In: Statistics and Probability
Examine the three samples obtained independently from three populations:
Population Data 3
Item | Group 1 | Group 2 | Group 3 |
---|---|---|---|
1 | 14 | 17 | 17 |
2 | 13 | 16 | 14 |
3 | 12 | 16 | 15 |
4 | 15 | 18 | 16 |
5 | 16 | 14 | |
6 | 16 |
A single factor ANOVA is used in excel to test the null hypothesis that all the means are equal. The hypothesis is defined as,
Null Hypothesis: At least one mean differ significantly.
Alternative Hypothesis: All the means are equal,
The test is performed in excel by using following steps,
Step 1: Write the data values in excel. The screenshot is shown below,
Step 2: DATA > Data Analysis > ANOVA: single Factor > OK. The screenshot is shown below,
Step 3: Select Input Range: All the data values column, Alpha = 0.05. The screenshot is shown below,
The result is obtained. The ANOVA table is shown below,
From the ANOVA result summary,
Since the P-value is less than 0.05 at 5% significance level, the null hypothesis is rejected. Hence we can conclude that there is atleast one mean is significantly differ.
Post-Hoc test
Now, the Tukey-Kramer (T-K) multiple comparisons procedure is used to test all the pairwise comparison and identify which pair is significantly different.
The Tukey-Kramer method uses the formula,
The q value is obtained using the Studentized Range q table for significance level = 0.05, number of groups, k = 3, degree of freedom = N - k = 15 - 3 = 12.
The HSD value for group 1 and 2 comparison ,
The HSD value for group 1 and 3 comparison ,
The HSD value for group 2 and 3 comparison ,
Now,
The mean value for each groups are,
Groups | Average |
Group 1 | 14 |
Group 2 | 16.75 |
Group 3 | 15.33333 |
There are 3 possible comparison as follows,
Comparison | Difference | HSD | ||
2.75 | > | 2.3153 | Significant | |
1.333333 | < | 2.0899 | Not Significant | |
1.416667 | < | 2.2279 | Not Significant |
The mean difference in Group 1 vs Group 2 is significant.