In: Statistics and Probability
In an experiment to determine the effectiveness of sleep-inducing drugs, 18 insomniacs were randomly assigned to three treatments:
1. Placebo (no drug)
2. Standard drug
3. New experimental drug
The data shown is the average number of sleep per night for a week. Use α = 0.05 (use excel)
c. Use Dunnett’s method and perform comparisons to the placebo. Fully explain your conclusions.
| Treatments | ||
| 1 | 2 | 3 | 
| 5.6 | 8.4 | 10.6 | 
| 5.7 | 8.2 | 6.6 | 
| 5.1 | 8.8 | 8 | 
| 3.8 | 7.1 | 8 | 
| 4.6 | 7.2 | 6.8 | 
| 5.1 | 8 | 6.6 | 
(a) The model is:
Sleep per night = 4.983 + 0.0 Treatments_1 + 2.967 Treatments_2 + 2.783 Treatments_3
(b) The hypothesis being tested is:
| Null hypothesis | All means are equal | 
| Alternative hypothesis | Not all means are equal | 
| Significance level | α = 0.05 | 
Equal variances were assumed for the analysis.
Factor Information
| Factor | Levels | Values | 
| Treatments | 3 | 1, 2, 3 | 
Analysis of Variance
| Source | DF | Adj SS | Adj MS | F-Value | P-Value | 
| Treatments | 2 | 33.16 | 16.582 | 15.04 | 0.000 | 
| Error | 15 | 16.54 | 1.102 | ||
| Total | 17 | 49.70 | 
Since the p-value (0.000) is less than the significance level (0.05), we can reject the null hypothesis.
Therefore, we can conclude that Not all means are equal.
(c) Using the Dunnett Method and 95% Confidence
| Treatments | N | Mean | Grouping | 
| 3 (control) | 6 | 7.767 | A | 
| 2 | 6 | 7.950 | A | 
| 1 | 6 | 4.983 | 
Means not labelled with the letter A are significantly different from the control level mean.