Question

A researcher is interested in testing two different dosages of a new sleeping medication in a phase II clinical trial against a control group receiving a placebo. Suppose the following table observes the hours of sleep reported the preceding evening for subjects randomly assigned to each of the three groups. α = .05

Dosage One 8.5, 7.9, 8.6, 8.4, 7.6, 9.1.

Dosage Two- 8.1, 7.9 , 7.9, 7.6, 7.4.

Control Group- 5.2, 5.6, 5.7, 5.9, 6.2, 6.4, 6.1.

A) Calculate the MSB and MSW for a one-way analysis of variance procedure.

B) Using the information above, calculate an F statistic, and provide an interpretation of your p-value. (Continue as though all assumptions for ANOVA are met). How would you solve this without using excel?

Answer 1

treatment	1	2	3
count, ni =	6	5	7
mean , x̅ i =	8.350	7.78	5.87
std. dev., si =	0.532	0.277	0.407
sample variances, si^2 =	0.283	0.077	0.166
total sum	50.1	38.9	41.1		130.1	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				7.23
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	1.259	0.305	1.840
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	7.556	1.525	12.878		21.95883
SS(within ) = SSW = Σ(n-1)s² =	1.415	0.308	0.994		2.717

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   18
df within = N-k =   15
  
mean square between groups , MSB = SSB/k-1 = 21.9588/2 = 10.9794
  
mean square within groups , MSW = SSW/N-k = 2.717/15 = 0.1812

b)

F-stat = MSB/MSW = 10.9794/ 0.1812 = 60.6087

P value =   0.0000

anova table
	SS	df	MS	F	p-value
Between:	21.959	2	10.979	60.61	0.0000
Within:	2.717	15	0.181
Total:	24.676	17