States the oneway ANOVA hypothesis then calculate the Sum of

States the one-way ANOVA hypothesis, then calculate the Sum of Squares (SST), sum of squares between (SSB), and sum of squares within (SSW) using the information below that was recorded by 15 customers satisfaction ratings for Wal-Mart, Target, and TJ Max using a scale 1 to 10.

Wal-Mart = 7,7,6,5,3

Target = 8,9,7,6,9

TJ Max = 9,7,8,10,8

Then perform a hypothesis test to determine whether there is difference in the customer satisfaction rating of the scores at a 0.05 significance level using the mean square between (MSB) and then the mean square within (MSW).

Solution

Here we set up the null hypothesis

H₀   : To test whether there is a significant difference in the customer satisfaction rating of the score

Against the alternative hypothesis

H_{1    :} To test whether there is no significant difference in the customer satisfaction rating of the score

Raw sum of squares (R.S.S) = ( X_ij )²

                                       = (7)²+(7)²+ (6)²+(5)²+(3)²+(8)²+(9)²+(7)²+(6)²+(9)²+(9)²+(7)²+(8)²+(10)²+(8)²

                                   = 837

Grand total(G) = ( X_ij )

                        =7+7+6+5+3+8+9+7+6+9+9+7+8+10+8

                         =109

Correction Factor(C.F) = (G)²/N

                                    = (109)²/15

                                = 792.0667

Total sum of squares(SST)   = R.S.S   -    C.F

                                                = 837 -792.0667

                                                =44.9333

Treatment(Between) Sum of squares (SSB) = ( -   C.F

                                                = + +    -    792.0667

                                                =   21.7333

Error(within) sum of square   =   T.S.S - Tr.S.S

                                                           =44.9333 – 2107333

                                                =23.2

Sources of variation

Degrees of freedom

Sum of squares

Mean sum of square

F Ratio

Between( Treatment)

5-1 = 4

21.7333

S²_tr =

   =5.4333

F_(N-1,N-K)= S²_tr / S²_e

Within( Error)

15-5 = 10

23.2

S²_e =

   = 2.32

    = 5.4333/2.32

    =2.3419

Total

15-1 = 14

44.9333

F_tab (4,10) at 0.05 level of significance is 3.48

F_cal (4,10) is 2.3419

Since F_cal   F_tab we accept the null hypothesis at 0.05 level of significance

Therefore there is a signicficance difference in the customer satisfaction rating of the scores.

Sources of variation	Degrees of freedom	Sum of squares	Mean sum of square	F Ratio
Between( Treatment)	5-1 = 4	21.7333	S2tr =    =5.4333	F(N-1,N-K)= S2tr / S2e
Within( Error)	15-5 = 10	23.2	S2e =    = 2.32	= 5.4333/2.32     =2.3419
Total	15-1 = 14	44.9333