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

And we have this property  

The degrees of freedom for the numerator on this case is given by where k =4 represent the number of groups.  

The degrees of freedom for the denominator on this case is given by .  

And the total degrees of freedom would be  

We can find the

And

And the best answer would be:

d. 2,250

Step-by-step explanation:

We want to test the following null hypothesis:

If we assume that we have groups and on each group from we have individuals on each group we can define the following formulas of variation:  

And we have this property  

The degrees of freedom for the numerator on this case is given by where k =4 represent the number of groups.  

The degrees of freedom for the denominator on this case is given by .  

And the total degrees of freedom would be  

We can find the

And

And the best answer would be:

d. 2,250

22 dimes and 9 nickels

step-by-step explanation: