Question

Some commercial airplanes recirculate approximately 50% of the cabin air in order to increase fuel efficiency. The researchers studied 1100 airline passengers, among which some traveled on airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 517 passengers who flew on planes that did not recirculate air, 106 reported post-flight respiratory symptoms, while 113 of the 583 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 α = 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 p_{do not recirculate} − p_{do recirculate}. Round your test statistic to two decimal places and your P-value to four decimal places.)

z	=
P	=

Answer 1

given

sample size for planes that did not recirculate air=n1=517

no. of passenger on planes that did not recirculate air reported post-flight respiratory symptoms =x1=106

sample size for planes that recirculate air=n1=583

no. of passenger on planes that recirculate air reported post-flight respiratory symptoms =x1=113

let P1 is proportion of passenger on planes that did not recirculate air reported post-flight respiratory symptoms

P2 on planes that recirculate air reported post-flight respiratory symptoms

now we have to test

H0: P1-P2=0

H1:P1-P20

so

now

as under H0: P1=P2 hence pooled proportion is given by

now Z statistics is given by

Since Z statistics is positive and its two tailed test hence

P-Value=2*P(Z>0.46) =2*(1-P(Z<0.46))

            =2*(1-0.6772) =2*0.3228 =0.6456

where using following excel formula we can get value of P(Z<0.46)

P(Z<0.46)=NORM.DIST(0.46,0,1,TRUE)=0.6772