In: Statistics and Probability
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 | 154 | |||
Error | 570 | 114 | ||
Total | 864 |
(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 |
154 | |||
Error | 570 | 114 | ||
Total | 864 |
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?
Compute pairwise comparisons for the length factor.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 technology factor.
You may need to use the appropriate table in Appendix C to answer
this question.
https://www.webassign.net/priviterastats3/priviterastats3_appendix_c.pdf
a) Level of Factor A, a = 2
Level of Factor B, b = 3
Number of repetition, r = 20
df(A) = a-1 = 1
df(B) = b-1 = 2
df(AB) = (a-1)*(b-1) = 2
df(total) = 114+1+2+2 = 119
SSB = SST -SSA - SSAB- SSE = 130
MSA = SSA/df(A) = 10
MSB = SSB/df(B) = 65
MSAB = SSAB/df(AB) = 77
MSE = SSE/df(error) = 5
F for factor A :
F = MSA/MSE = 2
p-value = F.DIST.RT(2, 1, 114) = 0.1600
F for factor B :
F = MSB/MSE = 13
p-value = F.DIST.RT(13, 2, 114) = 0.0000
F For Interaction :
F = MSAB/MSE = 15.4
p-value = F.DIST.RT(15.4, 2, 114) = 0.0000
ANOVA | |||||
Source of Variation | SS | df | MS | F | P-value |
Factor A | 10 | 1 | 10 | 2 | 0.1600 |
Factor B | 130 | 2 | 65 | 13 | 0.0000 |
Interaction | 154 | 2 | 77 | 15.4 | 0.0000 |
Error | 570 | 114 | 5 | ||
Total | 864 | 119 |
State the decision for the main effect of length.
Retain the null hypothesis.
State the decision for the main effect of technology.
Reject the null hypothesis.
State the decision for the interaction effect.
Reject the null hypothesis.
b)
Compute pairwise comparisons for the technology factor.