Question

1. A sales analyst wants to determine whether there is difference in the mean monthly sales (in RM) of cosmetic company of six sales region in Johor. Several salespersons from each region are chosen and their monthly sales were recorded in Table Q2.

Monthly sale (RM)
Company A	Company B	Company C	Company D	Company E	Company F
4250	3930	3960	4350	4150	3970
3980	4010	4020	4450	3890	4160
4020	4050	4130	3880	4568	4370
4130	4220	4350	4574	4480	4290
4050	4177		4130	4430	4580
	4250		4375		4058
	4090				4148

Determine whether there is evidence of a difference in the monthly sales of the six cosmetics company at the 0.05 level of significance.

Answer 1

A sales analyst wants to determine whether there is difference in the mean monthly sales (in RM) of cosmetic company of six sales region in Johor.

Given data

Monthly sale (RM)
Company A	Company B	Company C	Company D	Company E	Company F
4250	3930	3960	4350	4150	3970
3980	4010	4020	4450	3890	4160
4020	4050	4130	3880	4568	4370
4130	4220	4350	4574	4480	4290
4050	4177		4130	4430	4580
	4250		4375		4058
	4090				4148

To determine whether there is evidence of a difference in the monthly sales of the six cosmetics company at the 0.05 level of significance.

Let aj denotes monthly sales of the 'j th' cosmetics company where j = 1,2,3,4,5,6.

To Test :-

H0 : aj = 0        { treament mean do not differ i.e monthly sales of the six cosmetics company may be same }

H1 : aj 0 , for atleast one i    { there may be difference in monthly sales of the six cosmetics company }

Test Statistics F:-

F = MSRes / MSE

Criteria :-

Reject null hypothesis if calculated test statistics F-value is greater than F-Critical value .

Calculations :

MSRes = SSRes / (k-1)

MSE = SSE / (n-k)

where

SSRes = Sum of squre due to Residual or Between sum of squares

SSE =    n

SSRes =

SSE =

and k - number of companies )

       nj = number of observation in each company

[ Note that we will find F-value using above formula . Also for calculation purpose we will use Excel or R , if manuall calculations are required can ask for that in comment box ]

Here n = 34 ( Total observations )

Now =

        = ( 4250 + 3980 + 4020 + 4130 + 4050 + 3930 + 4010 + ............. + 4580 + 4058 + 4148 ) / 34

        =4190.294118                 ( Overall Mean )               ... (i)

To find mean of each group

Formula =

For j = 1

= = ( 4250 + 3980 + 4020 + 4130 + 4050 ) / 5

= 4086

For j =2

= = ( 3930 + 4010 + 4050 + 4220 + 4177 + 4250 + 4090 ) / 7

= 4103.857143

Similary we calculate for j = 3,4,5,6

Thus

= 4086 , = 4103.857143 , = 4115.000

= 4293.167 , = 4303.600 , = 4225.143                              ... (ii)

also n1 =5 , n2=7 , n3 =4 , n4=6 , n5=5 , n6 =7                                 ... (iii)

For (i) , (ii) and (iii)

SSRes =

     = 5*(4086 - 4190.2941)² + 7*(4086 - 4190.2941)² +....+ 5*(4303.600- 4190.2941)²+ 6*( 4225.143   - 4190.2941)²

SSRes = 265551.3

Thus MSRes = SSRES / ( k-1)

                     = 265551.3 / ( 6-1)     

         MSRes = 53110.26

And

SSE =

        = (4250 - 4086)² + (3980 - 4086)² + (4020 - 4086)² +.... + (4580 -4225.143)² +(4058 -4225.143)² + (4148- 4225.143)²

After calculation

SSE =   45720 + 82324.85714 + 88500 + 310720.8333 + 311659.2 + 254342.8571

SSE = 1093267.747619

Thus MSE = SSE / ( n-k)

                     = 1093267.747619 / ( 34-6)     

         MSE = 39045.28

Also

Total sum of square TSS = SSRes + SSE =   265551.3 + 1093267.747619

TSS = 1358819

To calculate test statistics F-Value

F = MSRes / MSE

   = 53110.26 / 39045.28

F = 1.360222

Hence calculated test statistics value is F-Value = 1360222

ANOVA Table :-

Source	SS	df	MS	F
Company (Treatment)	265551.3	k-1 = 5	53110.26	1.36
Error	1093267.7476	n-k = 28	39045.28
Total	1358819	n-1 = 33

To find F-critical value

Here given level of significance = 0.05

df1 = k-1 = 5    and   df2 = n-k = 28

Critical F-Value with df1=5 and df2=28 at = 0.05 , can be obtained from statistical book or from any software like R/Excel .

From R

> qf(1-0.05,df1=5,df2=28)
[1] 2.558128

Thus

= 2.558128   

Conclusion :-

{ Reject null hypothesis if calculated test statistics F-value is greater than F-Critical value }

Since calculated F-Value = 1.36 is less that F-Critical value = 2.558128

i.e F-Value = 1.36 < 2.558128 ( )

We fail to reject null hypothesis

So , there is not sufficient evidence of a difference in the monthly sales of the six cosmetics company at the 0.05 level of significance.

Hence there may not be any difference in the monthly sales (in RM) of the six cosmetics company .