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Part 1

Four different types of fertilizers are used on raspberry plants. The number of raspberries on each randomly selected plant is given to the right. Find the test statistic F to test the claim that the type of fertilizer makes no difference in the mean number of raspberries per plant.

Start 7 By 4 Matrix 1st Row 1st Column Fertilizer 1 2nd Column Fertilizer 2 3rd Column Fertilizer 3 4st Column Fertilizer 4 2nd Row 1st Column 6 2nd Column 5 3rd Column 6 4st Column 3 3rd Row 1st Column 7 2nd Column 8 3rd Column 3 4st Column 5 4st Row 1st Column 6 2nd Column 5 3rd Column 4 4st Column 3 5st Row 1st Column 5 2nd Column 5 3rd Column 2 4st Column 4 6st Row 1st Column 7 2nd Column 5 3rd Column 3 4st Column 4 7st Row 1st Column 6 2nd Column 6 3rd Column 3 4st Column 5 EndMatrix

Fertilizer 1

 Fertilizer 2

 Fertilizer 3

 Fertilizer 4

6

5

6

3

7

8

3

5

6

5

4

3

5

5

2

4

7

5

3

4

6

6

3

5

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Part 1

A.

7.123

B.

8.357

C.

8.123

D.

6.91

To determine whether there is a significant difference in the mean number of raspberries per plant based on the type of fertilizer, we'll conduct an Analysis of Variance (ANOVA) test. The test statistic \( F \) is calculated by comparing the variance between the group means (fertilizer types) to the variance within the groups.

Here is the procedure for calculating the \( F \) statistic:

### Step 1: Organize the Data
We have four groups (fertilizer types) with the following observations:

- **Fertilizer 1:** 6, 7, 6, 5, 7, 6
- **Fertilizer 2:** 5, 8, 5, 5, 5, 6
- **Fertilizer 3:** 6, 3, 4, 2, 3, 3
- **Fertilizer 4:** 3, 5, 3, 4, 4, 5

### Step 2: Calculate the Group Means and Overall Mean

- **Mean of Fertilizer 1:** \(\bar{X}_1 = \frac{6+7+6+5+7+6}{6} = 6.17\)
- **Mean of Fertilizer 2:** \(\bar{X}_2 = \frac{5+8+5+5+5+6}{6} = 5.67\)
- **Mean of Fertilizer 3:** \(\bar{X}_3 = \frac{6+3+4+2+3+3}{6} = 3.50\)
- **Mean of Fertilizer 4:** \(\bar{X}_4 = \frac{3+5+3+4+4+5}{6} = 4.00\)

- **Overall Mean (\(\bar{X}\)):** \(\bar{X} = \frac{6.17 + 5.67 + 3.50 + 4.00}{4} = 4.835\)

### Step 3: Calculate the Sum of Squares

1. **Sum of Squares Between (SSB):**

\[
SSB = n(\bar{X}_1 - \bar{X})^2 + n(\bar{X}_2 - \bar{X})^2 + n(\bar{X}_3 - \bar{X})^2 + n(\bar{X}_4 - \bar{X})^2
\]

Where \( n = 6 \) (since each group has 6 observations).

\[
SSB = 6[(6.17 - 4.835)^2 + (5.67 - 4.835)^2 + (3.50 - 4.835)^2 + (4.00 - 4.835)^2] 
\]

\[
SSB = 6[(1.335)^2 + (0.835)^2 + (-1.335)^2 + (-0.835)^2] = 6(1.7822 + 0.6972 + 1.7822 + 0.6972) = 6(4.9588) = 29.7528
\]

2. **Sum of Squares Within (SSW):**

\[
SSW = \sum_{i=1}^{4} \sum_{j=1}^{6} (X_{ij} - \bar{X}_i)^2
\]

Calculate this for each group:

- **Fertilizer 1:** \((6-6.17)^2 + (7-6.17)^2 + \ldots + (6-6.17)^2 = 3.17\)
- **Fertilizer 2:** \((5-5.67)^2 + (8-5.67)^2 + \ldots + (6-5.67)^2 = 8.67\)
- **Fertilizer 3:** \((6-3.50)^2 + (3-3.50)^2 + \ldots + (3-3.50)^2 = 17.50\)
- **Fertilizer 4:** \((3-4.00)^2 + (5-4.00)^2 + \ldots + (5-4.00)^2 = 8.00\)

\[
SSW = 3.17 + 8.67 + 17.50 + 8.00 = 37.34
\]

### Step 4: Calculate the \( F \) Statistic

1. **Mean Square Between (MSB):**
\[
MSB = \frac{SSB}{df_{between}} = \frac{29.7528}{3} = 9.9176
\]

2. **Mean Square Within (MSW):**
\[
MSW = \frac{SSW}{df_{within}} = \frac{37.34}{20} = 1.867
\]

3. **\( F \) Statistic:**
\[
F = \frac{MSB}{MSW} = \frac{9.9176}{1.867} \approx 5.31
\]

None of the provided answers match exactly, but the closest \( F \) value from typical rounding errors would be between the typical answers available, suggesting you might need to adjust calculations (such as slight rounding or recheck), as some variants might yield these options.

Would you like a more detailed step-by-step or any questions? 

---

1. What assumptions must be met for conducting an ANOVA test?
2. How do you interpret the \( F \)-statistic in an ANOVA test?
3. What is the meaning of degrees of freedom in the context of ANOVA?
4. How would you calculate the p-value associated with the \( F \)-statistic?
5. What happens if the assumptions of ANOVA are violated?
6. What are post-hoc tests, and when should they be used after ANOVA?
7. How does ANOVA differ from a t-test?
8. How would you modify this analysis for a repeated-measures design?

**Tip:** In hypothesis testing, the \( F \)-statistic compares the variance between groups to the variance within groups. A higher \( F \) value suggests a significant difference between groups.

Learn how to conduct an Analysis of Variance (ANOVA) test to compare the mean number of raspberries per plant based on different types of fertilizers. Calculate the F-statistic and interpret its significance. Suitable for advanced high school and college students studying statistics and hypothesis testing.

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