An SRS of 100 flights of a large airline airline 1 showed th

An SRS of 100 flights of a large airline (airline 1) showed that 64 were on time An SRS of 100 nights of another large airline (airline 2) showed that 80 were on time . Let p1 and p2 be the proportion of all flights that are on time for these two airlines. Find a 95% confidence interval for the difference p1 -p2.

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 flights on time( X1 )=64
No.Of Observed (n1)=100
P1= X1/n1=0.64
Proportion 2
No. of airline 2 on time(X2)=80
No.Of Observed (n2)=100
P2= X2/n2=0.8
C.I = (0.64-0.8) ±Z a/2 * Sqrt( (0.64*0.36/100) + (0.8*0.2/100) )
=(0.64-0.8) ± 1.96* Sqrt(0.004)
=-0.16-0.122,-0.16+0.122
=[-0.282,-0.038]

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 =64, n1 =100, P1= X1/n1=0.64
Sample 2 : X2 =80, n2 =100, P2= X2/n2=0.8
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.72
Q^ Value For Proportion= 1-P^=0.28
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.64-0.8)/Sqrt((0.72*0.28(1/100+1/100))
Zo =-2.52
| Zo | =2.52
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =2.52 & | 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 != -2.5198 ) = 0.0117
Hence Value of P0.05 > 0.0117,Here we Reject Ho