Question

Two antidibetic drugs (Amylin and Metformin) are compared to check more effectiveness. Amylin was intake by 12 patients [Note: The average glucose level=120 mg/dL and standard deviation=15 mg/dL] and Metformin was intake by 30 patients [Note: The average glucose level=105 mg/dL and standard deviation=20 mg/dL].

Can we say that Amylin is more effective than Metformin at 95% CL.?

Answer 1

To Test :-

H0 :- µ1 = µ2
H1 :- µ1 > µ2

Test Statistic :-
t = (X̅1 - X̅2) / SP √ ( ( 1 / n1) + (1 / n2))



t = ( 120 - 105) / 18.7583 √ ( ( 1 / 12) + (1 / 30 ) )
t = 2.3411

Test Criteria :-
Reject null hypothesis if t > t(α, n1 + n2 - 2)
Critical value t(α, n1 + n1 - 2) = t( 0.05 , 12 + 30 - 2) = 1.684
t > t(α, n1 + n2 - 2) = 2.3411 > 1.684
Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 2.3411 ) = 0.0121
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0121 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that  Amylin is more effective than Metformin at 95% CL.