Question

Tomato weights and Fertilizer: Carl the farmer has three fields of tomatoes, on one he used no fertilizer, in another he used organic fertilizer, and the third he used a chemical fertilizer. He wants to see if there is a difference in the mean weights of tomatoes from the different fields. The sample data is given below. The second table gives the results from an ANOVA test. Carl claims there is a difference in the mean weight for all tomatoes between the different fertilizing methods.

Tomato-Weight in Grams

											x
No Fertilizer	123	119	118	120	117	120	114	118	129	128	120.6
Organic Fertilizer	112	127	138	133	140	114	126	134	123	144	129.1
Chemical Fertilizer	115	141	143	134	129	134	135	129	113	148	132.1

ANOVA Results

F	P-value
4.040	0.0292

The Test: Complete the steps in testing the claim that there is a difference in the mean weight for all tomatoes between the different fertilizing methods.

(a) What is the null hypothesis for this test?

H₀:  μ₁ = μ₂ = μ₃.H₀:  μ₁ ≠ μ₂ ≠ μ₃.    H₀: At least one of the population means is different from the others.H₀:  μ₃ > μ₂ > μ₁.

(b) What is the alternate hypothesis for this test?

H₁: At least one of the population means is different from the others.H₁:  μ₁ = μ₂ = μ₃.    H₁:  μ₁ ≠ μ₂ ≠ μ₃.H₁:  μ₃ > μ₂ > μ₁.

(c) What is the conclusion regarding the null hypothesis at the 0.01 significance level?

reject H₀fail to reject H₀    

(d) Choose the appropriate concluding statement.

We have proven that all of the mean weights are the same.There is sufficient evidence to conclude that the mean weights are different.    There is not enough evidence to conclude that the mean weights are different.

(e) Does your conclusion change at the 0.10 significance level?

YesNo    

Answer 1

Step 1: State null and alternative hypotheses
H_o: μ₁= μ₂= μ₃
H₁: At least one mean is different from the others


Step 2: Find the critical value
k = 3
N = total number of data values from all groups = 33
d.f.N. = k -1 = 2
d.f.D. = N -1 = 30
α = 0.1
Critical Value = 2.488

Step 3: Calculate F Test Value
ΣX = Sum of all data values = 1326.6+1420.1+1453.1 = 4199.79
N = 33
Grand Mean = X_GM = 4199.79/33 = 127.266

Calculate between-group variance, denoted by S_B2:

SS_B = 11(120.6 - 127.26)²+ 11(129.1 - 127.26)²+ 11(132.1 - 127.26)²

SS_B = 782.83

S²_B = 782.83 / 2 = 391.416

Calculate within-group variance, denoted by S_W2:

SS_W = (11- 1)(20.44)+ (11- 1)(103.09)+ (11- 1)(114.29)

SS_W = 2378.20

N - k = 33 - 3 = 30

S_W2 = SS_W/(N - k) = 2378.20/30 = 79.27

Calculate F test value:

F = 391.41 / 79.273 = 4.9375

Step 4: Make Decision:
Compare F test value = 4.937557816836262 with Critical Value = 2.4887

The decision is to reject the null hypothesis since 4.9375 > 2.4887



Analysis of Variance Summary Table

Source	Sum of Squares	Degrees of Freedom (df)	Mean Square	F test value
Between	SSB = 782.83	k - 1 = 2	MSB = 391.416	F = 4.9375
Within (error)	SSW = 2378.20	N - k = 30	MSW = 79.273
Total	3161.03	32