Some commercial airplanes recirculate approximately 50% of the cabin air in order to increase fuel efficiency. The researchers studied 1096 airline passengers, among which some traveled on airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 515 passengers who flew on planes that did not recirculate air, 106 reported post-flight respiratory symptoms, while 112 of the 581 passengers on planes that did recirculate air reported such symptoms. Is there sufficient evidence to conclude that the proportion of passengers with post- flight respiratory symptoms differs for planes that do and do not recirculate air? Test the appropriate hypotheses using a = 0.05. You may assume that it is reasonable to regard these two samples as being independently selected and as representative of the two populations of interest. (Use a statistical computer package to calculate the P-value. Use Pdo not recirculate - Pdo recirculate- Round your test statistic to two decimal places and your P-value to four decimal places.) Solution Null , There Is No Significance between them Ho: p1 = p2 Alternate , There Is Significance between them H1: p1 != p2 Test Statistic Sample 1 : X1 =106, n1 =515, P1= X1/n1=0.206 Sample 2 : X2 =112, n2 =581, P2= X2/n2=0.193 Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2) P^=0.199 Q^ Value For Proportion= 1-P^=0.801 we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2)) Zo =(0.206-0.193)/Sqrt((0.199*0.801(1/515+1/581)) Zo =0.54 | Zo | =0.54 Critical Value The Value of |Z | at LOS 0.05% is 1.96 We got |Zo| =0.54 & | Z | =1.96 Make Decision Hence Value of |Zo | < | Z | and Here we Do not Reject Ho P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 0.5403 ) = 0.589 Hence Value of P0.05 < 0.589,Here We Do not Reject Ho No sufficient evidence to condude that the proportion of passengers with post-flight respratory symptoms differs for planes that do and do not recirculate air ANS: Z = 0.54 P = 0.589.

1. Some commercial airplanes recirculate approximately 50% of the cabin air in order to increase
fuel efficiency. The researchers studied 1096 airline passengers, among which some traveled on
airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 515
passengers who flew on planes that did not recirculate air, 106 reported post-flight respiratory
symptoms, while 112 of the 581 passengers on planes that did recirculate air reported such
symptoms. Is there sufficient evidence to conclude that the proportion of passengers with post-
flight respiratory symptoms differs for planes that do and do not recirculate air? Test the
appropriate hypotheses using a = 0.05. You may assume that it is reasonable to regard these two
samples as being independently selected and as representative of the two populations of interest.
(Use a statistical computer package to calculate the P-value. Use Pdo not recirculate - Pdo
recirculate- Round your test statistic to two decimal places and your P-value to four decimal
places.)
Solution
Null , There Is No Significance between them Ho: p1 = p2
Alternate , There Is Significance between them H1: p1 != p2
Test Statistic
Sample 1 : X1 =106, n1 =515, P1= X1/n1=0.206
Sample 2 : X2 =112, n2 =581, P2= X2/n2=0.193
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.199
Q^ Value For Proportion= 1-P^=0.801
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.206-0.193)/Sqrt((0.199*0.801(1/515+1/581))
Zo =0.54
| Zo | =0.54
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =0.54 & | Z | =1.96
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 0.5403 ) = 0.589
Hence Value of P0.05 < 0.589,Here We Do not Reject Ho
No sufficient evidence to condude that the proportion of passengers
with post-flight respratory symptoms differs for planes that do and do not recirculate air

2. ANS:
Z = 0.54
P = 0.589