An enzymatic method for determining alcohol in wine is evaluated by comparison with a gas-chromatography (GC) method. the same sample is analyzed several times by both methods with the following results (% ethanol). Enzymatic method: 13.1, 12.7, 12.6, 13.3, 13.3. GC method: 13.5, 13.3, 13.0, 12.9. Does the enzymatic method give the same value as the GC method at the 95% confidence level?
Answer
Good
To determine if the enzymatic method gives the same value as the GC method at the 95% confidence level, we can perform a statistical analysis.
Step 1: Calculate the mean for each method
For the enzymatic method:
Mean = (13.1 + 12.7 + 12.6 + 13.3 + 13.3) / 5 = 12.8
For the GC method:
Mean = (13.5 + 13.3 + 13.0 + 12.9) / 4 = 13.175
Step 2: Calculate the standard deviation for each method
For the enzymatic method:
Standard Deviation = √((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) / 4
= √(0.01 + 0.01 + 0.04 + 0.25 + 0.25)
= √0.56
≈ 0.748
For the GC method:
Standard Deviation = √((13.5 - 13.175)² + (13.3 - 13.175)² + (13.0 - 13.175)² + (12.9 - 13.175)²) / 3
= √(0.081 + 0.015 + 0.031 + 0.052)
= √0.179
≈ 0.423
Step 3: Calculate the t-value using the formula
t = (Mean1 - Mean2) / √((SD1²/n1) + (SD2²/n2))
t = (12.8 - 13.175) / √((0.748²/5) + (0.423²/4))
≈ -0.375 / √(0.264 + 0.044)
≈ -0.375 / √0.308
≈ -0.375 / 0.555
≈ -0.675
Step 4: Calculate the critical t-value from the t-distribution table
For a 95% confidence level with 7 degrees of freedom (5+4-2), the critical t-value is approximately 2.365.
Step 5: Compare the calculated t-value with the critical t-value
|-0.675| ≤ 2.365
Since |-0.675| ≤ 2.365, the calculated t-value falls within the range of the critical t-values for a 95% confidence level, indicating that there is no significant difference between the enzymatic method and the GC method at the 95% confidence level.
Therefore, we can conclude that the enzymatic method gives the same value as the GC method at the 95% confidence level.
To determine if the enzymatic method gives the same value as the GC method at the 95% confidence level, we need to perform a hypothesis test.
Here's how you can do it:
Step 1: State the null and alternative hypotheses:
Null hypothesis (H0): The mean alcohol percentage measured by the enzymatic method is equal to the mean alcohol percentage measured by the GC method.
Alternative hypothesis (Ha): The mean alcohol percentage measured by the enzymatic method is not equal to the mean alcohol percentage measured by the GC method.
Step 2: Set the significance level:
In this case, we want to test at the 95% confidence level, which means the significance level (alpha) is 0.05.
Step 3: Calculate the test statistic:
To compare the means of the two methods, we can use a two-sample t-test. The calculation involves finding the t-value and the degrees of freedom (df). The formula for the t-value is:
t = (x̄1 - x̄2) / √((s1^2 / n1) + (s2^2 / n2))
where:
x̄1 and x̄2 are the sample means,
s1 and s2 are the sample standard deviations, and
n1 and n2 are the sample sizes for the two methods.
Step 4: Determine the critical value:
Look up the critical value for a two-tailed test at the 95% confidence level with the appropriate degrees of freedom. This critical value will define the rejection region for the null hypothesis.
Step 5: Compare the test statistic with the critical value:
If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 6: Interpret the result:
If we reject the null hypothesis, it means that there is enough evidence to suggest that the mean alcohol percentages measured by the enzymatic method and the GC method are different at the 95% confidence level. If we fail to reject the null hypothesis, it means that we do not have enough evidence to conclude that there is a significant difference between the means.
Performing the calculations, we find that the test statistic is approximately -0.934, and the critical value at the 95% confidence level with 7 degrees of freedom is approximately ±2.365.
Since the absolute value of the test statistic (-0.934) is smaller than the critical value (2.365), we fail to reject the null hypothesis.
Therefore, at the 95% confidence level, we do not have enough evidence to conclude that the enzymatic method gives a different value from the GC method.