This is One-Way ANOVA analysis because we have two variables,
namely, relationship status and whether people own a pet or
not.
The hypothesis being tested is:
H0: µ1 = µ2 = µ3 =
µ4
Ha: At least one means is not equal for all
groups
The output is:
|
|
Mean |
n |
Std. Dev |
|
|
|
|
7.3 |
4 |
0.96 |
Group 1 |
|
|
|
3.3 |
4 |
0.96 |
Group 2 |
|
|
|
4.0 |
4 |
0.82 |
Group 3 |
|
|
|
3.0 |
4 |
0.82 |
Group 4 |
|
|
4.4 |
16 |
1.93 |
Total |
|
|
|
|
|
|
|
| ANOVA
table |
|
|
|
|
|
|
Source |
SS |
  df |
MS |
F |
  p-value |
|
Treatment |
46.25 |
3 |
15.417 |
19.47 |
.0001 |
|
Error |
9.50 |
12 |
0.792 |
|
|
|
Total |
55.75 |
15 |
|
|
|
Since the p-value (0.0001) is less than the significance level
(0.05), we can reject the null hypothesis.
Therefore, we can conclude that at least one group of
relationship status has a significant difference mean of the level
of loneliness from the other 3 groups.
