In our first worked example we revisit the study by Meyer et al. (1995) on the occurrence of portosystemic shunt in Irish wolfhounds. The authors described the distribution of ammonia concentrations in the venous blood as 'essentially normal', but performed no statistical (or graphical) assessments.
If we had the raw data available, then one of the cumulative rank methods or ShapiroWilks would have been the preferred method of testing normality. Since we only have access to the categorized distribution, we will use Pearson's chi square test  but with the reservation that we are loosing information by categorizing a continuous variable. The first figure below shows observed frequencies with the lowest and highest frequencies pooled in order to eliminate small expected frequencies.
Ammonia 
f_{i} 
_{i} 
<25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 >120 
7 6 15 26 38 43 32 68 69 105 127 121 94 97 85 34 25 21 13 1 17

6.2 6.3 11.2 18.8 29.3 42.9 58.8 75.4 90.4 101.6 106.8 105.1 96.9 83.5 67.4 51.0 36.1 23.9 14.8 8.6 9.0 
{Fig. 1}

The second figure shows a normal distribution with the same mean (73.8), standard deviation (19.4), interval width (5) and number of observations (1044). Expected probabilities were obtained using R by obtaining the cumulative probability up to each upper bound, subtracting the cumulative probability up to the upper bound below that, and multiplying by sample size.
Pearson's X^{2} can then be calculated using the general formula:
Using
X^{2} = 
(7 − 6.205)^{2} 
+ .... 
(17 − 9.002)^{2} 
= 61.28 


6.205 
9.002 
The number of degrees of freedom is given by [number of classes  1  number of parameters estimated] which in this case is [2112] =18. Hence P = 0.00000127. We can therefore reject the null hypothesis and conclude that a normal distribution does not provide a good fit to the data.
If preferred, a likelihood ratio Gtest could be used instead.
Using
G = 2 × [ 7 ln ( 
7 
) 
+ .... 
17 ln ( 
17 
)]
 = 66.6 


6.205 
9.002 
The number of degrees of freedom is again given by [number of classes  1  number of parameters estimated] = 18. Hence P = .000000166. Again we reject the null hypothesis and conclude that observed distribution deviates significantly from a normal distribution.