A common assumption is that sleep deprivation influences aggressive behavior. To test this assumption,
volunteer subjects were randomly assigned to one of two conditions. One group (Group 1) was not
deprived of sleep while another on (Group 2) was kept awake for 72 hours. The number of aggressive
acts was counted during the test period for each group. Group 1 committed a mean of 7 aggressive acts
with a standard deviation of 2 while Group 2 committed a mean of 10 aggressive acts with a standard
deviation of 2.2 aggressive acts. There were 12 participants in each group.
1) State the null and alternative hypothesis.
2) What is the decision rule at the .05 level?
3) Choose the appropriate statistic, calculate its value and state your conclusion in terms of the
problem.
A: 1) null hypothesis: x = y common assumption is not correct
alternative hypothesis: x > y common assumption is correct
where x is the mean aggressive acts of Group 2, y is the mean aggressive acts of
Group 1.
2): let  =0.05, since Z = Z0.05 = 2.575, thus if calculated value of test statistics
greater than Z0.05, then reject null hypothesis else accept null hypothesis.
3): use test statistics Z = (mean(x) – mean(y))/(x2/n + y2/n )1/2 where n is number of
participants in each group.
So Z = (10-7)/(2.22/12 + 22/12)1/2 = 3/(8.84/12)1/2 = 3.495
Since Z > Z0.05, null hypothesis is rejected, that is, the common assumption that sleep
deprivation influences aggressive behavior is acceptable.
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