Question

lesson 40 homework study of different in tow population proportions

Sam claims that the proportion of all males who have been locked out of their houses is less than the proportion of females who have been locked out of their houses in the last three years. Sam found that 135 out of 300 randomly selected males had locked themselves out of their houses within the last three years. Also, he found that 220 out of 400 randomly selected females had locked themselves out of their houses in the last three years.

Test Sam's claim at the 1% significance level.

Answer 1

P1 :-  Proportion of all males who have been locked out of their houses

P2 :-   Proportion of all females who have been locked out of their houses

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

p̂1 = 135 / 300 = 0.45
p̂2 = 220 / 400 = 0.55

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 = ( 135 + 220 ) / ( 300 + 400 )
p̂ = 0.5071
q̂ = 1 - p̂ = 0.4929
Z = ( 0.45 - 0.55) / √( 0.5071 * 0.4929 * (1/300 + 1/400) )
Z = -2.6189 ≈ - 2.62

Test Criteria :-
Reject null hypothesis if Z < -Z(α)
Critical value Z(α) = Z(0.01) = 2.326 ( From Z table )
Z < -Z(α) = -2.6189 < -2.326, hence we reject the null hypothesis
Conclusion :- We Reject H0

Decision based on P value
P value = P ( Z < -2.6189 ) = 0.0044
Reject null hypothesis if P value < α = 0.01
Since P value = 0.0044 > 0.01, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that proportion of all males who have been locked out of their houses is less than the proportion of females who have been locked out of their houses in the last three years.