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	15	9	17	41
Yearling	10	8	15	33
Adult	31	24	37	92
Column Total	56	41	69	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?


State the null and alternate hypotheses.

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


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

uniformchi-square    normalbinomialStudent's t

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 > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-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) What is the level of significance?

The level of significance is given as α = 0.05.

State the null and alternate hypotheses.

H₀: Age distribution and location are independent.

H₁: Age distribution and location are not independent.

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

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	Lamar	Nez	Firehole	Total
Calf	15	9	17	41
Yearling	10	8	15	33
Adult	31	24	37	92
Total	56	41	69	166

Expected Frequencies
	Column variable
Row variable	Lamar	Nez	Firehole	Total
Calf	13.83133	10.12651	17.04217	41
Yearling	11.13253	8.150602	13.71687	33
Adult	31.03614	22.72289	38.24096	92
Total	56	41	69	166

(O - E)
1.168675	-1.12651	-0.04217
-1.13253	-0.1506	1.283133
-0.03614	1.277108	-1.24096

(O - E)^2/E
0.098747	0.125316	0.000104
0.115214	0.002783	0.12003
4.21E-05	0.071778	0.040271

Chi square = ∑[(O – E)^2/E] = 0.574285

Chi square test statistic = 0.574

Are all the expected frequencies greater than 5?

Yes

What sampling distribution will you use?

Chi square

What are the degrees of freedom?

We are given

Number of rows = r = 3

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 2*2 = 4

Degrees of freedom = 4

(c) Find or estimate the P-value of the sample test statistic.

P-value = 0.966

(by using Chi square table)

p-value > 0.100

(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.

(e) Interpret your conclusion in the context of the application.

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