Worked example 1
We will base our first example on a trial of a deworming drug against cyathostome worms in horses. The researchers infected each of twelve pony foals with 150,000 thirdstage cyathostome larvae. Twentyone days after infection the animals were assigned to replicates on the basis of their bodyweight. One animal from each replicate was then given a dose of ivermectin. The other animal of each replicate was left untreated as a control.
Adult worms in lumen 
Replicate 
Untreated 
Treated 
Difference 
1 2 3 4 5 6 
30520 20220 24120 14200 17620 15660 
300 0 100 0 0 0 
30220 20220 24020 14200 17620 15660 
Mean difference =  20323 
 
The worm burdens at post mortem examination, 56 days after infection, are given here.
Since observations are paired by weight category, the paired t test is the appropriate statistical test. The null hypothesis is that the treatment has no effect, so the mean difference between paired observations would be zero. The alternative hypothesis is that treatment reduces the number of worms so we will be carrying out a onetailed test.
But do your data meet the assumptions for the paired t test?
For this you need to ask if the differences between treated and untreated are normally distributed. We have assessed this using QQplots (rankit plots would have been a better choice given the small sample size, but they are not available on most software packages)
The first plot suggests that the distribution is skewed. This was expected given that parasite numbers are generally overdispersed. Square root (√(Y+0.5)) and log (log_{10}(Y+1)) transformations indicate that only the square root transformation will normalize this distribution  a log transformation will actually make the situation worse! This is problematic because the confidence interval for a difference cannot be interpreted on the detransformed scale for a square root transformation. Hence since the deviation from normality is not very great, we will carry out the analysis on both the untransformed data and square root transformed data.
Estimate t with (n − 1) degrees of freedom by subtracting the known parametric mean (in this case zero) from the sample mean, and standardizing the difference by dividing by the standard error.
t_{df = 5} = 
20323 − 0 
= 8.34 

5969.4/√6   
A onetailed comparison gives a Pvalue of 0.0002. If we apply the same test to square root transformed differences we get an even lower Pvalue of 0.000001.
Hence our result with untransformed data is conservative. Notice that, if we double these Pvalues, both 2tailed tests are still highly significant  and 2tailed tests are less controversial than 1tailed tests.
To avoid needless controversy, let us calculate 2sided 95% confidence limits  on the basis that each limit is equivalent to a 1sided 97.5% limit, and compatible with our 1tailed alternate hypothesis (treatment reduces burden).
The normal approximation 95% confidence limits of the untransformed difference (20323) are given by:
20323 − (2.571 × 2437) = 14059
20323 + (2.571 × 2437) = 26588
The lower limit of which is equivalent to a onesided 97.5% limit of 14059 (compared to the 1sided 95% limit of 15412.65, given above).
None of our differences being negative, a squareroot transformation is reasonable. The detransformed mean difference between pairs is 19978, and its detransformed normal approximation 95% confidence limits are 14396 to 26472.
In conclusion we can say the ivermectin treatment was associated with a highly significant reduction in the number of worms.
Although this statistical test is valid, it is only one way of looking at the data. We could for example use a quite different variable here, for example the proportion of ponies with any worms (a dichotomous variable). Here we would be comparing a proportion of 1.0 (for untreated) with 0.33 (for treated), a less dramatic improvement than if we use number of worms as the variable.