The following randomly selected sample data comes from two independent populations for which ANOVA is appropriate.
Treatment One - 4.1, 3.6, 4.1, 7.1, 3.8
Treatment Two - 3.8, 3.5, 4, 4.1, 4.3
Give all answers accurate to at least 3 places after the decimal point.
With technology (such as a TI-83/84), use
ANOVA to test the claim that the population means are equal
(i.e. "u1 = u2").
F test statistic:
P-value:
Now perform a hypothesis test of the same claim using a t test statistic (using methods from Chapter 6).
Hint: Think carefully about the required conditions of ANOVA. Since we are told that it is appropriate to
use ANOVA to test the given claim for this data, we know that the three necessary conditions/ criteria are
satisfied. This will help you determine whether you should pool the sample variances or not in this t test.

To analyze this problem, we'll conduct both an ANOVA test and a two-sample t-test to test the claim that the population means are equal (\( \mu_1 = \mu_2 \)). 

### **Step 1: Conduct ANOVA Test**

The ANOVA test involves comparing the variance between the groups to the variance within the groups to determine if the population means are different.

The steps to calculate the ANOVA test statistic are:
1. Calculate the mean for each treatment group.
2. Find the overall mean.
3. Compute the "Between Group" and "Within Group" Sum of Squares.
4. Calculate the Mean Squares and the F-statistic.

Let me perform these calculations.### **ANOVA Test Results:**
- **F test statistic:** 0.823
- **P-value:** 0.391

Since the p-value (0.391) is greater than the typical significance level (e.g., 0.05), we do not reject the null hypothesis. There isn't enough evidence to suggest that the population means are different based on the ANOVA test.

### **Step 2: Conduct a Two-Sample t-Test**

We will now perform a two-sample t-test. Since ANOVA is deemed appropriate, we can assume equal variances between the two treatments, and thus we can pool the variances.

Let’s calculate the t-test statistic using the pooled variance approach.### **Two-Sample t-Test Results:**
- **t test statistic:** 0.907
- **P-value:** 0.391

### **Conclusion:**
For both the ANOVA and t-test, the p-value (0.391) is greater than the common significance level (e.g., 0.05). Therefore, we fail to reject the null hypothesis in both cases, suggesting there is not enough evidence to conclude that the population means are different.

Would you like a detailed explanation of any part of this analysis or have any questions?

### 5 Related Questions:
1. How does the F statistic relate to the t statistic in this context?
2. Why is it appropriate to assume equal variances when using the t-test in this case?
3. What are the conditions required for ANOVA to be valid?
4. How would the results change if the sample sizes were larger?
5. What alternative tests can be used if the assumption of equal variances is violated?

### Tip:
Always check the assumptions of ANOVA or the t-test before proceeding, as violations of these assumptions can lead to incorrect conclusions.

The following randomly selected sample data comes from two independent populations for which ANOVA is appropriate. Treatment One - 4.1, 3.6, 4.1, 7.1, 3.8 Treatment Two - 3.8, 3.5, 4, 4.1, 4.3. Give all answers accurate to at least 3 places after the decimal point. With technology (such as a TI-83/84), use ANOVA to test the claim that the population means are equal (i.e. 'u1 = u2'). F test statistic: P-value: Now perform a hypothesis test of the same claim using a t test statistic.

Learn how to use ANOVA and a two-sample t-test to compare two independent population means based on sample data. This problem covers statistical concepts including the F-statistic, P-value, and pooled variance t-test. Suitable for university-level statistics students.

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