In a study about dieting and weight loss, N=80 subjects were gathered to participate. All of the participants were approximately 20-25% overweight (from their ideal weight), with no habits of exercising. The researchers randomly assigned these participants into four different categories: participants in Category I (N=20) ate the traditional low-caloric diet, participants in Category II (N=20) ate the ketogenic diet, participants in Category III (N=20) were told to eat whatever they want as if they were not on a diet at all, and participants in Category IV were ate an intermittent fasting diet. The amount of participant’s weight loss (in percentage) was recorded after 6 weeks of dieting period.
mean | 1.72 | 4.11 | -3.3 | 2.9 |
SD | 1.513 | 2.141 | 1.437 | 2.682 |
SEM | 0.303 | 0.428 | 0.287 | 2.75 |
The study design was a Completely Randomized design and the study is analyzed using an one way ANOVA.
a. The Weight loss was seen highest in Category II ie who are on Ketogenic diet followed by intermittent fasting and then the low calorie diet. As is expected, the control group had seen an overall increase in weight. The variation (SD) is maximum in Category IV and the lowest in Category III.
b. Now we shall run an one way ANOVA using the data given by following the steps:
1. Calculate the overall mean :
Since all are equal =20, and we have k=4 groups(categories) we have
2. Calculate the between categories sum of squares:
3. Error sum of squares:
3. Total sum of squares:
4. Degrees of freedom: Total:80-1=79
Categories :4-1=3
Error (n-k) :80-4=76
We shall now construct the ANOVA table:
Source | SS | df | MSS | F-ratio | F-critical |
Categories | 635.5854 | 3 | 211.8618 | 52.5347 | 2.7249 |
Error | 306.4917 | 76 | 4.0328 | ||
Total | 942.0771 | 79 |
Since, the calculated value of F(52.5347)>the critical value, we reject the Null hypothesis. Hence, we conclude that at least one pair of means of categories are different.
Post-hoc analysis:
Since, the Categories is significant, we shall calculate the Critical difference using Fisher's LSD using the formula at 5% level of significance:
Where t is calculated for error df and MSe is the mean square for error.
We now form a table showing the comparisons:
Mean | Comparison | Abs(Diff) | LSD | Significant? | |
Cat I | 1.72 | Cat I Vs. Cat II | 2.39 | 1.2648 | Yes |
Cat II | 4.11 | Cat I Vs. Cat III | 5.02 | 1.2648 | Yes |
Cat III | -3.3 | Cat I Vs. Cat IV | 1.18 | 1.2648 | No |
Cat IV | 2.9 | Cat II Vs. Cat III | 7.41 | 1.2648 | Yes |
Cat II Vs. Cat IV | 1.21 | 1.2648 | No | ||
Cat III Vs. Vat IV | 6.2 | 1.2648 | Yes |
We can see that the comparisons Category I with II and III are significant
Category II and III is significant
Category III and IV is significant.
In a study about dieting and weight loss, N=80 subjects were gathered to participate. All of...