1. Park managers need to know how resistant different vegetative types are to tr
ID: 3124245 • Letter: 1
Question
1. Park managers need to know how resistant different vegetative types are to trampling so that the number of visitors can be controlled in sensitive areas. The experiment deals with alpine meadows in the White Mountains of New Hampshire. Twenty lanes were established, each 0.5m wide and 1.5m long. These twenty lanes were randomly assigned to five treatments: 0, 25, 75, 200 or 500 walking passes. Each pass consists of a 70-kg individual wearing lug-soled boots walking in a natural gait down the lane. The response measured is the average height of the vegetation along the lane one year after trampling.
Number of passes: 0 Height (cm):20.7, 15.9, 17.8, 17.6
Number of passes:25 Height (cm):12.9, 13.4, 12.7, 9.0
Number of passes: 75 Height (cm):11.8, 12.6, 11.4, 12.1
Number of passes:200 Height (cm):7.6, 9.5, 9.9, 9.0
Number of passes: 500 Height (cm):7.8, 9.0, 8.5, 6.7
a. Identify the elements of this experiment: experimental units, response variable, factor and treatment levels.
b. What assumptions are necessary for conducting a one-way ANOVA test.? Are they satisfied?
c. State the null and alternative hypotheses of the one-way ANOVA test for this experiment.
d. Complete the following ANOVA table (Show your work):
Source | df | SS | MS | F | p-value|
Treatment| | 243.16 (under SS, next to treatment) |
Error| | 30.93 (under SS, next to error)) |
Total|
e. What is your conclusion for this test?
f. How many pairwise comparisons are there for this experiment? Show your work.
g. What should the individual confidence level be in order to use Bonferroni’s method with a family confidence level of 94 percent? Show your work.
h. Compute the margin of error (ME) for Bonferroni’s method using a family confidence level of 94 percent. Show your work.
i. Make a diagram summarizing the results of the pair-wise comparisons using Bonferroni’s method. Order the sample means of each treatment and then draw lines to connect treatments for which there is not a significant difference. Interpret these results in context of the problem.
Explanation / Answer
a) the variables are lanes and heights
b)
The assumptions are:
c)
Null hypothesis: H0 : All the treatmentsare equal
Alternative hypothesis: Atleast one treatment is not equal
Levl of significancce: 5%
d) From Excel
ANOVA TABLE
e) Conclusion:
P value=0.066 > 0.05, we accept the null hypothesis.
Anova: Single Factor SUMMARY Groups Count Sum Average Variance Heights 5 800 160 42062.5 1 5 60.8 12.16 28.353 2 5 60.4 12.08 8.187 3 5 60.3 12.06 12.783 4 5 54.4 10.88 17.797 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 87864.4976 4 21966.12 2.606969 0.066635 2.866081 Within Groups 168518.48 20 8425.924 Total 256382.9776 24Related Questions
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