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.)

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.

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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