A researcher wants to determine whether or not a given drug has any effect on the scores of human subjects performing a task of ESP sensitivity. He randomly assigns his subjects to one of two groups. Nine hundred subjects in group 1 (the experimental group) receive an oral administration of the drug prior to testing. In contrast, 1000 subjects in group 2 (control group) receive a placebo (dummy). The results of the study found the following: For the drug group, the mean score on the ESP test was 9.78, std. deviation was 4.05, n = 900 For the no-drug group (placebo), the mean = 15.10, std. deviation was 4.28, n= 1000
a. State the Null and Alternate hypothesis
b. What is your conclusion after calculating Z and determining Z critical?
(A)
Null hypothesis
(Drug has no effect on the scores of human subjects performing a
task of ESP sensitivity)
Alternate hypothesis
(Drug has significant effect on the scores of human subjects
performing a task of ESP sensitivity)
(B)
Calculation for Z statistics
we have x1 = 9.78,
=4.05, n1 = 900 and x2 = 15.10,
=4.28, n2 = 1000
using the formula
Z statistics =
setting the given values
we get
z statistics=
= -27.83
Z critical for 0.05 significance level is -1.96 to +1.96
so, we know that any values outside the region (-1.96, 1.96) will result in rejection of null hypothesis.
It is clear that calculated z statistics -27.83 is outside the region (-1.96, 1.96), thus we can reject the null hypothesis
We can conclude that Drug has significant effect on the scores of human subjects performing a task of ESP sensitivity at 0.05 significance level.
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