First, we need to calculate the mean and standard deviation for each method.
Enzymatic method:
Mean = (13.1 + 12.7 + 12.6 + 13.3 + 13.3) / 5 = 12.8
Standard Deviation = √((1/(n-1))*((13.1-12.8)^2 + (12.7-12.8)^2 + (12.6-12.8)^2 + (13.3-12.8)^2 + (13.3-12.8)^2)) = 0.27
GC method:
Mean = (13.5 + 13.3 + 13.0 + 12.9 + 13.0) / 5 = 13.14
Standard Deviation = √((1/(n-1))*((13.5-13.14)^2 + (13.3-13.14)^2 + (13.0-13.14)^2 + (12.9-13.14)^2 + (13.0-13.14)^2)) = 0.21
Next, we need to calculate the t-value to determine if the means of the two methods differ significantly at 95% confidence level.
t = (13.14 - 12.8) / (√((0.21^2 / 5) + (0.27^2 / 5))) = 1.676
The calculated t-value of 1.676 is less than the critical t-value of 2.776 at a 95% confidence level with 4 degrees of freedom. Therefore, we can conclude that the means of the two methods do not differ significantly at a 95% confidence level.