Question

Tomato weights and Fertilizer (Raw Data, Software Required):
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 for tomato-weights in grams is given below. Carl claims there is a difference in the mean weight for all tomatoes between the different fertilizing methods.

No Fertilizer	Organic Fertilizer	Chemical Fertilizer
123	112	115
119	127	141
118	138	143
120	133	134
117	140	129
120	114	134
114	126	135
118	134	129
129	123	113
128	146	150

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

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

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

(c) Use software to get the P-value of the test statistic ( F ). Round to 4 decimal places unless your software automatically rounds to 3 decimal places.
P-value =

(d) What is the conclusion regarding the null hypothesis at the 0.05 significance level?

reject H₀ fail to reject H₀    

(e) 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.

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

Yes No    

Answer 1

Solution:

Install analysis tool pak in excel

go to data>Data analysis >ANOVA Single factor

You will get

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
No Fertilizer	10	1206	120.6	22.71111
Organic Fertilizer	10	1293	129.3	121.5667
Chemical Fertilizer	10	1323	132.3	134.4556
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	738.6	2	369.3	3.974767	0.030687	3.354131
Within Groups	2508.6	27	92.91111
Total	3247.2	29

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

H₀: μ₁ = μ₂ = μ₃.

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

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

(c) Use software to get the P-value of the test statistic ( F ). R

P-value =0.0307

(d) What is the conclusion regarding the null hypothesis at the 0.05 significance level?

p<0.05

Reject Ho.

reject H₀

(e) Choose the appropriate concluding statement.

There is sufficient evidence to conclude that the mean weights are different.  

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

p=0.0307

alpha=0.10

p<alpha

reject Ho.
Conclusion wont change

NO