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)

(i)

the level of significance = = 0.05

(ii)

Correct option:

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

(b)

(i)

the value of the chi-square statistic for the sample. is got as follows:

Observed (O)	Expected (E)	(O - E)2/E
108	320 X 0.32 = 102.400	0.306
113	320 X 0.387 = 123.840	0.949
39	320 X 0.12 = 38.400	0.009
32	320 X 0.093 = 29.760	0.169
28	320 X 0.08 = 25.600	0.225
Total = =		1.658

So,

the value of the chi-square statistic for the sample = 1.658

(ii)

Correct option:

Yes

(iii)

Correct option:

chi-square

(iv)

the degrees of freedom = 5 - 1 = 4

(c)
By Technology, p - value = 0.7983

So,

Correct option:

P-value > 0.100

(d)

Correct option:

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

(e)

Correct option:

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