QSO 510 Discussion 7 - m Average Variance ´lavor 1...

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A marketing research firm tests the effectiveness of three new flavorings for a leading beverage using a sample of 30 people, divided randomly into three groups of 10 people each. Group 1 tastes flavor 1, group 2 tastes flavor 2 and group 3 tastes flavor 3. Each person is then given a questionnaire which evaluates how enjoyable the beverage was. I would like to determine whether there is a perceived significant difference between the three flavorings. The sample data are as follows: Flavor 1 Flavor 2 Flavor 3 13 12 7 17 8 19 19 6 15 11 16 14 20 12 10 15 14 16 18 10 18 9 18 11 12 4 14 16 11 11 My null hypothesis is that any difference between the three flavors is due to chance. H0: μ1 = μ2 = μ3 By inputting this data into my Excel spreadsheet and using my Data Analysis tool for a single factor ANOVA with an alpha of 0.05. The results: SUMMARY Groups Coun t Su
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Unformatted text preview: m Average Variance ´lavor 1 10 150 15 13.3333333 3 ´lavor 2 10 111 11.1 18.7666666 7 ´lavor 3 10 135 13.5 14.0555555 6 ANOVA Source of VariaTon SS df MS F P-value F crit BeTween Groups 77.4 2 38.7 2.51540683 7 0.099596 3 3.35413082 9 WiThin Groups 415. 4 27 15.3851851 9 ³oTal 492. 8 29 In these results, the null hypothesis states that the mean likability values of 3 diFerent ±avors are equal. Because the p-value is 0.995963, which is larger than the signi²cance level of 0.05, I will fail to reject the null hypothesis and conclude that some of the ±avors have nearly the means. However, I can run the risk of making a Type II Error which is the failure to reject a false null hypothesis. I may want to increase the power of my test by using a larger sample or use a higher alpha....
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