Respuesta :
Answer:
a) Null hypothesis: [tex] \mu_A =\mu_B =\mu C[/tex]
Alternative hypothesis: [tex] \mu_i \neq \mu_j, i,j=A,B,C [/tex]
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5[/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333[/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667[/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=3-1=2[/tex] where k =3 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=3*6-3=15[/tex].
And the total degrees of freedom would be [tex]df=N-1=3*6 -1 =15[/tex]
The mean squares between groups are given by:
[tex] MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166[/tex]
And the mean squares within are:
[tex] MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544[/tex]
And the F statistic is given by:
[tex] F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326[/tex]
And the p value is given by:
[tex] p_v= P(F_{2,15} >11.326) = 0.00105[/tex]
So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.
b) [tex] (\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}[/tex]
The degrees of freedom are given by:
[tex] df = n_B +n_C -2= 6+6-2=10[/tex]
The confidence level is 99% so then [tex] \alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex] and the critical value would be: [tex] t_{\alpha/2}=3.169[/tex]
The confidence interval would be given by:
[tex] (43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321[/tex]
[tex] (43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345[/tex]
Step-by-step explanation:
Previous concepts
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"
Part a
Null hypothesis: [tex] \mu_A =\mu_B =\mu C[/tex]
Alternative hypothesis: [tex] \mu_i \neq \mu_j, i,j=A,B,C [/tex]
If we assume that we have [tex]3[/tex] groups and on each group from [tex]j=1,\dots,6[/tex] we have [tex]6[/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 =20.5[/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333[/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667[/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=3-1=2[/tex] where k =3 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=3*6-3=15[/tex].
And the total degrees of freedom would be [tex]df=N-1=3*6 -1 =15[/tex]
The mean squares between groups are given by:
[tex] MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166[/tex]
And the mean squares within are:
[tex] MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544[/tex]
And the F statistic is given by:
[tex] F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326[/tex]
And the p value is given by:
[tex] p_v= P(F_{2,15} >11.326) = 0.00105[/tex]
So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.
Part b
For this case the confidence interval for the difference woud be given by:
[tex] (\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}[/tex]
The degrees of freedom are given by:
[tex] df = n_B +n_C -2= 6+6-2=10[/tex]
The confidence level is 99% so then [tex] \alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex] and the critical value would be: [tex] t_{\alpha/2}=3.169[/tex]
The confidence interval would be given by:
[tex] (43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321[/tex]
[tex] (43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345[/tex]
