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%	95
Rabbit brush	38.7%	131
Salt brush	12%	41
Service berry	9.3%	27
Other	8%	26

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?

(B) State the null and alternate hypotheses:

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

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

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

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

(C)  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.)

(D) Are all the expected frequencies greater than 5? (Yes/No)

(E) What sampling distribution will you use? (uniform / Student's t / chi-square / binomial / normal)

(F) What are the degrees of freedom?

(G) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

(H) Based on your answers in parts (a) to (g), will you reject or fail to reject the null hypothesis of independence?

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.

(I) 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

Following table shows the calculations:

	O	p	E=p*320	O-E	(O-E)^2	(O-E)^2/E
	95	0.32	102.4	-7.4	54.76	0.53476563
	131	0.387	123.84	7.16	51.2656	0.41396641
	41	0.12	38.4	2.6	6.76	0.17604167
	27	0.093	29.76	-2.76	7.6176	0.25596774
	26	0.08	25.6	0.4	0.16	0.00625
Total	320	1	320			1.38699144

(A)

Hypotheses are:

H0: The published deer feeding pattern follows the natural distribution of browse.

Ha: The published deer feeding pattern does not follow the natural distribution of browse.

Correct option:

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

(C)

The test statistics is

(d)

Yes

(e)

chi-square

(f)

Degree of freedom: df =n-1=4

(g)

The p-value using excel function "=CHIDIST(1.387,4)" is 0.846.

(h)

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

(l)

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