Question

The time it takes to contain wildfires with and without aerial support was studied. Containment time in hours) was estimated by fire management personnel for a particular wildfire scenario that had no aerial support. Fire management personnel were classified according to one of three primary roles: ground, office, or air support Data for 21 fire managers (simulated, based on information provided in the article) are listed in the table. One objective is to compare the estimated containment times of the three groups of fire managers. Complete parts a through f. Click the icon to view the estimated containment times, a. Why is the ANOVA F-test to compare multiple treatment means in a completely randomized design inappropriate for analyzing these data? Use graphs to support your answer. Select all that apply. A. Frequency histograms of the data for the groups indicate that the data are not independent B. Frequency histograms of the data for the groups indicate that this is not actually a completely randomized design C. Frequency histograms of the data for the groups indicate that the variances of the underlying population distributions are not equal D. Frequency histograms of the data for the groups indicate that the underlying population distributions are nonnormal b. What is the appropriate nonparametric test to apply? State Ho and H, for the test. What is the appropriate nonparametric test to apply? Click to select your answers).

Answer 1

(a) The answer for this question is option D. This is because the data given is not normal which can be identified from frequency histograms.

Frequency histograms of the data for the groups indicate that the underlying population distributions are non normal.

(b) The appropriate nonparametric test to be used here is Kruskal Wallis test .i.e. option A.

This is because when the assumptions of ANOVA are not satisfied, we use its nonparametric test which is the Kruskal Wallis test.

The hypotheses will be

H0: The population distributions of the three groups are identical

Ha: At least two of the three population distributions differ in location.

(c) Ranks

Office	ranks		Ground	ranks		Air	ranks
16.7	20		3.9	10		3.3	7
2.7	3		4.6	12		2.8	4.5
20.1	21		5.9	14.5		2.5	1
5.9	14.5		2.8	4.5		2.6	2
3.2	6		4.3	11		3.6	9
9.5	16		5.4	13
13.7	18
15.8	19
10.2	17
3.4	8

Sum of ranks of the observations from the office group is R1= 142.5

Sum of ranks of the observations from the ground group is R2= 65

Sum of ranks of the observations from the air group is R3= 23.5

(d)

The test statistic used is H is calculated and is distributed as $1587808511643_blob.png$ with 2 degrees of freedom H - 12 ΙΣΠ 3(1 + 1) where T; = Sum of ranks of treatments i ni = Number of observations for treatment i n=) n; = Total number of observations k = Number of treatments

So the calculation will be

12 (142.52 + 652 + 23.52 21(21 + 1) | 1+1) 12 5x [2030.625 + 704.167 + 110.45) – 3 x 22 = 73.902 - 66 = 7.902

(e) The rejection region for the test using 1% level of significance is 9.210.

The answer is option C. H < 9.210

(f) Since H=7.902 < 9.210, we do not reject H0. It means that the three groups are identical.

So the answer is option B.

The conclusion will be There is insufficient evidence to indicate the time to contain a fire for at least two groups have different locations at α=0.01.

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