In an experiment,1919 telephone callers to a company were put on hold with an ad
ID: 3241935 • Letter: I
Question
In an experiment,1919 telephone callers to a company were put on hold with an advertisement, comedy comma Musak commacomedy, Musak,or classical music in the background. Data was then collected on telephone holding times for each combination of type and repeat time. Also, the message randomly assigned to each caller would repeat after fivefive or ten minutes.
Use this information to complete parts a through c.
The results of a two-way ANOVA.
Source
DF
SS
MS
F
P
MessageMessage
33
57.0657.06
19.0219.02
1.591.59
0.2370.237
Repeat
11
119.12119.12
119.12119.12
9.969.96
0.0070.007
Error
1414
167.44167.44
11.9611.96
Total
1818
343.62343.62
State the null hypothesis to which the F test statistic in the MessageMessage row refers.
Choose the correct null hypothesis below.
A. The type of messageThe type of message has no effect on customer hold time.
B. The length of the messageThe length of the message has no effect on customer hold time.
C. The length and type of message have an effect on customer hold time.
D. The length and type of message have no effect on customer hold time.
b. How is the F test statistic for the message main effect constructed?
A. F= mean square for repeat/mean square for error
B. F=mean square for message/mean square for repeat
C. F=mean square for message/mean square for error
D. F=mean square for repeat/mean square for message
What is the P-value for the message main effect?
P=
Interpret the P-value for the message main effect. Use a 0.05 level of significance
The results of a two-way ANOVA.
Source
DF
SS
MS
F
P
MessageMessage
33
57.0657.06
19.0219.02
1.591.59
0.2370.237
Repeat
11
119.12119.12
119.12119.12
9.969.96
0.0070.007
Error
1414
167.44167.44
11.9611.96
Total
1818
343.62343.62
Explanation / Answer
Answer:
In an experiment,1919 telephone callers to a company were put on hold with an advertisement, comedy comma Musak commacomedy, Musak,or classical music in the background. Data was then collected on telephone holding times for each combination of type and repeat time. Also, the message randomly assigned to each caller would repeat after fivefive or ten minutes.
Use this information to complete parts a through c.
The results of a two-way ANOVA.
Source
DF
SS
MS
F
P
MessageMessage
3
57.06
19.02
1.59
0.237
Repeat
1
119.12
119.12
9.96
0.007
Error
14
167.44
11.96
Total
18
343.62
State the null hypothesis to which the F test statistic in the MessageMessage row refers.
Choose the correct null hypothesis below.
Answer: A. The type of message The type of message has no effect on customer hold time.
B. The length of the messageThe length of the message has no effect on customer hold time.
C. The length and type of message have an effect on customer hold time.
D. The length and type of message have no effect on customer hold time.
b. How is the F test statistic for the message main effect constructed?
A. F= mean square for repeat/mean square for error
B. F=mean square for message/mean square for repeat
Answer: C. F=mean square for message/mean square for error
D. F=mean square for repeat/mean square for message
What is the P-value for the message main effect?
P=0.237
Interpret the P-value for the message main effect. Use a 0.05 level of significance
Calculated P=0.237 > 0.05 level of significance.
Ho is not rejected.
We conclude that the type of message has no effect on customer hold time.
The results of a two-way ANOVA.
Source
DF
SS
MS
F
P
MessageMessage
3
57.06
19.02
1.59
0.237
Repeat
1
119.12
119.12
9.96
0.007
Error
14
167.44
11.96
Total
18
343.62
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