Researchers want to determine whether or not there is a difference in systolic blood pressure based on how many hours a person exercises per week. They divide a sample of 72 people into 3 groups based on how many hours they exercise per week. Group 1 exercises less than 2 hours per week, Group 2 exercises between 2 and 5 hours per week, and Group 3 exercises more than 5 hours per week. Researchers measure and record the systolic blood pressure for each participant. They choose α = 0.05 level to test their results. For your convenience, I have prepared an excel file with the data titled: data_homework10_BP groups. Use this data to run a One-way Anova.

1. What is the between groups degrees of freedom for this study?

a. 2
b. 3
c. 72
d. 69

2. This finding is statistically significant.

a. True
b. False

3. Based on this information, the researcher should make the decision to ___________.

a. reject the null hypothesis
b. fail to reject the null hypothesis

Respuesta :

Answer:

(1) The between groups degrees of freedom is 2.

(2) TRUE.

(3) The correct option is (a).

Step-by-step explanation:

(1)

The between groups degrees of freedom for the study is:

[tex]\text{df}_{B}=k-1\\=3-1\\=2[/tex]

Thus, the between groups degrees of freedom is 2.

(2)

The hypothesis for he one-way ANOVA is:

H₀: All the means are equal.

Hₐ: At least one of the mean is not equal.

The output of the ANOVA test is attached below.

The p-value of the test is 0.00005.

p-value = 0.00005 < α = 0.05

Thus, the result is statistically significant.

The statement is TRUE.

(3)

As the p-value of the test is less than the significance level, the researcher should make the decision to reject the null hypothesis.

The correct option is (a).

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