Question

Solve this problem with one of these methods. Critical value method, P-value method, or confidence interval estimate method.

A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. Before treatment with zopiclone 16 subjects had a mean wake time of 102.8 min. After treatment with zopiclone the 16 subjects had a mean wake time of 98.9 min and a standard deviation of 42.3 min. Assume that the 16 sample values appear to be from a normally distributed population, and test the claim that after treatment with zopiclone, subjects will have a mean wake time of less than 102.8 min. Does zopiclone appear to be effective. Please show all steps.

Answer 1

Considering α = 0.05

To Test :-

H0 :- µ = 102.8
H1 :- µ < 102.8

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 98.9 - 102.8 ) / ( 42.3 / √(16) )
t = -0.3688
Test Criteria :-
Reject null hypothesis if t < -t(α, n-1)
Critical value t(α, n-1) = t(0.05 , 16-1) = 1.753
t > -t(α, n-1) = -0.3688 > - 1.753
Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 0.3688 ) = 0.3587
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.3587 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

There is insufficient evidence to support the claim that after treatment with zopiclone, mean wake time is less than 102.8 min.

Zopiclone does not appear to be effective.