Why do we divide by the standard error when we evaluate sample statistics?

I'm reading Experimental Design and Data Analysis for Biologists by Professor Quinn and Profesoor Keough and, on page 20, they write -

We can use the methods just described to reliably determine standard errors for statistics (and confidence intervals for the associated parameters) from a range of analyses that assume normality, e.g. regression coefficients. These statistics, when divided by their standard error, follow a t distri- bution and, as such, confidence intervals can be determined for these statistics (confidence interval = t * standard error).

What I understand is two things:

a) Our sample statistics follows is normally distributed, because of the central limit theorem, but we need to use the t-distribution because we don't know the standard deviation of the parameter.

b) We use the t-distribution to calculate the confidence interval so that we capture 95%, or some percent, of the possible values for the parameter.

I have two questions: Is that accurate and, why do we divide our statistic by the standard error? Why not just use the distribution of the sample statistic?

• You divide by the standard error $\text{SE} = \frac{s}{\sqrt{n}}$ because you want a measure of significance that increases with sample size and decreases with variance. However, I am not sure I follow your question: How would you use the distribution of the sample statistic? The $t$-distribution is the distribution of the $t$-value, which is the difference in means divided by the standard error. – Frans Rodenburg Jun 3 at 0:27
• If the parameter is a fixed value that represents the value for the population, then how does it have a standard deviation? – ivan Jun 3 at 0:35
• If you want a statistic to be normally or t-distributed, then $Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}}$ and $T = \frac{\bar X - \mu}{S/\sqrt{n}}$ are often reasonable choices. // It's estimators of parameters that have distributions, not the parameters themselves (unless you're taking a Bayesian approach). – BruceET Jun 3 at 3:04
• The $t$-value is a random variable, a function of your sample. The parameter is the difference in means you're estimating. Since you only have an estimate, it comes with uncertainty, expressed through the standard error. – Frans Rodenburg Jun 3 at 3:05
• My dog weighs $4.$ Is she large or small? It depends on whether the $4$ is in units of pounds, kilograms, stones, or something else. Likewise, when you are examining any sample statistic, its numerical value is meaningless without a similar standard unit to serve as a basis for comparison. If the number is meaningless, how useful could its distribution possibly be? – whuber Jun 3 at 13:40