Question

SSTR = 6,750 H0: μ1=μ2=μ3=μ4
SSE = 8,000 Ha: at least one mean is different
nT = 20
Refer to Exhibit 10-16. The mean square between treatments (MSTR) equals
a. 400
b. 500
c. 1,687.5
d. 2,250

Answer 1

Answer:

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

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

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

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 $MSTR=\frac{6750}{3}=2250$

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

And the best answer would be:

d. 2,250

Step-by-step explanation:

We want to test the following null hypothesis:

$ H0: \mu_1 =\mu_2 =\mu_3 =\mu_4$

If we assume that we have $4$ groups and on each group from $j=1,\dots,k$ we have $k$ 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 $MSTR=\frac{6750}{3}=2250$

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

And the best answer would be:

d. 2,250