Question

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.    

Answer 1

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.