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%	100
Rabbit brush	38.7%	133
Salt brush	12%	40
Service berry	9.3%	23
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 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.

(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?

YesNo     

What sampling distribution will you use?

Student's tnormal     uniformchi-squarebinomial

What are the degrees of freedom?


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

P-value > 0.1000.050 < P-value < 0.100     0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-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 (5%)

State the null and alternate hypotheses.

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

(b)

Test Statistic

O : Observed Frequency

E : Expected Frequency

Given:

Type of Browse	O :Observed Number of Deer Feeding on This Plant
Sage brush	100
Rabbit brush	133
Salt brush	40
Service berry	23
Other	24

Expected Frequency : E = Total number of deers in the sample * Plant composition %

Type of Browse	Plant Composition	E : Expected Frequency	E
Sage brush	32%	32%*320	102.4
Rabbit brush	38.7%	38.7%*320	123.84
Salt brush	12%	12%*320	38.4
Service berry	9.3%	9.3%*320	29.76
Other	8%	8%*320	25.6
	Total		320

O	E	O-E	(O-E)2	(O-E)2/E
100	102.4	-2.4	5.76	0.0563
133	123.84	9.16	83.9056	0.6775
40	38.4	1.6	2.56	0.0667
23	29.76	-6.76	45.6976	1.5355
24	25.6	-1.6	2.56	0.1
			Total	2.436

value of the chi-square statistic for the sample :

Are all the expected frequencies greater than 5?

Yes

What sampling distribution will you use?

chi-square

Degrees of freedom = Number of types of browse - 1 = 5-1 =4

What are the degrees of freedom?

4

(c)

For 4 degrees of freedom

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

P-value > 0.100

(d)

p-value : 0.656 > Level of significance : : 0.05 ; Fail to reject the null hypothesis;

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.

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

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