i need to make ANOVA ON EXCEL

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A karate school Master Instructor has recently promoted a student to the rank of black belt. The Master Instructor tasked one of the senior black belts to train the new black belt on how to judge forms (kata) competitions. These are where students perform their form in front of several blackbelt judges, and the judges vote on the students’ performances, using a scale from 0 to 10 points.The scores for each student are then averaged, providing a final score for each student.

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CHAPTER 11 ANOVA CASE STUDY AND GRADING RUBRIC A karate school Master Instructor has recently promoted a student to the rank of black belt. The Master Instructor tasked one of the senior black belts to train the new black belt on how to judge forms (kata) competitions. These are where students perform their form in front of several black belt judges, and the judges vote on the students’ performances, using a scale from 0 to 10 points. The scores for each student are then averaged, providing a final score for each student. In this competition, ten students are performing their forms in front of four judges, one of them being the new black belt. If the new black belt was properly trained, she would be scoring the students comparably with the other experienced judges. If her training was insufficient, then her scoring would be significantly different than the other three. Perform a statistical test to indicate whether the new black belt judge has scored the student competitors comparably with the other judges. The new black belt is Judge 3. Below are the scores recorded. Assume all conditions required for ANOVA are met. Examinee Alpha Bravo Charlie Delta Echo Foxtrot Golf Hotel India Juliet Judge 1 8.2 8.1 7.2 6.8 9.1 8.6 9.3 7.6 8.8 9.3 JUDGE SCORING Judge 2 Judge 3 8.1 7.6 8 7.5 7 6.5 7 7 9 8.4 8.7 9.1 9.4 7.6 7.7 8.2 8.7 7.4 9.2 7.1 Judge 4 8.4 8.4 7.3 6.9 8.8 8.3 9.2 7.4 8.9 9.4 At the 5% level of significance, test whether the means for the three treatments are equal. a. State the null hypothesis. 3 pts. b. State the alternate hypothesis. 3 pts. c. Use Excel to compute the ANOVA table for this problem. Show the table in an Excel file with the corresponding data. 4 pts. 3. Draw your conclusion. Justify your answers. 5 pts. Nonbrow Light ser Browser 4 5 5 6 6 5 3 4 3 7 4 4 5 6 4 5 Heavy Browser 5 7 5 7 4 6 5 7 Nonbrow Light ser Browser 4 5 5 6 6 5 3 4 3 7 4 4 5 6 4 5 Nonbrow Heavy ser Browser 4 5 5 7 6 5 3 7 3 4 4 6 5 5 4 7 Heavy Browser 5 7 5 7 4 6 5 7 Light Browser ANOVA Source of Variation Between Groups Within Groups Total Anova: Single Factor SUMMARY Groups Nonbrowser Heavy Browser ANOVA Source of Variation Between Groups Within Groups Total Anova: Single Factor SUMMARY Groups Nonbrowser Light Browser Heavy Browser ANOVA Source of Variation Count Sum 8 8 8 SS 34 42 46 df Average 4.25 5.25 5.75 MS Between Groups Within Groups 9.333333 24.5 2 4.666667 21 1.166667 Total 33.83333 23 Variance 1.071428571 1.071428571 1.357142857 F 4 Anova: Single Factor SUMMARY Groups Light Browser Heavy Browser ANOVA Source of Variation Between Groups Within Groups Total Count Sum Average 42 5.25 46 5.75 df 1 17 MS 1 1 14 1.214286 18 15 8 8 SS Variance 1.071428571 1.357142857 F 0.823529412 Anova: Single Factor SUMMARY Groups Nonbrowser Count Sum 8 Average 34 4.25 Variance 1.071428571 Light Browser ANOVA Source of Variation Between Groups Within Groups 8 SS Total 42 df 5.25 4 15 MS 1 4 14 1.071429 19 15 1.071428571 F 3.733333333 Anova: Single Factor SUMMARY Groups Nonbrowser Heavy Browser ANOVA Source of Variation Between Groups Within Groups Total Count Sum 8 8 SS 34 46 df Average 4.25 5.75 MS 9 17 1 9 14 1.214286 26 15 Variance 1.071428571 1.357142857 F 7.411764706 P-value F crit 0.033738 3.4668 P-value 0.379506 F crit 4.60011 P-value 0.07383 F crit 4.60011 P-value 0.016514 F crit 4.60011
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