Given the following contingency table, conduct a test for independence at the 5%
ID: 3060083 • Letter: G
Question
Given the following contingency table, conduct a test for independence at the 5% significance level Use T e 23 47 32 53 Critical value approach a-1. Choose the null and alternative hypotheses Mp The two vanables are independent: Hi The two variables are dependent. He The two variables are dependent, Ha The two variables are independent a-2. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final Test statistic a-3. Specify the decision ule. (Round your answer to 3 decimal places.) Reject Ho if Xa > a-4. What is your conclusion? Reject Ho there is enough evidence to support the claim that the two categories are not independent Reject He there is not enough evidence to support the claim that the two categories are not independent Do not reject He there is enough evidence to support the claim that the two categories are not independent Do not reject H-there is not enough evidence to support the claim that the two categories are not independent p-value approach b-1Approximate the p-value 0050a01 Type here toExplanation / Answer
Result:
a1).
H0: The two variables are independent. H1: The two variables are dependent
a2).
Test statistic = 0.385
a3).
Critical value = 3.841
Do not reject Ho., there is not enough evidence to support the claim.
b1).
P value > 0.10
b2).
No, since p value is more than .
Chi-Square Test
Observed Frequencies
Column variable
Calculations
Row variable
C1
C2
Total
fo-fe
R1
23
47
70
-1.8387
1.8387
R2
32
53
85
1.8387
-1.8387
Total
55
100
155
Expected Frequencies
Column variable
Row variable
C1
C2
Total
(fo-fe)^2/fe
R1
24.83871
45.16129
70
0.1361
0.0749
R2
30.16129
54.83871
85
0.1121
0.0617
Total
55
100
155
Data
Level of Significance
0.05
Number of Rows
2
Number of Columns
2
Degrees of Freedom
1
Results
Critical Value
3.841459
Chi-Square Test Statistic
0.384717
p-Value
0.535089
Do not reject the null hypothesis
Chi-Square Test
Observed Frequencies
Column variable
Calculations
Row variable
C1
C2
Total
fo-fe
R1
23
47
70
-1.8387
1.8387
R2
32
53
85
1.8387
-1.8387
Total
55
100
155
Expected Frequencies
Column variable
Row variable
C1
C2
Total
(fo-fe)^2/fe
R1
24.83871
45.16129
70
0.1361
0.0749
R2
30.16129
54.83871
85
0.1121
0.0617
Total
55
100
155
Data
Level of Significance
0.05
Number of Rows
2
Number of Columns
2
Degrees of Freedom
1
Results
Critical Value
3.841459
Chi-Square Test Statistic
0.384717
p-Value
0.535089
Do not reject the null hypothesis
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