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Testing Hypothesis at 0.05 Level

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Given the following sample information, test the hypothesis that the treatment means are equal at the .05 significance level.

Treatment 1 Treatment 2 Treatment 3

8 3 3
11 2 4
10 1 5
3 4
2

A. State the null hypothesis and the alternate hypothesis.

B. What is the decision rule?

C. Compute SST, SSE, SS total.

D. Complete an ANOVA table.

E. State your decision regarding the null hypothesis.

F. If Ho is rejected, can we conclude that treatment 1 and treatment 2 differ? Use the 95% level of confidence.

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Solution Summary

This solution contains stepwise calculations in testing the hypothesis and determines if the null hypothesis is accepted or rejected. All formulas and workings are shown clearly.

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Inferences based on 2 samples

Excess postexercise oxygen consumption (EPOC) describes the energy expended during the body's recovery period immediately following aerobic exercise. A journal published a study designed to investigate the effect of fitness level on the magnitude and duration of EPOC. Ten healthy young adult males volunteered for the study. Five of these were endurance trained and comprised the fit group; the other five were not engaged in any systematic training and comprised the sedentary group. Each volunteer engaged in a weight supported exercise on a cycle ergometer until 300 kilocalories were expended. The magnitude (in kilocalories) and duration (in minutes) of the EPOC of each exerciser were measured. The study results are summarized as:

VARIABLE FIT SEDENTARY p-value
(n=5) (n=5)

Magnitude (kcal) mean 12.2 12.2 0.998
Std. dev. 3.1 4.3

Duration (min) mean 16.6 20.4 0.344
Std. dev. 3.1 7.8

a.) Conduct a test of hypothesis to determine whether the true mean magnitude of EPOC differs for fit and sedentary young adult males. Use alpha=0.10

b.) The p-value for the test, part a, is given in the table. Interpret this value.

c.) Conduct a test of hypothesis to determine whether the true mean duration of EPOC differs for fit and sedentary young adult males. Use alpha=0.10

d.) The p-value for the test, part c, is given in the table. Interpret this value.

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