Question

Suppose you want to determine whether the brand of laundry detergent used and the temperature affects the amount of dirt removed from your laundry. To this end, you buy two different brand of detergent ( Super and Best) and choose three different temperature levels (cold, warm, and hot). Then you divide your laundry randomly into 6r piles of equal size and assign each r=4 piles into the combination of (Super and Best) and (cold,warm, and hot).

	super	best
cold	4,5,6,5	6,6,4,4
warm	7,9,8,12	13,15,12,12
hot	10,12,11,9	12,13,10,13

a.) Complete the ANOVA table (show work).

Source	DF	SS	MS	F
Detergent (A)		20.167
Temperature (B)		200.333
Interaction		16.333
Error				X
Total		273.833	X	X

b.) Use the ANOVA table to test for the global utility of the model. Clearly state all 5 steps.

c.) Test whether there is a significant interaction. Clearly state all 5 steps.

d.) Is it appropriate to test for main effect of factor A or B? Why? If yes, use hypothesis tests to determine which of the factors are significant and clearly state all 5 steps for each test.

Thank You for any insight you can provide!

Answer 1

I USED SPSS FOR ANALYSIS

C) NULL HYPOTHESIS H0: THERE IS NO INETERACTION

ALTERNATIVE HYPOTHESIS HA: THERE IS INTERACTION

ALPHA= 0.05

DEGREES OF FREEDOM= (3-1)*(2-1)= 2

F= 3.973

P value= 0.037

Since P value <0.05 hence significant.

Conclusion: Since P value is significant we reject null hypothesis and conclude that there is an interaction effect between temperatutre and detergent type.

d) Yes it is appropriate to test for main effects. Because to know how they act indepedently.

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS HA:

ALPHA=0.05

DEGREES OF FREEDOM= 2-1=1

F=9.811

P VALUE= 0.006 <0.05 HENCE SIGNIFICANT.

SINCE P VALUE IS SIGNIFICANT . WE REJECT NULL HYPOTHESIS AT 0.05 LEVEL OF SIGNIFICANCE.

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS HA: At least one mean is different.

alpah=0.05

degrees of freedom= n-1= 3-1=2

F= 48.730

P VALUE= 0.000<0.05. THEREFORE SIGNIFICANT AT 0.05

CONCLUSION: SINCE P VALUE IS SIGNIFICANT AT 0.05 LEVEL OF SIGNIFICANCE WE THERFORE REJECT NULL HYPOTHESIS AND COCNLUDE THAT AT LEAST ONE MEAN IS DIFFERENT.

SOURCE	SUM OF SQUARES	D.F	MEAN SUM OF SQUARES	F	P VALUE
TEMPERATURE	200.33	3-1=2	200.33/2=100.167	48.730	0.000
DETERGENT	20.167	2-1=1	20.167/1=20.167	9.811	0.006
TEMP*DETERGENT	16.333	2*1=2	16.333/2=8.167	3.973	0.037
ERROR	37.000	23-5=18
TOTAL	273.833	23