Question

Some commercial airplanes recirculate approximately 50% of the cabin air to increase fuel efficiency. The authors of the paper “Aircraft Cabin Air Recirculation and Symptoms of the Common Cold” (Journal of the American Medical Association [2002]: 483-486) studied 1100 airline passengers who flew from San Francisco to Denver between January and April 1999. Some passengers traveled on airplanes that recirculated air and others traveled on planes that did not recirculate the air. Of the 517 passengers who flew on planes that did not recirculate air, 108 reported postflight respiratory symptoms, whereas 111 of the 583 passengers on planes that did recirculate air reported such symptoms. Estimate the difference between passengers that did and did not receive recirculated air at ? = 0.01 significance level. Interpret your interval in the context of the problem.

Answer 1

Let p1 , p2 be the proportion of passengers who reported postflight respiratory symptoms on planes that did not and did recirculate air respectively.

p1 = 108 / 517 = 0.2088975

p2 = 111 / 583 = 0.1903945

Point estimate of difference in proportions = 0.2088975 - 0.1903945 = 0.018503

Pooled proportion, p = (108 + 111) / (517 + 583) = 0.1990909

Standard error mean difference =

= 0.02412316

Z value for ? = 0.01 significance level is 2.576

Margin of error = z * Std error = 2.576 * 0.02412316 = 0.06214126

99% confidence interval fo difference in proportion of passengers who reported postflight respiratory symptoms on planes that did not and did recirculate air is,

(0.018503 - 0.06214126, 0.018503 + 0.06214126)

(-0.0436,  0.0806)

We're 95% confident that the interval (-0.0436,  0.0806) captured the true difference in proportion of passengers who reported postflight respiratory symptoms on planes that did not and did recirculate air.