Question

A large company is conducting an experiment to determine if a new mental training program will benefit worker productivity. Because the company has a large number of employees it is more cost effective to randomly select employees to participate in the training program. Thirty-six employees are randomly se- lected for the experiment, with twenty employees assigned to the new program and the remainder assigned to the placebo. The company determine the scores for each group and notices there are no outliers within either group. The group receiving the new training has a mean productivity time of 72.26 minutes with a standard deviation of 20.80 minutes while the placebo group has a mean pro- ductivity time of 86.44 minutes with a standard deviation of 26.20 minutes. Due to the cost of retraining all employees, the company would like to be certain the program is working; therefore they are only willing to risk being wrong 1% of the time. Is the program working? Construct a 98% estimate for the difference in the two programs.

Answer 1

Let X1 :-  New mental training program

X2 :-  placebo program

To Test :-

H0 :-  

H1 :-  

Test Statistic :-


t = -1.7651

Test Criteria :-
Reject null hypothesis if

DF = 28

Result :- Fail to Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 1.7651 ) = 0.0885
Reject null hypothesis if P value < level of significance
P - value = 0.0885 > 0.01 ,hence we fail to reject null hypothesis

Conclusion :- We Accept H0

There is insufficient evidence to support the claim that new program is working at 1% level of significance.

Confidence interval :-



Lower Limit =
Lower Limit = -36.3782
Upper Limit =
Upper Limit = 8.0182
99% Confidence interval is ( -36.3782 , 8.0182 )

lies in the interval  ( -36.3782 , 8.0182 ) , hence we fail to reject null hypothesis.