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%	125
Salt brush	12%	46
Service berry	9.3%	25
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    

normal

uniform

Student's t

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

Solution:

Given: The types of browse favored by deer are shown in the following table.

Type of Browse	Plant Composition in Study Area	Observed Number of Deer Feeding on This Plant
Sage brush	32%	100
Rabbit brush	38.70%	125
Salt brush	12%	46
Service berry	9.30%	25
Other	8%	24

Level of significance = 5% = 0.05

Claim: the natural distribution of browse fits the deer feeding pattern.

Part a) What is the level of significance?

0.05

State the null and alternate hypotheses.

H₀: The distributions are the same.

H₁: The distributions are different.

Part b) Find the value of the chi-square statistic for the sample.

Chi square test statistic for goodness of fit

Where

O_i = Observed Counts

E_i =Expected Counts

To get Ei values,we multiply each % value by 320.

Thus we need to make following table:

Type of Browse	Expected % values: Plant Composition in Study Area	Oi: Observed Number of Deer Feeding on This Plant	Ei:	Oi2/Ei
Sage brush	32%	100	102.400	97.656
Rabbit brush	38.70%	125	123.840	126.171
Salt brush	12%	46	38.400	55.104
Service berry	9.30%	25	29.760	21.001
Other	8%	24	25.600	22.500
		N= 320

Thus we get:

Are all the expected frequencies greater than 5?

Yes

What sampling distribution will you use?

chi-square

What are the degrees of freedom?

df = k - 1 = 5 - 1 = 4

df =4

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

Look in Chi-square table for df = 4 row and find interval in which fall then find corresponding right tail area interval.

fall between 1.064 and 7.779

corresponding right tail area interval is: 0.100 and 0.900

thus P-value > 0.100

Thus correct option is: P-value > 0.100

Part 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 P-value > 0.100 which is also > 0.05 level of significance, hence we fail to reject null hypothesis.

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

Part 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