KRUSKAL - WALLIS TEST The following data are average yields in bushels per acre
ID: 3177152 • Letter: K
Question
KRUSKAL - WALLIS TEST
The following data are average yields in bushels per acre on four plots of land for four different varieties of corn Yield (bushels per acre) is the response variable and corn variety is the treatment factor with four levels Use these data for Questions on this task. Rank the combined data and calculate rank sums for each corn variety (treatment). 2. State the null and alternative hypotheses for the Kruskall-Wallis test 3. For o = 0.06. specify the critical region for the Kruskal-Wallis test. 4. Calculate the Kruskal-wallis test statistic. 5. Find the P-value and make a decision about Ho based on alpha = 0 05.Explanation / Answer
Solution:
First, all the data needs to be put together in one column as shown below:
Sample
Value
1
68.82
1
84.84
1
90.16
1
61.58
2
76.99
2
75.69
2
77.84
2
70.51
3
74.30
3
72.87
3
80.65
3
74.57
4
76.73
4
76.18
4
83.58
4
70.75
Now, the data needs to be organized in ascending order by value (keeping track of what sample the values belongs to). The results are shown below:
Sample
Value (In Asc. Order)
1
61.58
1
68.82
2
70.51
4
70.75
3
72.87
3
74.30
3
74.57
2
75.69
4
76.18
4
76.73
2
76.99
2
77.84
3
80.65
4
83.58
1
84.84
1
90.16
Now, we need to assign ranks to the values that are already organized in ascending order. Make sure that take the average of ranks in case of rank ties (Ex. If two values shared the first place in the list, instead of assigning rank 1 and rank 2 to them, assign rank 1.5 to both) The following ranks are obtained:
Sample
Value (In Asc. Order)
Rank
Rank (Adjusted for ties)
1
61.58
1
1
1
68.82
2
2
2
70.51
3
3
4
70.75
4
4
3
72.87
5
5
3
74.30
6
6
3
74.57
7
7
2
75.69
8
8
4
76.18
9
9
4
76.73
10
10
2
76.99
11
11
2
77.84
12
12
3
80.65
13
13
4
83.58
14
14
1
84.84
15
15
1
90.16
16
16
In order to compute the sum of rank for each sample, it is easier to organize the above table by samples. The following is obtained:
Sample
Value
Rank (Adjusted for ties)
1
61.58
1
1
68.82
2
1
84.84
15
1
90.16
16
2
70.51
3
2
75.69
8
2
76.99
11
2
77.84
12
3
72.87
5
3
74.30
6
3
74.57
7
3
80.65
13
4
70.75
4
4
76.18
9
4
76.73
10
4
83.58
14
With the information provided we can now easily compute the sum of ranks for each of the samples:
R1 = 1 + 2 + 15 + 16 = 34
R2 = 3 + 8 + 11 + 12 = 34
R3 = 5 + 6 + 7 + 13 = 31
R4 = 4 + 9 + 10 + 14 = 37
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: The samples come from populations with equal medians
Ha: The samples come from populations with medians that are not all equal
The above hypotheses will be tested using the Kruskal-Wallis test.
(2) Rejection Region
The significance level is alpha = 0.05, and the number of degrees of freedom is (df = 4 - 1 = 3). But, since at least one sample size is less than 5, we cannot use normal approximation, so then, the H-statistic cannot longer be approximated by the Chi-Square distribution.
Using the appropriate for table for H values when normal approximation cannot be used, it is found that the critical H-value is (H_c = 7.235). The rejection region for this test is (R = 7.235).
(3) Test Statistics
KW=0.199
(4) Decision about the null hypothesis
Since it is observed that (H = 0.199 < 7.235), it is then concluded that the null hypothesis is not rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim some of the population medians are unequal, at the 0.05 significance level.
Sample
Value
1
68.82
1
84.84
1
90.16
1
61.58
2
76.99
2
75.69
2
77.84
2
70.51
3
74.30
3
72.87
3
80.65
3
74.57
4
76.73
4
76.18
4
83.58
4
70.75
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