How to understand t-value in R's lm()? 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!
 A: 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.
