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ANOVA Test Comparison

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Here is the problem and I need help I am in a group and no one is helping me. I have talked to the teacher and now i have to do it by myself and turn in the assignment. I have struggling bad in this class. PLEASE HELP ME.

In this regard, remember the goal will be to work through an ANOVA test to compare the means of three or more groups of data. If the article your group selected does not have enough data to complete an ANOVA calculation, please feel free to create the data as needed. Remember to also ensure that you begin by stating the null and alternative hypotheses. Your final answer should point out if you are rejecting or failing to reject the null and why. Please show as much work as possible.

I am attaching a copy of the summary of the article. I know that the data is hard to read in this format but here is the link to the article. http://www.thesportjournal.org/2003Journal/Vol6-No1/satisfaction.asp

The three groups of data that we have decided to go with are:
1. Male Athletic Trainer
2. Bachlors Degree
3. Trainers with 1-5 years

I do not know if I am doing this right or wrong. If I chose the right three groups or not.

I need help with stating the null hypothesis and the alternative hypothesis.

I also need help with the Z-TEST I do not know how to do it and if you could please help me do it and show me the steps.

Here is the data from the article:

Job Satisfaction among Athletic Trainers Based on Their Demographic Characteristics Variable General Satisfaction
Mean SD F p
Employment
152.875 .000
Program director 3.58 0.27
Faculty 3.47 0.26

Head athletic trainer 2.85 0.28

Assistant athletic trainer 2.47 0.25

Graduate assistant 1.40 0.50

Gender 20.401 .000

Female 2.21 0.59

Male 2.73 0.76

Age

17.709 .000

20's 2.11 0.73

30's 2.81 0.48

40's 2.94 0.57

50's 2.96 0.44

Education
3.149 .046

Bachelors 2.30 0.87

Masters 2.56 0.65

Doctorate 2.83 0.44

Experience
19.826 .000
1-5 years 2.00 0.78
6-10 years 2.78 0.50
11-15 years 2.74 0.42
16 or more years 2.85 0.52

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3. ANOVA test
a. Introduction: statement of the problem/situation being analyzed and source of data used.

For analyzing the job satisfaction using ANOVA technique, I have selected the study by Herrera et al (2003). This study analyzes the job satisfaction levels of certified athletic trainers in selected NCAA Division IAA institutions. The amount of responsibility and stress attached with the job of the trainers the authors found a compelling need for conducting this study to understand the job satisfaction level among the certified athletic trainers.

For the ANOVA test, we investigate whether the satisfaction level of the trainers vary with their education qualification. There are bachelor, masters and doctoral degree holders.

b. Null hypothesis: mu#1=mu#2=mu#3
c. Alternate hypothesis: The mean satisfaction level of three education qualification groups of trainers is not same.

Data:
Data recreated and presented below:

d. Testing that requirements of ANOVA are met:
(i) test/assumptions regarding normality: The complete data not available but the research articles says that the condition of ...

Solution Summary

This Solution contains data used to complete the given comparison using ANOVA analysis.

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ANOVA Problem

When we want to test two samples to determine if it is likely that the population means (estimated by the sample means) are different, we typically use a t-test. If the samples are large, we can also use a z-test. (Note that the formulas for computing s, t and/or z in the case of a two-sample test are different than the formulas for computing the same values in a one-sample test. Use Excel data analysis to conduct tests comparing two sample means.)

Using ANOVA (short for Analysis of Variance), however, we can test 3 or more sample means to determine if at least one of the sample means comes from a population with a mean that is significantly different from all of the others in the test. We actually do this by estimating a combined population variance two different ways and comparing the two estimates (the ratio of these two variance estimates follows the so-called "F distribution").

Question:

Why do we need a new test method to compare the means of 3 or more populations? Why can't we just use a series of z-tests or t-tests to compare all of the possible pairs of population means to see if one (or more) is different?

Most of the testing is to determine one or two things:

1. Is there a statistically significant difference between two or more population means? (based on comparison of 2 or more sample means)

2. Is there a statistically significant relationship between two or more variables? We can use regression analysis or chi-square tests to answer this second question.)

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