Question

A research paper describes an experiment in which 74 men were assigned at random to one of four treatments.

1. viewed slides of fit, muscular men
2. viewed slides of fit, muscular men accompanied by diet and fitness-related text
3. viewed slides of fit, muscular men accompanied by text not related to diet and fitness
4. did not view any slides

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	7	2	5
7	8	5	2
4	0	1	5
13	3	0	6
2	9	4	5
1	8	0	2
5	7	4	0
9	2	3	0
11	6	3	4
5	8	5	3
1	8	5	2
0	5	7	3
6	13	9	1
4	9	3	1
10	0	0
7	7	6
0	4	4
12	12
	5
	7
	10
	8
	7
	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 =

What can be said about the P-value for this test?

P-value > 0.1000.050 < P-value < 0.100    0.010 < P-value < 0.0500.001 < P-value < 0.010P-value < 0.001

What can you conclude?

Reject H₀. The data do not provide convincing evidence that the mean number of pretzels consumed is not the same for all four treatments.Fail to reject H₀. The data do not provide convincing evidence that the mean number of pretzels consumed is not the same for all four treatments.    Reject H₀. The data provide convincing evidence that the mean number of pretzels consumed is not the same for all four treatments.Fail to reject H₀. The data provide convincing evidence that the mean number of pretzels consumed is not the same for all four treatments.

You may need to use the appropriate table in Appendix A to answer this question.

Answer 1

usig minitab>stat>ANNOVA>One way anova

we have

One-way ANOVA: Treatment 1, Treatment 2, Treatment 3, Treatment 4

Method

Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
Significance level α = 0.05
Rows unused 17

Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values
Factor 4 Treatment 1, Treatment 2, Treatment 3, Treatment 4

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Factor 3 176.1 58.69 5.70 0.002
Error 70 720.9 10.30
Total 73 897.0

Model Summary

S R-sq R-sq(adj) R-sq(pred)
3.20918 19.63% 16.18% 10.99%

Means

Factor N Mean StDev 95% CI
Treatment 1 25 6.200 3.808 (4.920, 7.480)
Treatment 2 18 6.444 3.601 (4.936, 7.953)
Treatment 3 17 3.588 2.526 (2.036, 5.141)
Treatment 4 14 2.786 1.968 (1.075, 4.496)

Pooled StDev = 3.20918

the test statistic F =5.70

p value is 0.002

0.001 < P-value < 0.01

since p value is less than 0.05 so we reject H0

Reject H₀. The data provide convincing evidence that the mean number of pretzels consumed is not the same for all four treatments.