Sampling error vs. measurement error I have a dataset involving sensor measurements (GPS trajectories). I am using this dataset to estimate aggregate statistics such as the total distance travelled and related quantities. My understanding is that I can calculate confidence intervals for these quantities based on the sampling error using the CLT, which would give me a CI for the true population statistic.
However does this interval take into account measurement error, i.e. random error due to the uncertainty in sensor measurements? It seems to me that this would not be the case, if not how would I account for this?
 A: Since I feel you like to go more advanced, I edit my answer. 
Let's assume you have data for one week $\{y_1, y_2, \ldots, y_7\}$. As you pointed out, each data-point consist of two components: the "true" traveled distance per day $d_i$ and measurement error $m_i$. Thus, our model is  $Y = D + M$, where $D$ and $M$ are random variables and $d_i$ and $m_i$ are realisations of those for the $i^{th}$ day.  
In order to obtain inside, you have to make assumptions about the random variables. The standard assumptions are that $D_i \sim N(\mu, \sigma_d)$ and $M \sim N(0, \sigma_m)$. 
What you are interested in are the contributions $\mu$ and $\sigma_d$. However, only the point-estimator
$$
\hat{\mu} = \textrm{E}[Y] = \frac{1}{7}\sum_{i=1}^7 y_i
$$ is "simple" to obtain. In order to estimate the two so called variance components $\{\sigma^2_d, \sigma^2_m\}$ you need to define your set-up first. There exists extensive literature on measurement system analysis and you might want to look into that.
Once you estimated $\sigma_d$ the 95% confidence interval for the distance travelled per day is given by $\hat{\mu} \pm 1.96 \hat{\sigma_d}$.
