Question

Note: for all problems, assume a.05 1. A common assumption is that sleep deprivation influences aggression. To test this assumption, volunteer subjects are randomly assigned by a researcher to sleep-deprivation periods of 0, 24, 48, and 72 hours and subsequently tested for aggressive behavior in a controlled social situation. Aggressive scores signify the total number of different aggressive behaviors, such as "put downs," arguments, or verbal interruptions, as demonstrated by subjects during the test period. You are given the following data Hours of Deprivation 72 48 24 meani 1.625 mean2-3.125 mean3 5.0 mean 7.5 S2-1.98 n 8 a) State the null and alternative hypothesis and complete the appropriate b) Assume that the researcher was interested in comparing 0 hours of deprivation S2-4.0S2 10.57 S2 2.4 n 8 n 8 analysis. with 48 hours and 24 hours of deprivation with 72 hours. Carry out those two comparisons

Answer 1

a)

Ho: µ1=µ2=µ3
H1: not all means are equal

excel data analysis tool for one factor anova,steps are:

write data>menu>data>data analysis>anova :one factor>enter required labels>ok, and following o/p Is obtained,

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
0	8	13	1.625	1.98
24	8	25	3.125	2.4
48	8	40	5.00	4.00
72	8	60	7.50	10.57
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	154	3	51.38	10.84	0.000	2.95
Within Groups	133	28	4.74
Total	287	31

p value < α=0.05, reject Ho

conclusion :there is significant difference in four means

b)

Level of significance	0.0500
no. of treatments,k	4
DF error =N-k=	28
MSE	4.738
t-critical value,t(α/2,df)	2.048

Fishers LSD critical value=tα/2,df √(MSE(1/ni+1/nj))

if absolute difference of means > critical value,means are significnantly different ,otherwise not                      
                      

absolute mean difference		critical value			result
µ1-µ3	3.375	2.2293			means are different
µ2-µ4	4.375	2.2293			means are different