It appears that your data can only take on positive values. In this case, the hypothesis of normality is often rejected. Normally distributed random variables range from positive to negative infinity, so only positive values would violate this. You could try taking the log of the observations and seeing whether these are normally distributed.
If your data follow a normal distribution, then the points in your QQ-plot should lie on a 45-degree line through the origin. Your plots do not look like that at all.
The KS test is giving an error because the distributions being tested are presumed to be continuous. In this case, the probability of witnessing two observations with the exact same value is 0. Your data set contains ties, invalidating this assumption. When there are ties, an asymptotic approximation is used (you can read about this in the help file). The error that you are receiving has nothing to do with data sets with different sizes.
In your post, you never specified the question that you are trying to answer--with sufficient precision, anyway. Do you really want to test that the distributions are the same? Would it be sufficient to test that the means are the same?
Unless you are willing to assume that the variables follow some distribution, there isn't much of an alternative to the KS test if you want to test for the distributions being the same. But there are several ways to test for differences in means.
You're asking for something like an effect size (A "how big?" type question).
P-values don't measure that; at a given value of W, the p-value tends to go down as n goes up.
The Shapiro-Wilk statistic, W, is in some sense a measure of "closeness to what you'd expect to see with normality", akin to a squared correlation (if I recall correctly, the closely related Shapiro-Francia test is actually a squared correlation between the data and the normal scores, while the Shapiro Wilk tends to be slightly larger; I seem to recall that it takes into account correlations between order statistics).
Specifically values closer to 1 indicate "closer to what you'd expect if the distribution the data were drawn from is normal".
However, keep in mind it's a random variable; samples can exhibit random fluctuations that don't represent their populations, and summary statistics will follow suit.
It's not immediately clear that it necessarily makes sense to compare Shapiro-Wilk statistics across data-sets in order to declare one set "more normal" than another; even less so with very different variables and different sample sizes.
Further, choosing the one closest to 1 among a collection of samples may actually be choosing something other than values randomly selected from a normal distribution, for a variety of reasons. For example, goodness of fit tests generally tend to be biased tests; what makes their criterion "closest" isn't necessarily the thing the test is actually designed to pick up. (I don't know what sorts of small-sample biases the Shapiro-Wilk specifically may have, however.)
Finally, I don't see any useful point to such an exercise. What possible value can there be in such a procedure?
Best Answer
You can't really even compare the two since the Kolmogorov-Smirnov is for a completely specified distribution (so if you're testing normality, you must specify the mean and variance; they can't be estimated from the data*), while the Shapiro-Wilk is for normality, with unspecified mean and variance.
* you also can't standardize by using estimated parameters and test for standard normal; that's actually the same thing.
One way to compare would be to supplement the Shapiro-Wilk with a test for specified mean and variance in a normal (combining the tests in some manner), or by having the KS tables adjusted for the parameter estimation (but then it's no longer distribution-free).
There is such a test (equivalent to the Kolmogorov-Smirnov with estimated parameters) - the Lilliefors test; the normality-test version could be validly compared to the Shapiro-Wilk (and will generally have lower power). More competitive is the Anderson-Darling test (which must also be adjusted for parameter estimation for a comparison to be valid).
As for what they test - the KS test (and the Lilliefors) looks at the largest difference between the empirical CDF and the specified distribution, while the Shapiro Wilk effectively compares two estimates of variance; the closely related Shapiro-Francia can be regarded as a monotonic function of the squared correlation in a Q-Q plot; if I recall correctly, the Shapiro-Wilk also takes into account covariances between the order statistics.
Edited to add: While the Shapiro-Wilk nearly always beats the Lilliefors test on alternatives of interest, an example where it doesn't is the $t_{30}$ in medium-large samples ($n>60$-ish). There the Lilliefors has higher power.
[It should be kept in mind that there are many more tests for normality that are available than these.]