Question

A health psychologist knows the smoking population at the hospital where she works smokes an average of 18 cigarettes per day with a standard deviation of 7. She plans to conduct a program to reduce smoking. If she has 25 persons in her program, what is the power to detect a real effect of the program such that cigarette consumption is reduced, on the average, by 5 cigarettes? Assume population normality and use a = 0.05 1 tail .

a. Insufficient information

b. 0.9732

c. 0.9463

d. 0.8925

Answer 1

Ans:

left tailed test

critical z value=-1.645

sample mean cut off value=18-1.645*(7/SQRT(25))=15.70

when true mean=13

(i.e. reduced on avergae by 5 cigarettes, 18-5=13)

z=(15.7-13)/(7/SQRT(25))

z=1.93

Power=P(z<1.93)=0.9732

Option b is correct.

0.9732