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% 106 Rabbit brush 38.7% 123 Salt brush 12% 41 Service berry 9.3% 28 Other 8% 22 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 the same. H0: The distributions are different. 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? Student's t uniform normal chi-square binomial 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

level of significance 0.05

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

b)

	relative	observed	Expected	residual	Chi square
category	frequency	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
1	0.320	106	102.40	0.36	0.127
2	0.387	123	123.84	-0.08	0.006
3	0.120	41	38.40	0.42	0.176
4	0.093	28	29.76	-0.32	0.104
4	0.080	22	25.60	-0.71	0.506
total	1.000	320	320		0.919

chi-square statistic =0.919

Are all the expected frequencies greater than 5? Yes

What sampling distribution will you use? chi-square

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.