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 |
| 95 | 138 | 143 |
| 97 | 133 | 134 |
| 94 | 140 | 129 |
| 120 | 114 | 134 |
| 114 | 126 | 135 |
| 118 | 134 | 129 |
| 129 | 123 | 113 |
| 129 | 143 | 148 |
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: μ1 = μ2 = μ3.
H1: μ1 ≠ μ2 ≠ μ3.
H1: At least one of the population means is different from the others.
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
Ans:
a)
H0: μ1 = μ2 = μ3
b)
H1: At least one of the population means is different from the others.
c)p-value=0.0040
d)As,p-value<0.05,so reject H0
e)There is sufficient evidence to conclude that the mean weights are different.
f)No,as p-value<0.10,so reject H0
ANOVA :
| Group 1 | Group 2 | Group 3 | Total | |
| Sum | 1138 | 1290 | 1321 | 3749 |
| Count | 10 | 10 | 10 | 30 |
| Mean, Sum/n | 113.8 | 129 | 132.1 | |
| Sum of square, Ʃ(xᵢ-x̅)² | 1657.6 | 1002 | 1142.9 | |
| Standard deviation | 13.5712 | 10.5515 | 11.2689 |
| Number of treatment, k = | 3 |
| Total sample Size, N = | 30 |
| df(between) = k-1 = | 2 |
| df(within) = N-k = | 27 |
| df(total) = N-1 = | 29 |
| SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N = | 1918.4667 |
| SS(within) = SS1 + SS2 + SS3 = | 3802.5000 |
| SS(total) = SS(between) + SS(within) = | 5720.9667 |
| MS(between) = SS(between)/df(between) = | 959.2333 |
| MS(within) = SS(within)/df(within) = | 140.8333 |
| F = MS(between)/MS(within) = | 6.8111 |
| p-value = F.DIST.RT(6.8111, 2, 27) = | 0.0040 |
| Source of Variation | SS | df | MS | F | P-value |
| Between Groups | 1918.4667 | 2 | 959.2333 | 6.8111 | 0.0040 |
| Within Groups | 3802.5000 | 27 | 140.8333 | ||
| Total | 5720.9667 | 29 |
Tomato weights and Fertilizer (Raw Data, Software Required): Carl the farmer has three fields of tomatoes,...
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...
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