Question

Bromeliads are tropical flowering plants. Many are epiphytes that attach to trees and obtain moisture and nutrients from air and rain. Their leaf bases from cups that collect water and are home to larvae of many insects. As a preliminary to 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.)
(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 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

Y: number of new leaves of Bromeliads grown under different types of fertilization.

Factor: Type of fertilizer

Treatments: 1. Nitrogen, 2. Phosphorus, 3. Both, 4. Neither (control group)

To compare these four groups you have to conduct an ANOVA

The hypotheses are:

H₀: μ₁= μ₂= μ₃= μ₄

H₁: At least one population mean is different from the others.

α: 0.05

a)

The degrees of freedom of the F-statistic are:

$F_{I-1;N-I}= F_{4-1;31-4}= F_{3;27}$

I= number of treatments

N= total number of experimental units (in all groups)

b)

SSTreatments= 39.37

SSError= 72.18

MSTreatments= SStreatments/DfTreatments= 39.37/3= 13.12

MSError=SSError/DfError= 72.18/27= 2.67

$F= \frac{MS_{Treatments}}{MS_{Error}} = \frac{13.12}{2.67} = 4.91$

c)

This test is one-tailed to the right and so is the p-value:

P(F₃;₂₇≥4.91)= 1 - P(F₃;₂₇<4.91)= 1 - 0.9925= 0.0075

d)

The decision rule using the p-value is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value is less than α, the test is significant at 5% significance level.

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

I hope this helps!