At a gymnastics meet, three judges evaluate the balance beam performances of five gymnasts. The judges use a scale of 1 to 10, where 10 is a perfect score. A statistician wants to examine the objectivity and consistency of the judges. Assume scores are normally distributed. (You may find it useful to reference the q table.)
Judge 1 | Judge 2 | Judge 3 | |
Gymnast 1 | 7.8 | 7.8 | 8.4 |
Gymnast 2 | 7.4 | 8.2 | 7.9 |
Gymnast 3 | 9.3 | 9.8 | 8.9 |
Gymnast 4 | 8.4 | 9.3 | 9.3 |
Gymnast 5 | 9.4 | 8.3 | 9.5 |
Here, H0 : There is no significance difference in the average scores by the judges.
H1 : There is a significance difference in the average score by the judges.
a1) Step 1: Feed the data in excel worksheet.
Step 2: Click on data analysis in data tab.
Step 3: Select ANOVA-one factor from the dialog box.
Step 4: Select the values as the input the range.
Step 5: Now, since we have to answer two types of questions like hypothesis for gymnasts and hypothesis for judges, we will be finding anova table under the two categories that is grouped by columns and grouped by rows.
Select these two options one by one.
If grouped by rows i.e. grouped by gymnasts then
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Row 1 | 3 | 24 | 8 | 0.12 | ||
Row 2 | 3 | 23.5 | 7.833333 | 0.163333 | ||
Row 3 | 3 | 28 | 9.333333 | 0.203333 | ||
Row 4 | 3 | 27 | 9 | 0.27 | ||
Row 5 | 3 | 27.2 | 9.066667 | 0.443333 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 5.557333 | 4 | 1.389333 | 5.788889 | 0.011211 | 3.47805 |
Within Groups | 2.4 | 10 | 0.24 | |||
Total | 7.957333 | 14 |
|
If grouped by columns i.e. grouped by judges then
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Column 1 | 5 | 42.3 | 8.46 | 0.788 | ||
Column 2 | 5 | 43.4 | 8.68 | 0.697 | ||
Column 3 | 5 | 44 | 8.8 | 0.43 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 0.297333 | 2 | 0.148667 | 0.232898 | 0.795733 | 3.885294 |
Within Groups | 7.66 | 12 | 0.638333 | |||
Total | 7.957333 | 14 |
.a2) No, the average scores by the judges does not differ as the
p-value = 0.795733 is much more than the significance level
0.05.
When the p-value is more than the significance level, we accept the
null hypothesis
b) Yes, the average scores of the gymnasts differ as the p-value = 0.011 is much less than the significance level 0.05.
When the p-value is less than the significance level, we reject the null hypothesis.
c)
Gymnasts | Difference | n(group 1) | n(group 2) | SE | q | |
Gymnast 1 | Gymnast 2 | 0.16666667 | 3 | 3 | 0.282843 | 0.589256 |
Gymnast 1 | Gymnast 3 | 1.33333333 | 3 | 3 | 0.282843 | 4.714045 |
Gymnast 1 | Gymnast 4 | 1 | 3 | 3 | 0.282843 | 3.535534 |
Gymnast 1 | Gymnast 5 | 1.06666667 | 3 | 3 | 0.282843 | 3.771236 |
Gymnast 2 | Gymnast 3 | 1.5 | 3 | 3 | 0.282843 | 5.303301 |
Gymnast 2 | Gymnast 4 | 1.16666667 | 3 | 3 | 0.282843 | 4.12479 |
Gymnast 2 | Gymnast 5 | 1.23333333 | 3 | 3 | 0.282843 | 4.360492 |
Gymnast 3 | Gymnast 4 | 0.33333333 | 3 | 3 | 0.282843 | 1.178511 |
Gymnast 3 | Gymnast 5 | 0.26666667 | 3 | 3 | 0.282843 | 0.942809 |
Gymnast 4 | Gymnast 5 | 0.06666667 | 3 | 3 | 0.282843 | 0.235702 |
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