A Kruskal-Wallis test is conducted on and experiment that was testing three different levels of a fertilizer on the growth of the same variety of tomatoes. Site 1 received no fertilizer, Site 2 received low levels of fertilizer, Site 3 received high levels of fertilizer. All sites were treated identically in terms of the amount they are watered, time of planting, etc. A random sample of 9 mature tomatoes is selected from each site. The weights are recorded.
The output is given below. At a significance level of 0.15, what is your decision and interpretation?
Kruskal-Wallis Test: Weight versus Fertilizer Level
Descriptive Statistics
Fertilizer |
N |
Median |
Mean Rank |
Z-Value |
HIGH |
9 |
5 |
13.5 |
-0.23 |
LOW |
9 |
7 |
17.6 |
1.67 |
NO |
9 |
5 |
10.9 |
-1.44 |
Overall |
27 |
14.0 |
Test
Null hypothesis |
H₀: All medians are equal |
Alternative hypothesis |
H₁: At least one median is different |
Method |
DF |
H-Value |
P-Value |
Not adjusted for ties |
2 |
3.28 |
0.194 |
Adjusted for ties |
2 |
3.41 |
0.182 |
Group of answer choices
Fail to Reject Ho, because there is enough evidence to say the levels of fertilizers do matter.
Fail to Reject Ho, because there is not enough evidence to say the level of fertilizer matters.
Fail to Reject Ho, there is enough evidence to say that the level of fertilizer matters.
Fail to Reject Ho, there is not enough evidence to say the level of fertilizer matters.
Answer :
A Kruskal-Wallis test :
Hypothesis :
Null hypothesis
H₀: All medians are equal.
Alternative hypothesis
H₁: At least one median is different.
Analysis :
DF |
H-Value |
P-Value |
|
Not adjusted for ties |
2 |
3.28 |
0.194 |
Adjusted for ties |
2 |
3.41 |
0.182 |
Interpretation:
Here test statistic=H are10.432 and 3.41
and
p-values > level of significance ( 0.15 ), we accept H0 at 0.15 level of significance.
That is , Fail to Reject Ho, there is enough evidence to say that the level of fertilizer matters.
Result:
The independent variables on dependent variable represent population with equal medians.
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