To determine if the means of the two methods differ significantly at the 95% confidence level, we can use a two-sample t-test assuming equal variances.
Let's organize the data into two groups:
Enzymatic method: 12.7, 12.6, 13.3, 13.3, 13.1
GC method: 13.5, 13.3, 13.0, 12.9, 13.0
Calculating the means of each group:
Enzymatic method mean = (12.7 + 12.6 + 13.3 + 13.3 + 13.1) / 5 = 12.8
GC method mean = (13.5 + 13.3 + 13.0 + 12.9 + 13.0) / 5 = 13.14
Next, calculate the pooled standard deviation:
s_p = sqrt(((n1-1)s1^2 + (n2-1)s2^2) / (n1 + n2 - 2))
n1 = number of samples in group 1 = 5
n2 = number of samples in group 2 = 5
s1 = standard deviation of group 1
s2 = standard deviation of group 2
Using the data provided in the question,
s1 = sqrt(((12.7 - 12.8)^2 + (12.6 - 12.8)^2 + (13.3 - 12.8)^2 + (13.3 - 12.8)^2 + (13.1 - 12.8)^2) / (5-1)) = 0.253
s2 = sqrt(((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) / (5-1)) = 0.227
Plugging these values into the formula for s_p:
s_p = sqrt(((5-1)*0.253^2 + (5-1)*0.227^2) / (5 + 5 - 2)) = 0.236
Now, calculate the t-value:
t = (mean1 - mean2) / (s_p * sqrt(1/n1 + 1/n2))
t = (12.8 - 13.14) / (0.236 * sqrt(1/5 + 1/5)) = -1.43
Since the t-value (-1.43) is less than the critical t-value (-2.776), we fail to reject the null hypothesis that the means of the two methods differ significantly at the 95% confidence level.