In the figure 19-17 we see there are ten strains survived out of 18 samples and other eight strains died in the high temperatures.
To perform a chi-square test the samples are very small in size. With small sized samples, the expected and the observed outcome should differ greatly to get significant results.
If we take SNP (single nucleotide polymorphism) 1, if it does not have any effect on the traits, there will be random distribution of the 10 survivors between A and G variants. Here in this case, there are 12 A strains and 6 G strains. So, the A strain will be two parts out of three (2/3). The G strain will be 1/3 as there are 18 strains totally. So, out of 10 strains, which survived, 6.67 would be A strain and 3.33 would be G strain.
In this figure 19-17, there are 7 A survived strains and 3 G survived strains. This observed outcome is almost the same as the expected outcome. So, the null hypothesis can be accepted in this case.
The chi-square formula is ? (OE)2/E.
For this data, the chi-square test is 0.049 with 1 df, P>0.05. So, for this sample size, the data set comes within the normal variation range.
If we take SNP 6, there is equal number of A strain and G strain. So out of 10 survived strains 5 would be A strain and 5 would be G strain. This is the expected outcome. The observed outcome is, in G there were 8 and in A strain there were only 2.
If we use this chi-square formula with the observed and expected values, it is 3.6. With 1 df, the critical value of chi-square is 3.81. This comes within the normal variation range and it is close to significance. In SNP 6 we see that G strain appears to survive at high temperatures. As the sample size is small, the null hypothesis cannot be rejected. A large sample size is needed to resolve a difference of this magnitude.