Respuesta :
Answer:
C. 5 numerator and 114 denominator degrees of freedom
Step-by-step explanation:
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have [tex]6[/tex] groups and on each group from [tex]j=1,\dots,20[/tex] we have [tex]20[/tex] individuals on each group we can define the following formulas of variation:
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]
And we have this property
[tex]SST=SS_{between}+SS_{within}[/tex]
The degrees of freedom for the numerator on this case is given by [tex]df_{num}=df_{within}=k-1=6-1=5[/tex] where k =6 represent the number of groups.
The degrees of freedom for the denominator on this case is given by [tex]df_{den}=df_{between}=N-K=6*20-6=114[/tex].
And the total degrees of freedom would be [tex]df=N-1=6*20 -1 =119[/tex]
On this case the correct answer would be 5 degrees of freedom for the numerator and 119 degrees of freedom for the denominator.
C. 5 numerator and 114 denominator degrees of freedom
The required degrees of freedom 5 numerator and 114 denominator .
Given that,
An ANOVA procedure is applied to data obtained from 6 samples where each sample contains 20 observations.
We have to determine,
The degrees of freedom for the critical value of F are Select one.
According to the question,
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
The groups and on each group from individuals on each group this define the following formulas of variation:
[tex]S\ S_t_o_t_a_l = \sum_{j=1}^{p} \sum_{i=1}^{n_j} (x_i_j-x)^{2}\\\\S\S_b_e_w_e_e_n = S \ S_m_o_d_e_l = \sum_{j=1}^{p} {n_j} (x_i_j-x)^{2}\\\\S \ S_b_e_t_w_e_e_n = S \ S_e_r_r_o_r = \sum_{j=1}^{p} \sum_{i=1}^{n_j} (x_i_j-x_j)^{2}\\[/tex]
And by using this property,
SST = [tex]S\ S_b_e_t_w_e_e_n + S \ S_w_i_t_h_i_n[/tex]
The degrees of freedom for the numerator on this case is given by,
[tex]df_n_u_m = df_w_i_t_h_i_n = k-1 = 6-1 = k = 6 \\[/tex]
Where k =6 represent the number of groups.
The degrees of freedom for the denominator on this case is given by,
[tex]df_n_u_m = df_w_i_t_h_i_n = N - k = 6 \times 20 - 6 = 114\\[/tex]
And the total degrees of freedom would be ,
[tex]df = N -1 = 6 \times 120 -1 = 119[/tex]
The 5 degrees of freedom for the numerator and 119 degrees of freedom for the denominator.
Hence, The required degrees of freedom 5 numerator and 114 denominator .
To know more about Degrees of freedom click the link given below.
https://brainly.com/question/20376777
