Question

In: Statistics and Probability

The types of browse favored by deer are shown in the following table. Using binoculars, volunteers...

The types of browse favored by deer are shown in the following table. Using binoculars, volunteers observed the feeding habits of a random sample of 320 deer.

Type of Browse Plant Composition
in Study Area
Observed Number of Deer
Feeding on This Plant
Sage brush           32% 105                
Rabbit brush           38.7% 125                
Salt brush           12% 42                
Service berry             9.3% 24                
Other             8% 24                

Use a 5% level of significance to test the claim that the natural distribution of browse fits the deer feeding pattern.

(a) What is the level of significance?
______________

State the null and alternate hypotheses.

H0: The distributions are different.
H1: The distributions are different.H0: The distributions are different.
H1: The distributions are the same.    H0: The distributions are the same.
H1: The distributions are the same.H0: The distributions are the same.
H1: The distributions are different.


(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)
___________

Are all the expected frequencies greater than 5?

Yes

No    


What sampling distribution will you use?

chi-square

binomial    

uniform

Student's t

normal


What are the degrees of freedom?
____________

(c) Estimate the P-value of the sample test statistic.

P-value > 0.100

0.050 < P-value < 0.100    

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005


(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?

Since the P-value > α, we fail to reject the null hypothesis.

Since the P-value > α, we reject the null hypothesis.    

Since the P-value ≤ α, we reject the null hypothesis.

Since the P-value ≤ α, we fail to reject the null hypothesis.


(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is sufficient to conclude that the natural distribution of browse does not fit the feeding pattern.

At the 5% level of significance, the evidence is insufficient to conclude that the natural distribution of browse does not fit the feeding pattern.    

Solutions

Expert Solution

a) level of significance =0.05

H0: The distributions are the same.
H1: The distributions are different

b)

applying chi square test:

           relative observed Expected residual Chi square
category frequency Oi Ei=total*p R2i=(Oi-Ei)/√Ei R2i=(Oi-Ei)2/Ei
1 0.32 105 102.40 0.26 0.066
2 0.387 125 123.84 0.10 0.011
3 0.12 42 38.40 0.58 0.338
4 0.093 24 29.76 -1.06 1.115
5 0.08 24 25.60 -0.32 0.100
total 1.000 320 320 1.629

value of the chi-square statistic =1.629

Are all the expected frequencies greater than 5? :Yes

What sampling distribution will you use? chi square

degrees of freedom =4

c) P-value > 0.100

d) Since the P-value > α, we fail to reject the null hypothesis.

e) At the 5% level of significance, the evidence is insufficient to conclude that the natural distribution of browse does not fit the feeding pattern.


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