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Bromeliads are tropical flowering plants. Many are epiphytes that attach to trees and obtain moisture and nutrients from air and rain. Their leaf bases form cups that collect water and are home to larvae of many insects. As a preliminary study of changes in the nutrient cycle, Jacqueline Ngai and Diane Srivastava examined the effects of adding nitrogen, phosphorus, or both to the cups. They randomly assigned 8 bromeliads growing in Costa Rica to each of four treatment groups, including an unfertilized control group. A monkey destroyed one of the plants in the control group, leaving 7 bromeliads in that group. Here are the numbers of new leaves on each plant over the seven months following fertilization:

| Nitrogen | Phosphorus | Both | Neither |
|----------|------------|------|---------|
| 16 | 14 | 14 | 11 |
| 15 | 14 | 16 | 13 |
| 15 | 14 | 15 | 16 |
| 17 | 11 | 14 | 15 |
| 17 | 13 | 14 | 15 |
| 18 | 12 | 13 | 11 |
| 18 | 15 | 17 | 12 |
| 13 | 15 | 14 | |

(a) Give the degrees of freedom for the F statistic.

- Numerator degrees of freedom:
- Denominator degrees of freedom:

(b) Find the F-statistic. (Round your answer to two decimal places.)

- F =

(c) Find the associated P-value. (Round your answer to four decimal places.)

- P-value =

(d) State conclusions. Select the correct answer:

- We have significant evidence at the 5% level that the means are not all the same.
- We do not have significant evidence at the 5% level that the means are not all the same.

Answer :

Final answer:

The numerator degrees of freedom for the F statistic is 3, and the denominator degrees of freedom is 27. Without specific data values, such as means or variances, we cannot provide the F-statistic or associated P-value, and thus cannot draw conclusions about the significance of the treatment effects on bromeliad growth.

Explanation:

When assessing the effects of different treatments (nitrogen, phosphorus, both, and neither) on bromeliad growth by comparing the number of new leaves, an ANOVA test is typically used. For an ANOVA test, degrees of freedom for the numerator (between groups) is calculated as the number of groups minus one. In this case, there are four groups, so the numerator degrees of freedom is 3. For the denominator (within groups), it's the total number of observations minus the number of groups. With 8 plants in three groups and 7 in one (due to a monkey destroying one plant), there are a total of 31 observations across the four groups, hence the denominator degrees of freedom is 31 - 4 = 27.

To find the F-statistic, we would typically calculate the variance between the groups and within the groups, then divide the former by the latter. However, specific data values are required to perform this calculation, which are not provided in the prompt. Therefore, without the specific summations or means, I cannot provide the F-statistic value.

Without the F-statistic, we cannot directly calculate the P-value. We would typically use statistical software or an F-distribution table to find this value based on the F-statistic and the degrees of freedom.

The conclusion on whether there is significant evidence at the 5% level that the means are not all the same cannot be drawn without calculating the F-statistic and its associated P-value. If the P-value were found to be less than 0.05, it would indicate significant evidence; if not, there'd be no significant evidence.

Learn more about ANOVA here:

https://brainly.com/question/31809956

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