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% 108 Rabbit brush 38.7% 113 Salt brush 12% 39 Service berry 9.3% 32 Other 8% 28 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? Yes No What sampling distribution will you use? Student's t normal binomial uniform chi-square 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 goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
1	0.3200	108.0000	102.40	0.55	0.306
2	0.3870	113.0000	123.84	-0.97	0.949
3	0.1200	39.0000	38.40	0.10	0.009
4	0.0930	32.0000	29.76	0.41	0.169
5	0.0800	28.0000	25.60	0.47	0.225
total	1.000	320	320		1.6581
test statistic X2 =		1.658

value of the chi-square statistic =1.658

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.