standard deviation from a single sample.