A university financial aid office polled an SRS of undergraduate students to study their summer employment. Not all students were employed the previous summer. Here are the results for men and women: Is there evidence that the proportion of male students employed during the summer differs from the proportion of female students who were employed? State Ho and Ha, compute the test statistic, give the critical region, and state your conclusion. Give a 99% confidence interval for the difference between the proportion or male and female students who were employed during the summer. Solution a) Null Hypothesis, There Is No Significance between them Ho: p1 = p2 Alternate, proportion of male employed is differ from female proportion, H1: p1 != p2 Test Statistic Sample 1 : X1 =718, n1 =797, P1= X1/n1=0.901 Sample 2 : X2 =593, n2 =732, P2= X2/n2=0.81 Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2) P^=0.857 Q^ Value For Proportion= 1-P^=0.143 we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2)) Zo =(0.901-0.81)/Sqrt((0.857*0.143(1/797+1/732)) Zo =5.071 | Zo | =5.071 Critical Value The Value of |Z | at LOS 0.01% is 2.576 We got |Zo| =5.071 & | Z | =2.576 Make Decision Hence Value of | Zo | > | Z | and Here we Reject Ho P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 5.071 ) = 0 Hence Value of P0.01 > 0,Here we Reject Ho We conclude that proportion of male employed is differ from female proportion b) CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 ) Proportion 1 Probability Succuses( X1 )=718 No.Of Observed (n1)=797 P1= X1/n1=0.901 Proportion 2 Probability Succuses(X2)=593 No.Of Observed (n2)=732 P2= X2/n2=0.81 C.I = (0.901-0.81) ±Z a/2 * Sqrt( (0.901*0.099/797) + (0.81*0.19/732) ) =(0.901-0.81) ± 2.58* Sqrt(0) =0.091-0.046,0.091+0.046 =[0.044,0.137].

1. A university financial aid office polled an SRS of undergraduate students to study their summer
employment. Not all students were employed the previous summer. Here are the results for men
and women: Is there evidence that the proportion of male students employed during the summer
differs from the proportion of female students who were employed? State Ho and Ha, compute
the test statistic, give the critical region, and state your conclusion. Give a 99% confidence
interval for the difference between the proportion or male and female students who were
employed during the summer.
Solution
a)
Null Hypothesis, There Is No Significance between them Ho: p1 = p2
Alternate, proportion of male employed is differ from female proportion, H1: p1 != p2
Test Statistic
Sample 1 : X1 =718, n1 =797, P1= X1/n1=0.901
Sample 2 : X2 =593, n2 =732, P2= X2/n2=0.81
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.857
Q^ Value For Proportion= 1-P^=0.143
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.901-0.81)/Sqrt((0.857*0.143(1/797+1/732))
Zo =5.071
| Zo | =5.071
Critical Value
The Value of |Z | at LOS 0.01% is 2.576
We got |Zo| =5.071 & | Z | =2.576
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 5.071 ) = 0
Hence Value of P0.01 > 0,Here we Reject Ho
We conclude that proportion of male employed is differ from female proportion
b)
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
Probability Succuses( X1 )=718
No.Of Observed (n1)=797

2. P1= X1/n1=0.901
Proportion 2
Probability Succuses(X2)=593
No.Of Observed (n2)=732
P2= X2/n2=0.81
C.I = (0.901-0.81) ±Z a/2 * Sqrt( (0.901*0.099/797) + (0.81*0.19/732) )
=(0.901-0.81) ± 2.58* Sqrt(0)
=0.091-0.046,0.091+0.046
=[0.044,0.137]