Question

Please answer this question based on the following data. SSTR = 6,750 H0: μ1= μ2=μ3=μ4 SSE = 8,000 Ha: at least one mean is different Nτ = 20 The null hypothesis is to be tested at the 5% level of significance. The p-value is less than .01 between .01 and .025 between .025 and .05 between .05 and .10

Answer 1

Answer:

$P_v =P(F_{3,16}>4.5)=0.0180$

So on this case the best option for the answer would be:

between .01 and .025

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 $4$ groups and on each group from $j=1,\dots,5$ we have $5$ individuals on each group we can define the following formulas of variation:  

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$  

$SS_{between=Treatment}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =6750$  

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8000$  

And we have this property  

$SST=SS_{between}+SS_{within}=6750+8000=14750$  

The degrees of freedom for the numerator on this case is given by $df_{num}=k-1=4-1=3$ where k =4 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=20-4=16$.

And the total degrees of freedom would be $df=N-1=20 -1 =19$

We can find the $MSR=\frac{6750}{3}=2250$

And $MSE=\frac{8000}{16}=500$

And the we can find the F statistic $F=\frac{MSR}{MSE}=\frac{2250}{500}=4.5$

And with that we can find the p value. On this case the correct answer would be 3 for the numerator and 16 for the denominator.  

$P_v =P(F_{3,16}>4.5)=0.0180$

So on this case the best option for the answer would be:

between .01 and .025