Relationship between point estimation and confidence interval Suppose I have a toy example for linear regression
set.seed(0)

n_data=50
true_coeff_b1=0.37
noise_ratio=0.13
conf_int_level=0.77

x=runif(n_data)
y=true_coeff_b1*x+noise_ratio*rnorm(n_data)
fit=lm(y~x-1)
summary(fit)
confint(fit)

Whats the relationship between Std. Error and Confidence Interval (as shown in red boxes in the figure?) 

I remember from some book, they are different things, one is point estimation another is a random interval estimation (population parameter is fixed and unknown). But is there any relationship between them?
 A: My attempt to answer (thanks GeoMat22 !!)
Confidence Interval minus point estimation divided by standard error of point estimation will satisfy T distribution with corresponding degree of freedom.
Here is the verification !
> point_est=fit$coefficients
> point_est_se=coef(summary(fit))[, "Std. Error"]
> c95_ci=as.vector(confint(fit))
> t_bnd=(c95_ci-point_est)/point_est_se
> pt(t_bnd,df=49)
[1] 0.025 0.975

OR
> point_est+qt(0.025,df=49)*point_est_se
        x 
0.2859806 
> point_est+qt(0.975,df=49)*point_est_se
        x 
0.3944128 
> c95_ci
[1] 0.2859806 0.3944128

A: In normal/Gaussian distribution, the confidence intervals can be derived as follows:
$$\text{95% CI lower limit} = \text{mean} - 1.96 \times \text{SE}$$
$$\text{95% CI upper limit} = \text{mean} + 1.96 \times \text{SE}$$
In your case, your mean estimate (or better, your estimated mean for the coefficient) is $0.34020$ and your standard error (SE)) is $0.02698$
Plug that into the equation above, and you get (roughly) the same values of your confidence interval. 
I assume (but may be wrong) that the small difference in some of the digits after the decimal are due to the output rounding the model results to certain number of digits, while the confint(fit) probably uses the actual number (with as many digits as it has "inside"). 
