Question

In an effort to promote a new product, a marketing firm asks participants to rate the effectiveness of ads that varied by length (short, long) and by type of technology (static, dynamic, interactive). Higher ratings indicated greater effectiveness.

Source of Variation	SS	df	MS	F
Length	10
Technology
Length × Technology	142
Error	570	114
Total	802

(a) Complete the F-table and make a decision to retain or reject the null hypothesis for each hypothesis test. (Assume experimentwise alpha equal to 0.05.)

Source of Variation	SS	df	MS	F
Length	10
Technology
Length × Technology	142
Error	570	114
Total	802

Answer 1

SOURCE OF VARIATION	S.S	D.F	M.S	F CAL	P value
LENGTH	10	=2-1=1	10/1=10	10/5=2	0.160025
TECHNOLOGY	80	=3-1=2	80/2=40	40/5=8	0.000561
LENGTHX TECHNOLOGY	142	2*1=2	142/2=71	71/5=14.2	0.00000
ERROR	570	114	570/114=5
TOTAL	802	119

alpha= 0.05

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS HA:

F= M.S(LENGTH)/M.S(ERROR)

= 10/5= 2

P VALUE (F= 2, DF1=1 AND DF2= 114) = 0.160025

P value >0.05 henec not significant at0.05.

CONCLUSION: SINCE P VALUE NOT SIGNIFICANT AT 0.05 LEVEL OF SIGNIFICANCE DO NOT REJECT NULL HYPOTHESIS H0.

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS HA: At least one mean is not equal.

alpha= 0.05

F= M.S(TECHNOLOGY)/M.S(ERROR)

= 40/5= 8

P VALUE (F= 8, DF1=2 AND DF2= 114) =0.000561

P value <0.05 henec significant at0.05.

CONCLUSION: SINCE P VALUE SIGNIFICANT AT 0.05 LEVEL OF SIGNIFICANCE REJECT NULL HYPOTHESIS H0. Therefore at least one mean is different.

NULL HYPOTHESIS H0: There is no interaction between length and technology.

ALTERNATIVE HYPOTHESIS HA: There is interaction between length and technology.

alpha= 0.05

F= M.S(LENGTHXTECHNOLOGY)/M.S(ERROR)

= 71/5= 14.2

P VALUE (F= 14.2, DF1=2 AND DF2= 114) =0.00000

P value <0.05 henec significant at0.05.

CONCLUSION: SINCE P VALUE IS SIGNIFICANT THEREFORE WE REJECT NULL HYPOTHESIS AND COCNLUDE THAT THERE IS AN INTERACTION BETWEEN LENGTH AND TECHNOLOGY.