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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?

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    $\begingroup$ 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. $\endgroup$
    – Frans Rodenburg
    Commented Jun 3, 2019 at 0:27
  • $\begingroup$ If the parameter is a fixed value that represents the value for the population, then how does it have a standard deviation? $\endgroup$
    – ivan
    Commented Jun 3, 2019 at 0:35
  • $\begingroup$ 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). $\endgroup$
    – BruceET
    Commented Jun 3, 2019 at 3:04
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    $\begingroup$ 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. $\endgroup$
    – Frans Rodenburg
    Commented Jun 3, 2019 at 3:05
  • $\begingroup$ 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? $\endgroup$
    – whuber ♦
    Commented Jun 3, 2019 at 13:40

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