Question

Let n1 = 100, X1 = 43, n2 = 100, and X2 = 30. at 0.05 level of significance is there evidence of a significant difference between the two population proportions

Answer 1

To Test :-
H0 :- P1 = P2
H1 :- P1 ≠ P2

p̂1 = 43 / 100 = 0.43
p̂2 = 30 / 100 = 0.3

Test Statistic :-
Z = ( p̂1 - p̂2 ) / √( p̂ * q̂ * (1/n1 + 1/n2) ))
p̂ is the pooled estimate of the proportion P
p̂ = ( x1 + x2) / ( n1 + n2)
p̂ = ( 43 + 30 ) / ( 100 + 100 )
p̂ = 0.365
q̂ = 1 - p̂ = 0.635
Z = ( 0.43 - 0.3) / √( 0.365 * 0.635 * (1/100 + 1/100) )
Z = 1.91

Test Criteria :-
Reject null hypothesis if Z > Z(α/2)
Z(α/2) = Z(0.05/2) = 1.96
Z < Z(α/2) = 1.9094 < 1.96, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Decision based on P value
P value = 2 * P ( Z < 1.91 )
P value = 0.0562
Reject null hypothesis if P value < α = 0.05
Since P value = 0.0562 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There is no sufficient evidence to support the claim that there is significant difference between the two population proportions.