Question

A study was conducted to explore the effects of ethanol on sleep time. Fifteen rats were randomized to one of three treatments. Treatment 1 got only water (control). Treatment 2 got 1g of ethanol per kg of body weight, and treatment 3 got 2g ethanol per kg of body weight. The amount of REM sleep in a 24hr period was recorded, in minutes. Data are below: Treatment 1: 63, 54, 69, 50, 72

Treatment 2: 45, 60, 40, 56

Treatment 3: 31, 40, 45, 25, 23, 28

(f) Calculate 90% confidence intervals for all pairwise comparisons of treatment means using the uncorrected method. Create a letter code table to summarize your results, then interpret the results in context.

(g) Now calculate 90% confidence intervals for all pairwise comparisons using the Bonferroni correction. Create a letter code table, and interpret your results in context. Do any of the results differ from part (f)?

Answer 1

from the given data...

Treatment 1(X1')	Treatment 2(X2')	Treatment 3(X3')	X1-X1'	(x1 - X1')^2	(x2 - x2')	(x2 - x2')^2	(X3-x3')	(X3 - X3')^2
63	45	31	1.4	1.96	-5.25	27.5625	-1	1
54	60	40	-7.6	57.76	9.75	95.0625	8	64
69	40	45	7.4	54.76	-10.25	105.0625	13	169
50	56	25	-11.6	134.56	5.75	33.0625	-7	49
72		23	10.4	108.16			-9	81
		28					-4	16
308	201	192		357.2		260.75		380

mean of treatment x1' = sum(X1)/ n1 = 308/5 = 61.6

Mean for Treatment X2' = sum(X2)/ n2 = 260.75/4 = 50.25

Mean of treatment X3' = sum(X3') / n3 = 192/6 = 32

b.

BY USING SPSS software to obtain 90% confidence Intervals for all pairwise comparisions using the Bonferroni correction to the following data.

Treatments(i)	Treatments(j)	I - j	std error	sig	95% CI
					Lower bound	Upper Bound
1	2	11.35	6.117	0.265	-5.65	28.35
3	29.600*	5.522	0.001	14.25	44.95
2	1	-11.35	6.117	0.265	-28.35	5.65
3	18.250*	5.887	0.028	1.89	34.61
3	1	-29.6	5.522	0.001	-44.95	-14.25
2	-18.25	5.887	0.028	-34.61	-1.89

Mean sleep tiem of treatment1 & treatment3 are significantly different because the p-value < level of significance.

Mean sleep time of treatment2 & treatment3 are significantly different because the p-value is less than level significance.