Compute the sum of squares between treatments.

Compute the mean square between treatments.

Compute the sum of squares due to error.

Compute the mean square due to error (to 1 decimal).

Set up the ANOVA table for this problem.

Source of Variation Sum of Squares Degrees of Freedom Mean Square F
Treatments
Error
Total

At the = .05 level of significance, test whether the means for thethree treatments are equal.

Calculate the value of the test statistic (to 2 decimals).

The degrees of freedom for the numerator on this case is given by where k =3 represent the number of groups.

For this case since the p value is lower than the significance level of 0.05 we can reject the null hypothesis that the means are equal.

Source variation      SS       df        MS           F          pv

__________________________________________

Treatments           1488      2         744       5.50   0.016

Within groups       2030    15        135.3    

_________________________________________

Total                      3518     17          

Step-by-step explanation:

For this case we assume the following data:

A: 162, 142, 165, 145, 148, 174

B: 142, 156, 124, 142, 136, 152

C: 126, 122, 138, 140, 150, 128

We have the following statistics given:

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

For this case the grand mean is given by:

After replace we got that

After replace we got:

And we have this property  

So then we can solve for the sum of squares within like this:

The degrees of freedom for the numerator on this case is given by where k =3 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

Now we can calculate the mean squares between and within like this:

The F statistic is calculated from:

And we ca calculate the p value like this since we knwo the degrees of freedom 2 for the numerator and 15 for the denominator:

Then we can conplete the table:

Source variation      SS       df        MS           F          pv

__________________________________________

Treatments           1488      2         744       5.50   0.016

Within groups       2030    15        135.3    

_________________________________________

Total                      3518     17          

For this case since the p value is lower than the significance level of 0.05 we can reject the null hypothesis that the means are equal.