#4
A research paper describes an experiment in which 74 men were assigned at random to one of four treatments.
The participants then went to a room to complete a questionnaire. In this room, bowls of pretzels were set out on the tables. A research assistant noted how many pretzels were consumed by each participant while completing the questionnaire. Data consistent with summary quantities given in the paper are given in the accompanying table.
Treatment 1 | Treatment 2 | Treatment 3 | Treatment 4 |
---|---|---|---|
9 | 6 | 1 | 5 |
7 | 8 | 5 | 1 |
4 | 0 | 2 | 5 |
13 | 4 | 0 | 7 |
1 | 9 | 4 | 5 |
2 | 8 | 0 | 1 |
5 | 6 | 4 | 0 |
9 | 2 | 3 | 0 |
12 | 7 | 3 | 3 |
5 | 8 | 5 | 4 |
2 | 8 | 5 | 1 |
0 | 5 | 7 | 4 |
6 | 14 | 8 | 2 |
4 | 9 | 3 | 2 |
10 | 0 | 0 | |
7 | 6 | 6 | |
0 | 3 | 4 | |
11 | 12 | ||
5 | |||
6 | |||
10 | |||
8 | |||
6 | |||
2 | |||
10 |
Do these data provide convincing evidence that the mean number of pretzels consumed is not the same for all four treatments? Test the relevant hypotheses using a significance level of 0.05.
Calculate the test statistic. (Round your answer to two decimal places.)
F =
H0: μ1 = μ2 = μ3=μ4 |
H1: There is at least one mean difference among the populations. |
Applying ANOVA from excel: (Data-data analysis:ANOVA:single factor:input data array)
Source of Variation | SS | df | MS | F | P-value |
Between Groups | 171.8525 | 3 | 57.2842 | 5.5452 | 0.0018 |
Within Groups | 723.1340 | 70 | 10.3305 | ||
Total | 894.9865 | 73 |
from above
test statistic F =5.55
since p value <0.05 , we reject Ho and conclude that mean number of pretzels consumed is not the same for all four treatments
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