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= for x4 at 95% confidence level). -2.776
Entymatic method ()13.1
12.7 12.6 13.3 13.3
DC method()
13.5
13.3
13:0
12.9
13.0
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.