Some commercial airplanes recirculate approximately 50% of the cabin air in orde

Question

Some commercial airplanes recirculate approximately 50% of the cabin air in order to increase fuel efficiency. The researchers studied 1104 airline passengers, among which some traveled on airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 519 passengers who flew on planes that did not recirculate air, 106 reported post-fight respiratory symptoms, while 111of the 585 passengers on planes that did air reported such symptoms. Is there sufficient evidence to conclude that the proportion of passengers with post-flight respiratory symptoms differs for that do and do not recirculate air? Test the appropriate hypotheses using alpha = 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 rho-value. Use. Round your test statistic to two decimal places and your rho-value to four decimal places). Conclusion: Yes, there is sufficient evidence. No, there is not sufficient evidence.

Explanation / Answer

Solution:-

pdo not recirculate = 106/519 = 0.20424

n do not recirculate = 519

pdo recirculate = 111/585 = 0.18974

n do recirculate = 585

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: Pdo not recirculate = Pdo recirculate

Alternative hypothesis: Pdo not recirculate Pdo recirculate

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (pdo not recirculate * n do not recirculate  + pdo recirculate * n do recirculate ) / (n do not recirculate+ n do recirculate  )

p = 0.19656

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = 0.02396

z = (pdo not recirculate - pdo recirculate ) / SE

z = 0.60518

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -0.61 or greater than 0.61.

Thus, the P-value = 0..27 + 0.27 = 0.5418

Interpret results. Since the P-value (0.542) is more than the significance level (0.05), we have to accept the null hypothesis.

Conclusion:

So we can conclude that we do not have sufficient evidence in the favor of the claim that there is difference between two groups.

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