The following table shows age distribution and location of a random sample of 16
ID: 3265638 • Letter: T
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 17 12 41 Yearling 9 13 11 33 Adult 32 32 28 92 Column Total 53 62 51 166 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? .05 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 not independent. H1: Age distribution and location are 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. (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? Yes No What sampling distribution will you use? chi-square Student's t uniform normal binomial What are the degrees of freedom? 4 (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 independeThe following table shows age distribution and location of a random sample of 166 buffalo in a national park.
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 not independent.
H1: Age distribution and location are 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.
(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?
chi-squareStudent's t uniformnormalbinomial
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 independent.
nt.
Age Lamar District Nez Perce District Firehole District Row Total Calf 12 17 12 41 Yearling 9 13 11 33 Adult 32 32 28 92 Column Total 53 62 51 166Explanation / Answer
(a) The level of significance=0.05
The null and alternate hypotheses are
H0: Age distribution and location are independent.
H1: Age distribution and location are not independent.
(b) From following results the value of the chi-square statistic =1.039
Are all the expected frequencies greater than 5-Yes
What sampling distribution will you use-chi-square
What are the degrees of freedom-df=4
(c) The P-value of the sample test statistic=0.904
(d) We fail to reject the null hypothesis of independence
Since the P-value > , we fail to reject the null hypothesis
(e) Interpretation:
At the 5% level of significance, there is insufficient evidence to conclude that age distribution and location are not independent.
.
Chi-Square Test
Observed Frequencies
Column variable
Calculations
Age
Lamar District
Nez Perce District
Firehole District
Total
fo-fe
Calf
12
17
12
41
-1.09036
1.686747
-0.59639
Yearling
9
13
11
33
-1.53614
0.674699
0.861446
Adult
32
32
28
92
2.626506
-2.36145
-0.26506
Total
53
62
51
166
Expected Frequencies
Column variable
Age
Lamar District
Nez Perce District
Firehole District
Total
(fo-fe)^2/fe
Calf
13.09036
15.31325
12.59639
41
0.090822
0.185794
0.028236
Yearling
10.53614
12.3253
10.13855
33
0.223966
0.036934
0.073195
Adult
29.37349
34.36145
28.26506
92
0.234856
0.162287
0.002486
Total
53
62
51
166
Data
Level of Significance
0.05
Number of Rows
3
Number of Columns
3
Degrees of Freedom
4
Results
Critical Value
9.487729
Chi-Square Test Statistic
1.038576
p-Value
0.903891
Do not reject the null hypothesis
Expected frequency assumption
is met.
Chi-Square Test
Observed Frequencies
Column variable
Calculations
Age
Lamar District
Nez Perce District
Firehole District
Total
fo-fe
Calf
12
17
12
41
-1.09036
1.686747
-0.59639
Yearling
9
13
11
33
-1.53614
0.674699
0.861446
Adult
32
32
28
92
2.626506
-2.36145
-0.26506
Total
53
62
51
166
Expected Frequencies
Column variable
Age
Lamar District
Nez Perce District
Firehole District
Total
(fo-fe)^2/fe
Calf
13.09036
15.31325
12.59639
41
0.090822
0.185794
0.028236
Yearling
10.53614
12.3253
10.13855
33
0.223966
0.036934
0.073195
Adult
29.37349
34.36145
28.26506
92
0.234856
0.162287
0.002486
Total
53
62
51
166
Data
Level of Significance
0.05
Number of Rows
3
Number of Columns
3
Degrees of Freedom
4
Results
Critical Value
9.487729
Chi-Square Test Statistic
1.038576
p-Value
0.903891
Do not reject the null hypothesis
Expected frequency assumption
is met.
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