Data on graduation rates was collected from a local high sch

Data on graduation rates was collected from a local high school. The following information was gathered:

Gender                                Number of Students                      Number who graduated

Female                                               460                                                         391

Male                                                    650                                                         507

a. Find the interval estimation for the difference between the proportions of females versus males who graduate. Use ?=.05.

                b. Find the test statistic for testing for the difference between the two proportions.

Solution

Null Hypothesis, There Is No Significance between them Ho: p1 = p2
Alternate Hypothesis, There Is Significance between them H1: p1 != p2
Test Statistic
Sample 1 : X1 =391, n1 =460, P1= X1/n1=0.85
Sample 2 : X2 =507, n2 =650, P2= X2/n2=0.78
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.809
Q^ Value For Proportion= 1-P^=0.191
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.85-0.78)/Sqrt((0.809*0.191(1/460+1/650))
Zo =2.923
| Zo | =2.923
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =2.923 & | Z | =1.96
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 2.9227 ) = 0.0035
Hence Value of P0.05 > 0.0035,Here we Reject Ho