Provided that each visit has an equal chance of conversion and the visits are independent of one another, you could compute a (Wald) confidence interval using the formula: $$\hat{p}\pm{z_{1-\alpha/2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

where $\hat{p}$ is the proportion in your sample, $\alpha$ is the desired confidence level and $n$ is the sample size. For example if you want a 95% confidence interval then $z_{1-\alpha/2} = 1.96$ so the interval is:

$$ 0.01664 \pm{1.96}\sqrt{\frac{0.01664(1-0.01664)}{1382}}$$

$$ = (0.010, 0.023)$$

This has the interpretation that, if the sampling was repeated, then, on average, 95 times out of 100, the calculated interval would contain the true (population) value. That is, the interval will contain the true proportion with 95% probability if repeated a large number of times. A less technical interpretation is that we are 95% confident that the true population parameter is within the calculated interval.

Provided that each visit has an equal chance of conversion and the visits are independent of one another, you could compute a (Wald) confidence interval using the formula: $$\hat{p}\pm{z_{1-\alpha/2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

where $\hat{p}$ is the proportion in your sample, $z_{1-\alpha/2}$ is the standardized normal distribution critical value for a probability of $\alpha/2$ in each tail, $\alpha$ is the desired confidence level and $n$ is the sample size. For example if you want a 95% confidence interval then $z_{1-\alpha/2} = 1.96$ so the interval is:

$$ 0.01664 \pm{1.96}\sqrt{\frac{0.01664(1-0.01664)}{1382}}$$

$$ = (0.010, 0.023)$$

This has the interpretation that, if the sampling was repeated, then, on average, 95 times out of 100, the calculated interval would contain the true (population) value. That is, the interval will contain the true proportion with 95% probability if repeated a large number of times. A less technical interpretation is that we are 95% confident that the true population parameter is within the calculated interval.