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	12	11	18	41
Yearling	10	14	9	33
Adult	38	31	23	92
Column Total	60	56	50	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 not independent.H₀: Age distribution and location are not independent.
H₁: Age distribution and location are independent.    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.

(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?

uniformbinomial    Student's tnormalchi-square

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.)


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

Answer 1

We can run this analysis in excel as shoen below

Please note that each cell in the expected table is calculated as

rowtotal*coltotal/Total

State the null and alternate hypotheses.
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.   

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.

as can be seen from the expected table , the frequencies are all greater than 5

we shall use the students chi square test

note that the chisq.tst function in excel returns the p values of the test , as this p value is not less than 0.05 , hence we fail to reject the null hypothesis at 5% signficance level

since the P-value ? ?, we fail to reject the null hypothesis.

hence finally we conclude that

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