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.