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	10	9	14	33
Adult	32	24	36	92
Column Total	58	43	65	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 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.

(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?

binomialStudent's t    chi-squareuniformnormal

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 > ?, 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 independent.   

Answer 1

a)

We have to test whether age distribution and location are independent or not at the 0.05 level of significance.

Therefore, given that the level of significance = 0.05

The null and alternative hypothesis are

Null hypothesis, H₀ : Age distribution and location are independent.

Alternative hypothesis, H₁ :Age distribution and location are not independent.

b)

Given the values in the table are the observed frequency which are denoted by O_ij .

Let, E_ij denote the expected frequency and is given by,

Observed frequency (Oij)	Expected Frequency (Eij)
16	14.325	0.196
10	10.620	0.036
15	16.054	0.069
10	11.530	0.203
9	8.548	0.024
14	12.922	0.090
32	32.145	0.001
24	23.831	0.001
36	36.024	0.000

From the above table, it is clear that all expected frequencies are greater than 5.

Then the test statistic for testing above hypothesis is

where m = number of rows and n = number of columns

Here, the sampling distribution is chi-square distribution.

The value of the test statistic is,

Therefore test statistic = 0.620

Here, m = no. of rows = 3 and n = no. of columns = 3

Therefore, the degrees of freedom = (m-1)(n-1) = 2*2 = 4

c)

The above probability is obtained by using EXCEL with the function "CHIDIST(0.620,4)".

Therefore, the p-value = 0.961

Hence, p-value >0.1000

d)

Since, p-value > level of significance = 0.05, we fail to reject the null hypothesis at 5% level of significance.

e)

Since we fail to reject the null hypothesis at 5 % level of significance, we conclude that at the 5% level of significance, there is insufficient evidence to conclude that age distribution and location are not independent.