Do teachers find their work rewarding and satisfying The art
Solution
A. Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=224
No.Of Observed (n1)=395
P1= X1/n1=0.567
Proportion 2
No. of chances(X2)=126
No.Of Observed (n2)=266
P2= X2/n2=0.474
C.I = (0.567-0.474) ±Z a/2 * Sqrt( (0.567*0.433/395) + (0.474*0.526/266) )
=(0.567-0.474) ± 1.96* Sqrt(0.002)
=0.093-0.077,0.093+0.077
=[0.016,0.171]
B.
Null, There Is No Significance between them Ho: p1 = p2
Alternate, There Is Significance between them H1: p1 != p2
Test Statistic
Sample 1 : X1 =224, n1 =395, P1= X1/n1=0.567
Sample 2 : X2 =126, n2 =266, P2= X2/n2=0.474
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.53
Q^ Value For Proportion= 1-P^=0.47
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.567-0.474)/Sqrt((0.53*0.47(1/395+1/266))
Zo =2.359
| Zo | =2.359
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =2.359 & | 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 != 2.3594 ) = 0.0183
Hence Value of P0.05 > 0.0183,Here we Reject Ho
