Question

In: Statistics and Probability

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

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 14 11 16 41
Yearling 11 10 12 33
Adult 34 27 31 92
Column Total 59 48 59 166

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 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.)


Are all the expected frequencies greater than 5?

YesNo    


What sampling distribution will you use?

normaluniform    binomialchi-squareStudent'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.

Solutions

Expert Solution

level of significance =0.05

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

b)

Applying chi square test of independence:
Expected Ei=row total*column total/grand total Lamar Nez Perce Firehole Total
calf 14.57 11.86 14.57 41
Yearling 11.73 9.54 11.73 33
Adult 32.70 26.60 32.70 92
total 59 48 59 166
chi square    χ2 =(Oi-Ei)2/Ei Lamar Nez Perce Firehole Total
calf 0.0225 0.0617 0.1399 0.2241
Yearling 0.0453 0.0220 0.0063 0.0735
Adult 0.0518 0.0059 0.0883 0.1460
total 0.1196 0.0896 0.2344 0.4436
test statistic X2 = 0.444
Are all the expected frequencies greater than 5? :Yes
What sampling distribution will you use? chi-square
degrees of freedom =(row-1)*(column-1)=4

c)

p-value > 0.10

d)

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

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


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