How to report the ratio of two sets of experimental results? I have measured the time taken to solve a problem by algorithm $X$ and by algorithm $Y$. It takes a quite long time, so I have only 10 data for each algorithm:
$$
X : ( x_1, x_2, \dots  , x_{10}) \\
Y : ( y_1, y_2, \dots  , y_{10})
$$
EDIT:
The problem they are solving is randomized. I generated 10 instances of the problem using 10 different random seeds. The 10 computational times correspond to these 10 problem instances. In this sense, the data are paired.
The change of a seed does not change the difficulty of the problem very much.
END OF EDIT
I have computed the ratio of the averages:
$$
avg =  \frac{\sum_{k=1}^{10} x_k }{\sum_{k=1}^{10} y_k }
$$
This however, does not convey any information on how precise the ratio is. 
One possible way is to estimate the standard deviation.
According to this answer, the average of iid random variables is asymptotically normal and therefore the ratio has asymptotical Cauchy distribution, whose standard deviation is infinite. This does not satisfy me, especially since I have only 10 data.
Then according to this answer I should approximate the standard deviation using Taylor series. This answer looks better, but it still doesn't feel right. 
The distribution of a ratio is intuitively highly assymetrical around 1. (you have only the interval 
$(0; 1)$ to capture the fact that algorithm $X$ is faster, but the entire $(1 ; \infty)$ to capture the fact that $Y$ is faster). So even a well estimated standard deviation can be of little use.
It would be better to provide some sort of confidence interval. For example: the ratio is 1,5 with an assymetrical confidence interval of (1,3 ; 2,8). But I have no idea how to estimate this since I do not know the distribution of my data.
EDIT2:
Here are my data:
X       Y
111536  160134
111165  164850
112494  165844
115959  166409
121296  161755
119948  167781
119172  168666
117330  169766
116661  166518
129311  169884

EDIT3:
To answer the question (in comment) of D L Dahly 
why not just report that one algorithm is faster in all instances
For the brevity of the question I did not mention that I have actually 84 sets of data described in this question. 2 problems x 6 dimensions of the problem x 7 possible sizes of the problem. In some instances X is faster, in some Y is faster and in some instances the results are inconclusive.
I do not necessarily need confidence intervals or standard deviations. I just want to provide the reader with something richer than just averages. The reader should have a sense of how well the average represents the experimental results.
 A: I would symmetrize the problem and recognize the matching by working with the logs of the individual ratios, say $z_i = \ln(x_i/y_i)$, getting the limits of a $100(1-\alpha)$% confidence interval for the mean of $z$ the usual way as $\bar{z} \pm t_{9,1-\alpha/2}\,s_z/\sqrt{10}$. (I know it's not strictly justified, but with such a small $n$ I prefer it to the bootstrap.)
A: You ask a very interesting question. The key problem is, as you state, that the theoretical distribution of both $X$ and $Y$ is unknown. If it was known, however, it might be possible to derive the variance of the ratio and thus find a sample estimate of the standard error. 
Suppose for a moment that both random variables follow a known distribution. As you noted, the normal distribution is a possibility, so that following the central limit theory the ratio is Cauchy distributed. I also think that response times to solve tasks are sometimes modelled by exponential distributions. Therefore, one could also assume the r.v. $X$ and $Y$ are exponentially distributed and their sum is hypoexponential.
More generally, the ratio of $sum(X)$ and $sum(Y)$ is ratio distributed. It is a known problem with ratio distributions, unfortunately, that they often do neither have an existing expectation (mean) nor variance. Therefore the s.e. of the mean often does not exist. This is also the case for the Cauchy distribution and the ratio of two exponential variables, as it is for other ratio distributions.
Fortunately, there are also ratios of distributions that have well-defined means and variances. In the following, I will assume the population mean of the ratio exists and construct an example based on this assumption.
In this case, you still do not know the distribution of your r.v. in practice. One option to get to the s.e. of the mean then is by non-parametric bootstrap, which I will demonstrate by example.
Suppose $X$ and $Y$ follow a scaled chi-square distribution with 1 and 5 degrees of freedom respectively. Then the ratio $sum(X)/sum(Y)$ is F-distributed with 1*n and 5*n degrees of freedom, where n is the number of summed r.v.. In practice n is the sample size.
n=10^3
df1=1
df2=5
X<-rchisq(n,df=df1)/(df1) #Scaled Chi-square distribution with df=df1
Y<-rchisq(n,df=df2)/(df2) #Scaled Chi-square distribution with df=df2
ratio<-sum(X)/sum(Y) # F-distributed with df1*n and df2*n degrees of freedom

You may verify that the mean of the F-distribution is known.
ratio #sample mean of ratio
df2*n/(df2*n-2) #theoretical mean of ratio (mean of F-distribution with df1 and df2)

Now suppose we have a small $n=10$ sized sample from the same distribution. 
n=10
df1=1
df2=5
X<-rchisq(n,df=df1)/(df1) #Scaled Chi-square distribution with df=df1
Y<-rchisq(n,df=df2)/(df2) #Scaled Chi-square distribution with df=df2
ratio_sample<-sum(X)/sum(Y) # F-distributed
df2*n/(df2*n-2) #theoretical mean of ratio (mean of F-distribution with df1 and df2)

The bootstrap procedure samples with replacement from the data. I will draw 10,000 samples of size 10, respectively. I estimate the mean ratio, variance and s.d. of the bootstrapped distribution. The latter provides the s.e..
boot=10^4
data<-data.frame(X,Y)
bootsamples<-numeric()
for(i in 1:boot){
  temp <- data[sample(n,n,replace=T),]
  bootsamples[i]<-sum(temp$X)/sum(temp$Y)
  }
ratio_var<-var(bootsamples)
ratio_se<-sqrt(ratio_var)
ratio_mean<-mean(bootsamples)

To summarize the results we can consider the classical confidence interval based on normal theory, but this is not immediately advisable due to the small sample size.
c(ratio_mean-1.965*ratio_se,ratio_mean+1.965*ratio_se) #Classical 95% CI based on asymptotics

Alternatively you may consider the 2.5 and 97.5 percentile of the bootstrapped distribution of ratios.
quantile(bootsamples,probs=c(.025,.975)) #Bootstrapped 95% CI

You may verify again that the bootstrapped confidence interval covers the true mean.
df2*n/(df2*n-2) #True mean

Again, I should stress that the bootstrap will only work, if the expectation of the ratio and its variance exist, which is not certain for ratios. In that case the s.e. of your ratio $avg$ would not exist either and the problem could not be solved.
A: Why doesn't the Taylor expansion look right?  
If you want a symmetric statistic, you can try look at the difference instead $\bar{x}-\bar{y}$, it is easy enough to work out the variance of $\bar{x}-\bar{y}$ with any distribution (not just normal) for $X, Y$, provided the variance exist (uniform??). This difference should be symmetric around 0. And you can use t-test to do the test.
Back to the ratio, if you really want to stick to the ratio, you can use the permutation test to work out the interval estimate. In R, it would look something like:
N=10
x=runif(N,1,3)  # your data x
y=runif(N,10,30)   # your data y
ratio=mean(x)/mean(y)
NP=100
stat=rep(NA,NP)
for(i in 1:NP){
  id<-sample.int(2*N,size=N,replace=F)
  stat[i]=mean(c(x,y)[id])/mean(c(x,y)[-id])
}
CI=quantile(stat,c(0.025,0.975))
print(CI);print(ratio)
(ratio<CI[2])&(ratio>CI[1])

A: Given that you have so few data points, it hardly makes sense to use all those statistical assumptions.. why not just report the standard statistics: mean x_i/y_i, median x_i/y_i, percentile etc.
