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	152
Error	570	114
Total	822

(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	152
Error	570	114
Total	822

State the decision for the main effect of length.

Retain the null hypothesis.Reject the null hypothesis.     

State the decision for the main effect of technology.

Retain the null hypothesis.Reject the null hypothesis.     

State the decision for the interaction effect.

Retain the null hypothesis.Reject the null hypothesis.     

(b) Based on the results you obtained, what is the next step?

No further analysis is needed, because none of the effects are significant.Compute simple main effect tests for the significant interaction.     Compute pairwise comparisons for the length factor.Compute pairwise comparisons for the technology factor.

Answer 1

(a)

Source of Variation	SS	df	MS	F
Length	10	1	10	2
Technology	90	2	45	9
Length × Technology	152	2	76	15.2
Error	570	114	5
Total	822

SS Technology = SS Total - (SS Length + SS Length × Technology + SS Error)

= 822 - (10 + 152 + 570)

= 90

df Length = Number of levels of Length - 1 = 2 - 1 = 1

df Technology = Number of levels of Technology - 1 = 3 - 1 = 2

df Length × Technology = (Number of levels of Length - 1) * (Number of levels of Technology - 1)

= (2 - 1) * (3 - 1) = 2

MS for each cell is calculated as,

MS = SS / df

F for Length = MS Length / MS Error = 10/5 = 2

F for Technology = MS Technology / MS Error = 45/5 = 9

F for Length × Technology = MS Length × Technology / MS Error = 76/5 = 15.2

State the decision for the main effect of length.

For Length, df = df Length, df Error = 1, 114

Critical value of F at alpha = 0.05 and df = 1,114 is 3.92

Since the observed F is less than the critical value,

Retain the null hypothesis.

State the decision for the main effect of technology.

For technology, df = df technology, df Error = 2, 114

Critical value of F at alpha = 0.05 and df = 2,114 is 3.08

Since the observed F is greater than the critical value,

Reject the null hypothesis.     

State the decision for the interaction effect.

For Length × Technology, df = df Length × Technology, df Error = 2, 114

Critical value of F at alpha = 0.05 and df = 2,114 is 3.08

Since the observed F is greater than the critical value,

Reject the null hypothesis.     

(b)

Since the effects of Technology is significant and there is significant effect between length and technology,

Compute pairwise comparisons for the technology factor.