How to understand t-value in R's lm()?

Question

It's similar to the post Interpretation of R's lm() output.

lm(formula = iris$Sepal.Width ~ iris$Petal.Width)

however, just a point that I can't understand for the explanation of t-value.

It shows that, t-value is the ratio from the first two values

t-value = estimate_mean/std.error

Questions: Is this t-value exactly the t-score in student's t distribution?

Based on my understanding, from the definition, t-score is calculated as follows.

If assuming a null hypothesis that response residual mean is 0, the correct t-score in this lm() case, in my understanding, should be as follows.

 t-score given H_null = estimated_mean / (std.error/sqrt(n)) 
                      = sqrt(n) * estimated_mean/std.error

Therefore, t-score I derived is sqrt(n) times larger than t-value given by lm() .... Any one know which part is wrong above? Thanks!

Answer 1

The equation in the orange image is for a single sample. If a sample is drawn from a normal distribution with mean $\mu$ and standard deviation $\sigma$, then $\bar{x}$ is normal with mean $\mu$ and variance $\sigma/\sqrt{n}$.

Hence $\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ will have a standard normal distribution. However, if we estimate $\sigma/\sqrt{n}$ from the sample, by the estimated standard error of the mean, $s/\sqrt{n}$, (where $s$ is the standard deviation of the sample), then

$\frac{\bar{x}-\mu}{std.err(\bar{x})}=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ has a $t$ distribution with $n-1$ degrees of freedom.

The orange image is consistent with what I just said.

But then you said:

t-score given H_null = estimated_mean / (std.error/sqrt(n))

This is not correct. Note that the standard error of the mean is $s/\sqrt{n}$, where $s$ is the standard deviation of the data.

The t-statistic in regression is slightly different (though analogous in form; its $t=\frac{b-\beta}{s.e.(b)}$ where $b$ is the estimated coefficient).

It's important to understand the problem here with the more basic case first.