To solve this question, we first need to conduct an ANOVA (Analysis of Variance) test and then a t-test to compare the two treatment conditions. Next, we will assess whether the p-values obtained from both tests affect the conclusions drawn from the hypothesis tests at the [tex]\(\alpha = 0.05\)[/tex] significance level.
### Conduct ANOVA
1. Data Sets:
- Treatment One: [tex]\([6.3, 6.3, 8.1, 6.3, 6.8, 6.8, 7, 5.6]\)[/tex]
- Treatment Two: [tex]\([2.9, 5.4, 6.4, 3, 3.2, 3.8, 3.7, 6.2]\)[/tex]
2. ANOVA Calculation:
- Calculate the F-ratio and p-value for the data sets.
From our computations:
- F-ratio: 16.421
- p-value: 0.001
### Conduct t-Test
Next, we perform a t-test for the same data sets.
3. t-Test Calculation:
- Calculate the t-statistic and p-value for the data sets.
From our computations:
- t-statistic: 4.052
- p-value: 0.001
### Conclusion from p-values
Finally, we compare the p-values obtained from both tests to determine the outcome at a significance level of [tex]\(\alpha = 0.05\)[/tex].
Since the p-value for both tests is 0.001, which is less than the significance level [tex]\(\alpha = 0.05\)[/tex], we reject the null hypothesis in both cases. This indicates a significant difference between the two treatment conditions.
### Observing the p-values
Even though the p-values obtained from both tests are the same, it is noteworthy that both tests show a significant difference. Hence, there is no discrepancy affecting the hypothesis test outcomes.
### Summary
- F-ratio: 16.421
- p-value (ANOVA): 0.001
- t-statistic: 4.052
- p-value (t-test): 0.001
- Conclusion: The p-values indicate a significant difference between the two treatments. Therefore, despite the similar p-values, the hypothesis test conclusions are consistent.
By looking at the consistent p-values, we can assert that the result of the hypothesis tests is not affected by the slight differences in methodology.
