The following is a partial ANOVA table regarding the service

The following is a partial ANOVA table regarding the service satisfaction on five auto body shops. 35 customers were surveyed. Use

The following is a partial ANOVA table regarding the service satisfaction on five auto body shops. 35 customers were surveyed.

alpha = 0.05

We know that,

sum of squares between groups + sum of squares within groups = total sum of squares.

16.3 + sum of squares within groups = 29

sum of squares within groups = 29 - 16.3

sum of squares within groups = 12.7

d.f. for between groups = r - 1

where r is number of shops = 5

d.f. for between groups = 5 - 1 = 4

d.f. for total = n - 1

where n is total number of customers were surveyed = 35

d.f. for total = 35 - 1 = 34

d.f. for total = d.f. for between groups + d.f. for within groups

34 = 4 + d.f. for within groups

d.f. for within groups = 34 - 4 = 30

Mean square :

Mean square between groups(MSB) = sum of squares for between groups / d.f. for between groups

MSB = 16.3 / 4 = 4.075

Mean square within groups(MSE) = sum of squares for within groups / d.f. fo within groups

MSE = 12.7 / 30 = 0.4233

The null hypothesis is,

H0 : mu1=mu2=mu3

and the alternative hypothesis is,

H1: at least one of the means is different.

Test statistic is,

F = MSB / MSE = 4.075 / 0.4233 = 9.6260

The P-value we can calculate by using EXCEL.

syntax:

FDIST(x, deg_freedom1, deg_freedom_2)

where x is the test statistic value.

deg_freedom1 = d.f. for between groups = 4

deg_freedom2 = d.f. for within groups = 30

P-value = 0.0000

P-value < alpha

reject H0 at 5% level of significance.

Conclusion : at least one of the means is different.

The complete partial ANOVA table is,