Two random samples of student loans were collected: one from students at for-profit schools and another...

Part a)

H0 :- P1 ≤ P2
H1 :- P1 > P2

p̂1 = 25 / 238 = 0.105
p̂2 = 11 / 212 = 0.0519

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 = ( 25 + 11 ) / ( 238 + 212 )
p̂ = 0.08
q̂ = 1 - p̂ = 0.92
Z = ( 0.105 - 0.0519) / √( 0.08 * 0.92 * (1/238 + 1/212) )
Z = 2.07

Critical value   Z(α) = Z(0.1) = 1.28

Test Criteria :-
Reject null hypothesis if Z > Z(α)
Z(α) = Z(0.1) = 1.28
Z > Z(α) = 2.07 > 1.28, hence we reject the null hypothesis
Conclusion :- We Reject H0

Reject null hypothesis, there is sufficient evidence to support the claim.

P value = P ( Z > 2.0747 ) = 0.019

Reject null hypothesis if P value < α = 0.1
Since P value = 0.019 < 0.1, hence we reject the null hypothesis
Conclusion :- We Reject H0

P value is less than α, therefore, reject null hypothesis, the conclusion is same.

Part c)

(p̂1 - p̂2) ± Z(α/2) * √( ((p̂1 * q̂1)/ n1) + ((p̂2 * q̂2)/ n2) )
Z(α/2) = Z(0.1 /2) = 1.645
Lower Limit = ( 0.105 - 0.0519 )- Z(0.1/2) * √(((0.105 * 0.895 )/ 238 ) + ((0.0519 * 0.9481 )/ 212 ) = 0.012
upper Limit = ( 0.105 - 0.0519 )+ Z(0.1/2) * √(((0.105 * 0.895 )/ 238 ) + ((0.0519 * 0.9481 )/ 212 )) = 0.094
90% Confidence interval is ( 0.012 , 0.094 )
( 0.012 < ( P1 - P2 ) < 0.094 )