An ezymatic method for determining alcohol in wine is evaluated by comparison with gas chromatography method. The same sample was analyzed by two methods giving following results. Determine if the means of the two methods differ significantly at 95% confidence level. (t=2.776 for v=4 at 95% confidence level).
Entymatic method ()13.1
12.7 12.6 13.3 13.3
DC method()
13.5
13.3
13:0
12.9
13.0
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.