Question

The following table shows age distribution and location of a random sample of 166 buffalo in a national park.

Age	Lamar District	Nez Perce District	Firehole District	Row Total
Calf	16	10	15	41
Yearling	13	9	11	33
Adult	33	27	32	92
Column Total	62	46	58	166

Use a chi-square test to determine if age distribution and location are independent at the 0.05 level of significance.

(a) What is the level of significance?
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State the null and alternate hypotheses.

H₀: Age distribution and location are not independent.
H₁: Age distribution and location are not independent.H₀: Age distribution and location are independent.
H₁: Age distribution and location are independent.    H₀: Age distribution and location are independent.
H₁: Age distribution and location are not independent.H₀: Age distribution and location are not independent.
H₁: Age distribution and location are independent.

(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.)
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Are all the expected frequencies greater than 5?

Yes

No    

What sampling distribution will you use?

binomial

Student's t    

chi-square

normal

uniform

What are the degrees of freedom?
_________________

(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

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 of independence?

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, there is sufficient evidence to conclude that age distribution and location are not independent.

At the 5% level of significance, there is insufficient evidence to conclude that age distribution and location are not independent.    

Answer 1

a) Level of significance = 0.05

Null and alternate hypotheses:

H₀: Age distribution and location are independent.
H₁: Age distribution and location are not independent.

b) Expected frequency of a cell = sum of row*sum of column / total sum

Observed Frequencies
	Lamar	Nez Perce	Firehole	Total
Calf	16	10	15	41
Yearling	13	9	11	33
Adult	33	27	32	92
Total	62	46	58	166
Expected Frequencies
	Lamar	Nez Perce	Firehole	Total
Calf	15.3133	11.3614	14.3253	41
Yearling	12.3253	9.1446	11.53012	33
Adult	34.3614	25.4940	32.14458	92
Total	62	46	58	166
	(fo-fe)^2/fe
Calf	0.0308	0.1631	0.0318
Yearling	0.0369	0.0023	0.0244
Adult	0.0539	0.0890	0.0007

Test statistic:

χ² = ∑ ((fo-fe)²/fe) =  0.4329

Yes, are all the expected frequencies greater than 5.

Sampling distribution: chi-square

degrees of freedom = (3-1)(3-1) = 4

c) p-value = CHISQ.INV.RT(0.4329, 4) = 0.9797

p-value > 0.100

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

e) Conclusion:

At the 5% level of significance, there is insufficient evidence to conclude that age distribution and location are not independent.