Respuesta :
Answer:
Degrees of freedom for independence in chi-square statistic
ν = ( r-1) (s-1) =6
Step-by-step explanation:
Explanation:-
Given data chi-square test for independence is being used to evaluate the relationship between two variables
Given "A" is classified into 3 categories
Second 'B' is classified into 4 categories
In this chi-square test, we test if two attributes A and B under consideration are independent or not
We will assume that
Null Hypothesis : H₀: The two variables are independent
Degrees of freedom in chi-square test for independence
ν = ( r-1) (s-1)
Given data 'r' = 3 and 's' = 4
Degrees of freedom for independence
ν = ( r-1) (s-1) = ( 3-1) ( 4-1) = 2×3 =6
Test statistic
χ ² = ∑ [tex]\frac{(O-E)^{2} }{E}[/tex]
This question is based on Chi-square test. Therefore, the chi-square statistic for this test would have df equal to [tex]X^{2} = \sum \dfrac{(O-E)^2}{E}[/tex].
Given:
Chi-square test for independence is being used to evaluate the relationship between two variables . Given "A" is classified into 3 categories . Second 'B' is classified into 4 categories
According to the question,
In this chi-square test, we would be test if two attributes A and B under consideration are independent or not.
Let assumed that, null Hypothesis : H₀: The two variables are independent
Now, degrees of freedom in chi-square test for independence is,
⇒ ν = ( r-1) (s-1)
It is given that, 'r' = 3 and 's' = 4.
Thus, degrees of freedom for independence is,
ν = ( r-1) (s-1) = ( 3-1) ( 4-1) = 2×3 =6
Therefore, test statistic be,
[tex]X^{2} = \sum \dfrac{(O-E)^2}{E}[/tex]
Therefore, the chi-square statistic for this test would have df equal to [tex]X^{2} = \sum \dfrac{(O-E)^2}{E}[/tex].
For more details, prefer this link:
https://brainly.com/question/23879950
