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.