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One Way ANOVA In prior notes, we looked at data analysis strategies for when the researcher needs to compare two sample means. But, what if a researcher.

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Presentation on theme: "One Way ANOVA In prior notes, we looked at data analysis strategies for when the researcher needs to compare two sample means. But, what if a researcher."— Presentation transcript:

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2 One Way ANOVA In prior notes, we looked at data analysis strategies for when the researcher needs to compare two sample means. But, what if a researcher designs a study in which there are more than two samples (groups)? Using the Single Factor Analysis of Variance technique in Excel, we can determine if there is a different between averages of multiple groups or if the averages are the same. If we have one independent variable, then we use the One Way Analysis of Variance.

3 In this example, we have data on stress levels from patients. The patients were randomly assigned to three different groups, and one group was given music therapy, one group was given relaxation therapy, and the other was a control group. The data is presented below: MusicRelaxation TechniqueControl 015 646 2310 428 306

4 We’ve measured their stress levels on a scale from 1-10, and we’d like to see if the three groups are the same or different. Put the data into Excel. Step 1

5 You are choosing Single Factor because you only have one dependent variable (stress level ) and three categories of independent variables (therapy type). Step 2 Click on Data Data Analysis Anova: Single Factor.

6 Select the input range and if you include the top row then be sure to check Labels in First Row. Step 3

7 Anova: Single Factor SUMMARY GroupsCountSumAverageVariance Music Therapy51535 Relaxation Technique51022.5 Control53574 ANOVA Source of VariationSSdfMSFP-valueF crit Between Groups702359.1304347830.0038886523.885293835 Within Groups46123.833333333 Total11614 Click OK and the results of the ANOVA analysis are shown as below: Step 4

8 Our null hypothesis is that there is no difference between mean scores of the three groups. Our alternative hypothesis is that there is a difference between mean scores of the three groups. Based upon the P-Value presented here, we see that 0.003 is less than the alpha significance that we chose of 0.05 when running the ANOVA analysis. Since it is less, then we must reject the hypothesis and conclude that the alternative hypothesis is correct, which means that there is a significance difference between the groups.


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