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Fertilizing Bromeliads

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 the larvae of many insects. As a preliminary to a 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 4 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 |
|----------|------------|------|---------|
| 15 | 15 | 14 | 14 |
| 14 | 17 | 18 | 19 |
| 18 | 13 | 14 | 11 |
| 16 | 13 | 15 | 16 |
| 14 | 14 | 15 | 13 |
| 11 | 17 | 14 | 15 |
| 13 | 12 | 15 | 15 |

(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.)

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

Answer :

Answer:

Calculated value of F = 0.0535

The critical region is F >F ₀.₀₅ (6,21) = 2.575

Reject H0

Step-by-step explanation:

1. Null hypothesis

H0: µ Nitrogen = µ Phosphorus = µ Both = µ Neither

2. Alternative hypothesis

H1: Not all means are equal.

3. The degrees of freedom for the numerator of the F-ratio = k- 1= 7-1=6

4.The degrees of freedom for the denominator of the F-ratio = n-k= 28-7

= 21

5. The significance level is set at α-0.05

The critical region is F >F ₀.₀₅ (6,21) = 2.575

The test statistic to use is

F = sb²/ sw²

Which if H0 is true has an F distribution with v₁=k-1 and v₂= n-k degrees of freedom

Correction Factor = CF = Tj²/n = (410)²/28= 6003.57

Total SS ∑∑X²- C. F = 6108- 6003.57= 104.43

Between SS ∑T²j/r - C.F = 42036/ 7 - 6003.57 = 1.57286

Within SS = Total SS - Between SS= 104.43- 1.573= 102.86

The Analysis of Variance Table is

Source Of Sum of Mean Computed

Variation d.f Squares Squares F

Between

Samples 6 1.57286 0.2621 0.0535

Within

Samples 21 102.86 4.898

Calculated value of F = 0.0535

Pvalue = 2.575

Since it is smaller than 5 % reject H0.

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