Has the recent drop in airplane passengers resulted in bette
Solution
Set Up Hypothesis
Null, H0:P=0.92
Alternate, H1: P>0.92
Test Statistic
No. Of Success chances Observed (x)=153
Number of objects in a sample provided(n)=165
No. Of Success Rate ( P )= x/n = 0.9273
Success Probability ( Po )=0.92
Failure Probability ( Qo) = 0.08
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.92727-0.92/(Sqrt(0.0736)/165)
Zo =0.3444
| Zo | =0.3444
Critical Value
The Value of |Z | at LOS 0.05% is 1.64
We got |Zo| =0.344 & | Z | =1.64
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Right Tail - Ha : ( P > 0.34435 ) = 0.36529
Hence Value of P0.05 < 0.36529,Here We Do not Reject Ho
a.
H0:P=0.92 , H1: P>0.92
b.
Zo =0.3444
c.
Ha : ( P > 0.34435 ) = 0.36529
d.
We don\'t have evidence to support the airlines performence were improved
