67. An agricultural biologist wants to find the best type of fertilizer for increasing crop yield for yellow squash. The biologist grows 12 squash plants in a single field and applies three different fertilizers (A, B, and C) to the plants. Then the biologist counted the number of squash fruits that each plant produced, as shown in the table below. The tests of normality and equal variances for these data all had P > 0.05. Evaluate whether there were differences in squash yield in the treatment groups, using an α = 0.05.

| Fertilizer A | Fertilizer B | Fertilizer C |
|--------------|--------------|--------------|
| 7 | 12 | 16 |
| 4 | 15 | 18 |
| 8 | 13 | 15 |
| 3 | 14 | 13 |

- [tex]\bar{X}_A = 5.5[/tex]
- [tex]\bar{X}_B = 13.5[/tex]
- [tex]\bar{X}_C = 15.5[/tex]

A. State the appropriate test and explain why this test is appropriate.

B. State the null and alternative hypotheses.

C. Perform the appropriate test and report the test statistic here. Show all of your work for full credit and use at least two decimal places in your calculations.

D. What is the critical value based on α?

- Blank 1: notation
- Blank 2: value

E. Draw a conclusion based on the statistical results. (Reject or do not reject the null hypothesis)?

F. Biological relevance.

The appropriate test to evaluate whether there were differences in squash yield in the treatment groups is the one-way analysis of variance (ANOVA). This test is appropriate because we have three different fertilizers (A, B, and C) and we want to determine if there are any significant differences in the mean yield among these groups.

The null hypothesis ([tex]H_0[/tex]) states that there are no differences in the mean yield among the treatment groups, while the alternative hypothesis (Ha) states that there are differences in the mean yield among the treatment groups.

To perform the ANOVA, we need to calculate the sum of squares between groups (SSB), sum of squares within groups (SSW), and the test statistic, F.
Step 1: Calculate the mean yield for each group:
[tex]X_A[/tex] = 5.5
[tex]X_{B}[/tex] = 13.5
[tex]X_C[/tex] = 15.5

Step 2: Calculate the sum of squares between groups (SSB):
SSB = [tex](nA * (X_A - X_G)^2) + (nB * (X_B - X_G)^2) + (nC * (XC - XG)^2)\\ = (12 * (5.5 - 11.5)^2) + (12 * (13.5 - 11.5)^2) + (12 * (15.5 - 11.5)^2)\\ = 348[/tex]

where [tex]X_G[/tex] is the grand mean and is calculated as the average of all the yields:
[tex]X_G = (X_A * nA + X_B * nB + X_C * nC) / (nA + nB + nC)\\ = (5.5 * 12 + 13.5 * 12 + 15.5 * 12) / (12 + 12 + 12)\\ = 11.5[/tex]

Step 3: Calculate the sum of squares within groups (SSW):
[tex]SSW = (nA - 1) * \sigma A^2 + (nB - 1) * \sigma B^2 + (nC - 1) * \sigma C^2\\ = (11 * 0.5^2) + (11 * 2.5^2) + (11 * 0.5^2)\\ = 62.5[/tex]
where σA, σB, and σC are the variances within each group. Given that the tests of normality and equal variances for these data had P > 0.05, we can assume that the variances within each group are equal.

Step 4: Calculate the degrees of freedom for between groups (dfB) and within groups (dfW):
dfB = k - 1 = 3 - 1 = 2
dfW = N - k = 36 - 3 = 33

Step 5: Calculate the mean squares between groups (MSB) and within groups (MSW):
MSB = SSB / dfB = 348 / 2 = 174
MSW = SSW / dfW = 62.5 / 33 ≈ 1.89

Step 6: Calculate the test statistic, F:
F = MSB / MSW = 174 / 1.89 ≈ 92.06
Now we need to determine the critical value based on α = 0.05. The critical value for an α of 0.05 and dfB of 2 and dfW of 33 is approximately 3.40.

Since the calculated F-value (92.06) is greater than the critical value (3.40), we reject the null hypothesis. This means that there are significant differences in the mean yield among the treatment groups.

Biological relevance: The results of this statistical analysis indicate that there are significant differences in the yield of yellow squash among the three different fertilizers (A, B, and C). This information can be used by the agricultural biologist to determine which fertilizer is the best for increasing crop yield.

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