Question

It contains information on a two factor experiment involving two types of laundry detergent (Super and Best) and three washing temperatures (cold, warm and hot). The numbers given represent the quantity of dirt removed in each wash. Use Excel to conduct a two-factor ANOVA experiment to determine if differences exist in either the type of detergent or the washing temperature. Also determine if differences exist with respect to the interaction of the two factors. Use α = 0.05.

	Cold	Warm	Hot
Super	4	7	10
	5	9	12
	6	8	11
	5	12	9
Best	6	13	12
	6	15	13
	4	12	10
	4	12	13

Answer 1

Step.1 First copy and paste data in Excel sheet,

Step.2 Go to 'Data' menu. Select 'Data Analysis' and then select 'Anova : Two factor with replication'

Step.3 New window will pop-up on screen. Refer following screen shot. Enter information accordingly.

Excel output:

	Cold	Warm	Hot
Super	4	7	10		Anova: Two-Factor With Replication
	5	9	12
	6	8	11		SUMMARY	Cold	Warm	Hot	Total
	5	12	9		Super
Best	6	13	12		Count	4	4	4	12
	6	15	13		Sum	20	36	42	98
	4	12	10		Average	5	9	10.5	8.166667
	4	12	13		Variance	0.666667	4.666667	1.666667	7.787879
					Best
					Count	4	4	4	12
					Sum	20	52	48	120
					Average	5	13	12	10
					Variance	1.333333	2	2	15.27273
					Total
					Count	8	8	8
					Sum	40	88	90
					Average	5	11	11.25
					Variance	0.857143	7.428571	2.214286
					ANOVA
					Source of Variation	SS	df	MS	F	P-value	F crit
					Sample	20.16667	1	20.16667	9.810811	0.005758	4.413873
					Columns	200.3333	2	100.1667	48.72973	5.44E-08	3.554557
					Interaction	16.33333	2	8.166667	3.972973	0.037224	3.554557
					Within	37	18	2.055556
					Total	273.8333	23

P-value for type of detergent is 0.005758 which is less than 0.05, hence at 5% level of significance we reject null hypothesis and conclude that there exist difference between type of detergent.

P-value for washing temperature is almost zero which is less than 0.05, hence at 5% level of significance we reject null hypothesis and conclude that there exist difference between washing temperature.

P-value for interaction effect is 0.0372 which is less than 0.05, hence at 5% level of significance we reject null hypothesis and conclude that there exist difference in the interaction effect.