I'm going to take you at your word that the population really is a population in the right sense.
I'm more interested in comparing the mean of the smaller sample with the mean of the total population.
If you know the distribution of the total population, you know its mean. So you're in the position of trying to compare a sample mean to a population mean. Your sample is of size 348. This is a one sample test, not a two sample.
Points (corresponding to your numbering):
1) The CLT should give you approximate normality of the sample mean, unless the distribution is pretty extreme (heavily skewed, for example -- you can check the approximate normality of the distribution of the sample mean in several ways); by the same token, you should be able to ignore the uncertainty in the standard deviation. You don't need to worry about whether the one sample t-statistic has a t-distribution because a one sample z-statistic (basically the same quantity but taking $s$ to be $\sigma$) should be very close to a z-distribution.
2) The relevant rank test is not a Wilcoxon rank-sum test because you know the population. The one-sample equivalent is a Wilcoxon signed-rank test, but that assumes symmetry; it's also not testing the mean, but a slightly different measure of location. If you don't have near symmetry the obvious rank based one sample test would be the sign test, which is also not testing the mean. IF you consider only shift alternatives and you assume the only difference between sample and population is that possible shift, you could test against the population equivalent of the locations that they do test for, and if they're different, by direct implication the means differ by the same amount. In the case of the sign test, that would mean testing against the population median in order to also test for a difference in means. However, I suspect that the required assumption of identical shapes apart from location shift is not likely to be tenable - let me know if you think shift-only alternatives make sense.
4) Resampling approaches (either using randomization or bootstrapping) are possible.
Lets discuss some randomization tests:
a) If you could assume symmetry, a test immediately suggests itself, based on the same randomization idea that the Wilcoxon signed rank test uses, but based on the data, not the ranks - that is you take away the population mean from the sample values, count how many + and - signs there are, and randomly reassign the signs to those differences to get the randomization distribution.
b) You could randomly sample 348 values from the whole (6857+348) combined sample-plus-other-population and calculate the distribution of means, comparing your mean with that distribution (I'd lean toward this).
TL;DR:
You can’t.
What is typically done
The current “state of the art” in determining whether a network is a small world uses the following approach:
Calculate the mean shortest path length $L$ and the clustering coefficient $C$ of your network.
Generate an appropriate ensemble of null-model networks, such as Erdős–Rényi random graphs, or Maslov–Sneppen random graphs.
Calculate the average of the mean shortest path length $L_\text{r}$ over this ensemble of null-model networks; calculate $C_\text{r}$ analogously.
Calculate the normalised shortest path $λ := L/L_\text{r}$. and $γ := C/C_\text{r}$.
If $λ$ and $γ$ fulfil certain criteria (e.g., $λ≈1$ and $γ>1$), call the network a small-world network.
The idea behind this is that:
Small-world networks should have some spatial structure, which is reflected by a high clustering coefficient. By contrast, random networks have no such structure and a low clustering coefficient.
Small-world networks are efficient in communicating and similar and thus have a small shortest path length, comparable to that of random networks. By contrast, purely spatial networks have a high shortest path length.
Where the problems are
This does not say anything about how the mean shortest path scales with the network size. In fact, for real networks, the entire definition you quoted cannot be applied, as there is no such thing as the same network with a different number of nodes.
Suppose, we take some other definition of a small world that is not directly based on the values of $λ$ and $γ$, e.g.:
A small-world network is a spatial network with added long-range connections.
Then we still cannot make robust implications as to whether such a definition is fulfilled just using $λ$ and $γ$ (or in fact other network measures). The interpretation of many studies assumes that all networks are a realisation of the Watts–Strogatz model for some rewiring probability, which is not justified at all: We know many other network models, whose realisations are entirely different from the Watts–Strogatz model.
The above method is not robust to measurement errors. Small errors when establishing a network from measurements suffice to make, e.g., a lattice look like a small-world network, see e.g., Bialonski et al., Chaos (2010) and Papo et al., Front. Hum. Neurosci. (2016). In fact, I am not aware of a single study that claims that some empirical network is not a small-world network.
Sidenote: What would you gain?
I am not aware of any useful insight that can be derived from some network being a small world. The claim that some type of network is well described by a certain network model (e.g., the Watts–Strogatz model) may be useful for modelling studies, but that’s going much further than just claiming small-world-ness.
Full disclaimer: One of the above papers is from my direct academic vicinity.
Best Answer
I think you should make a bootstrapped confidence interval. Rank your 1000 randomly sampled asp distances. Your CI is from the 25th smallest to the 25th largest interval. If the asp for your n nodes of interest lies outside of the interval, then the asp for your n nodes is significantly different from that expected from a random selection of n nodes.
Alternatively, if you want a rough p-value, rank your asp of interest among the 1000 randomly sampled asps. If it is above the median, divide it's rank by 500. If it's below the median, subtract it's rank from 500, and divide the result by 500. The resulting number is your p-value.