In: Statistics and Probability
A mail-order catalog firm designed a factorial experiment to test the effect of the size of a magazine advertisement and the advertisement design on the number of catalog requests received (data in thousands). Three advertising designs and two different-size advertisements were considered. The data obtained follow.
Size of Advertisement | |||
---|---|---|---|
Small | Large | ||
Design | A | 8 | 12 |
12 | 8 | ||
B | 22 | 26 | |
14 | 30 | ||
C | 10 | 18 | |
18 | 14 |
Use the ANOVA procedure for factorial designs to test for any significant effects due to type of design, size of advertisement, or interaction. Use α = 0.05.
Find the value of the test statistic for type of design. (Round your answer to two decimal places.)
Find the p-value for type of design. (Round your answer to three decimal places.)
p-value =
State your conclusion about type of design.
Because the p-value ≤ α = 0.05, type of design is not significant.Because the p-value > α = 0.05, type of design is not significant. Because the p-value > α = 0.05, type of design is significant.Because the p-value ≤ α = 0.05, type of design is significant.
Find the value of the test statistic for size of advertisement. (Round your answer to two decimal places.)
Find the p-value for size of advertisement. (Round your answer to three decimal places.)
p-value =
State your conclusion about size of advertisement.
Because the p-value ≤ α = 0.05, size of advertisement is significant.Because the p-value > α = 0.05, size of advertisement is not significant. Because the p-value > α = 0.05, size of advertisement is significant.Because the p-value ≤ α = 0.05, size of advertisement is not significant.
Find the value of the test statistic for interaction between type of design and size of advertisement. (Round your answer to two decimal places.)
Find the p-value for interaction between type of design and size of advertisement. (Round your answer to three decimal places.)
p-value =
State your conclusion about interaction between type of design and size of advertisement.
Because the p-value > α = 0.05, interaction between type of design and size of advertisement is not significant.Because the p-value ≤ α = 0.05, interaction between type of design and size of advertisement is significant. Because the p-value > α = 0.05, interaction between type of design and size of advertisement is significant.Because the p-value ≤ α = 0.05, interaction between type of design and size of advertisement is not significant
using excel data analysis tool for two factor anova, following o/p Is obtained :
write data>menu>data>data analysis>anova :two factor with replication>enter required labels>ok, and following o/p Is obtained,
Anova: Two-Factor With Replication | ||||||
SUMMARY | Small | Large | Total | |||
A | ||||||
Count | 2 | 2 | 4 | |||
Sum | 20 | 20 | 40 | |||
Average | 10 | 10 | 10 | |||
Variance | 8 | 8 | 5.333333 | |||
B | ||||||
Count | 2 | 2 | 4 | |||
Sum | 36 | 56 | 92 | |||
Average | 18 | 28 | 23 | |||
Variance | 32 | 8 | 46.66667 | |||
C | ||||||
Count | 2 | 2 | 4 | |||
Sum | 28 | 32 | 60 | |||
Average | 14 | 16 | 15 | |||
Variance | 32 | 8 | 14.66667 | |||
Total | ||||||
Count | 6 | 6 | ||||
Sum | 84 | 108 | ||||
Average | 14 | 18 | ||||
Variance | 27.2 | 72 | ||||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Sample | 344 | 2 | 172 | 10.75 | 0.010386 | 5.143253 |
Columns | 48 | 1 | 48 | 3 | 0.133975 | 5.987378 |
Interaction | 56 | 2 | 28 | 1.75 | 0.251932 | 5.143253 |
Within | 96 | 6 | 16 | |||
Total | 544 | 11 |
a)
value of the test statistic for type of design=10.75
p value=0.010
Because the p-value ≤ α = 0.05, type of design is significant.
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value of the test statistic for size of advertisement=3.00
p-value=0.134
Because the p-value > α = 0.05, size of advertisement is not significant.
-------------------------
value of the test statistic for interaction between type of design and size of advertisement=1.75
p value=0.252
.Because the p-value > α = 0.05, interaction between type of design and size of advertisement is not significant