Question

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?

Answer 1

(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.